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A First Course in Mathematical Analysis

Mathematical Analysis (often called Advanced Calculus) is generally found by students

to be one of their hardest courses in Mathematics. This text uses the so-called sequential

approach to continuity, differentiability and integration to make it easier to understand

the subject.

Topics that are generally glossed over in the standard Calculus courses are given

careful study here. For example, what exactly is a ‘continuous’ function? And how

exactly can one give a careful definition of ‘integral’? This latter is often one of the

mysterious points in a Calculus course – and it is quite tricky to give a rigorous

treatment of integration!

The text has a large number of diagrams and helpful margin notes, and uses many

graded examples and exercises, often with complete solutions, to guide students

through the tricky points. It is suitable for self study or use in parallel with a standard

university course on the subject.

A First Course inMathematical

Analysis

D A V I D A L E X A N D E R B R A N N A N

Published in association with The Open University

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

First published in print format

ISBN-13 978-0-521-86439-8

ISBN-13 978-0-511-34857-0

© The Open University 2006

2006

Information on this title: www.cambridge.org/9780521864398

This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

ISBN-10 0-511-34857-6

ISBN-10 0-521-86439-9

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

hardback

eBook (EBL)

eBook (EBL)

hardback

http://www.cambridge.org

http://www.cambridge.org/9780521864398

To my wife Margaret

and my sons David, Joseph and Michael

Contents

Preface page ix

1 Numbers 1

1.1 Real numbers 2

1.2 Inequalities 9

1.3 Proving inequalities 14

1.4 Least upper bounds and greatest lower bounds 22

1.5 Manipulating real numbers 30

1.6 Exercises 35

2 Sequences 37

2.1 Introducing sequences 38

2.2 Null sequences 43

2.3 Convergent sequences 52

2.4 Divergent sequences 61

2.5 The Monotone Convergence Theorem 68

2.6 Exercises 79

3 Series 83

3.1 Introducing series 84

3.2 Series with non-negative terms 92

3.3 Series with positive and negative terms 103

3.4 The exponential function x 7! ex 122

3.5 Exercises 127

4 Continuity 130

4.1 Continuous functions 131

4.2 Properties of continuous functions 143

4.3 Inverse functions 151

4.4 Defining exponential functions 161

4.5 Exercises 164

5 Limits and continuity 167

5.1 Limits of functions 168

5.2 Asymptotic behaviour of functions 176

5.3 Limits of functions – using " and � 181

5.4 Continuity – using " and � 193

5.5 Uniform continuity 200

5.6 Exercises 203

vii

6 Differentiation 205

6.1 Differentiable functions 206

6.2 Rules for differentiation 216

6.3 Rolle’s Theorem 228

6.4 The Mean Value Theorem 232

6.5 L’Hopital’s Rule 238

6.6 The Blancmange function 244

6.7 Exercises 252

7 Integration 255

7.1 The Riemann integral 256

7.2 Properties of integrals 272

7.3 Fundamental Theorem of Calculus 282

7.4 Inequalities for integrals and their applications 288

7.5 Stirling’s Formula for n! 303

7.6 Exercises 309

8 Power series 313

8.1 Taylor polynomials 314

8.2 Taylor’s Theorem 320

8.3 Convergence of power series 329

8.4 Manipulating power series 338

8.5 Numerical estimates for p 346

8.6 Exercises 350

Appendix 1 Sets, functions and proofs 354

Appendix 2 Standard derivatives and primitives 359

Appendix 3 The first 1000 decimal places offfiffiffi

2p

, e and p 361

Appendix 4 Solutions to the problems 363

Chapter 1 363

Chapter 2 371

Chapter 3 382

Chapter 4 393

Chapter 5 402

Chapter 6 413

Chapter 7 426

Chapter 8 443

Index 457

viii Contents

Preface

Analysis is a central topic in Mathematics, many of whose branches use

key analytic tools. Analysis also has important applications in Applied

Mathematics, Physics and Engineering, where a good appreciation of the

underlying ideas of Analysis is necessary for a modern graduate.

Changes in the school curriculum over the last few decades have resulted

in many students finding Analysis very difficult. The author believes that

Analysis nowadays has an unjustified reputation for being hard, caused by

the traditional university approach of providing students with a highly polished

exposition in lectures and associated textbooks that make it impossible for

the average learner to grasp the core ideas. Many students end up agreeing with

the German poet and philosopher Goethe who wrote that ‘Mathematicians are

like Frenchmen: whatever you say to them, they translate into their own

language, and forthwith it is something entirely different!’

Since 1971, the Open University in United Kingdom has taught Mathematics

to students in their own homes via specially written correspondence texts, and

has traditionally given Analysis a central position in its curriculum. Its philo-

sophy is to provide clear and complete explanations of topics, and to teach these

in a way that students can understand without much external help. As a result,

students should be able to learn, and to enjoy learning, the key concepts of the

subject in an uncluttered way. This book arises from correspondence texts for

its course Introduction to Pure Mathematics, that has now been studied suc-

cessfully by over ten thousand students.

This book is therefore different from most Mathematics textbooks! It adopts

a student-friendly approach, being designed for study by a student on their own

OR in parallel with a course that uses as set text either this text or another text.

But this is the text that the student is likely to use to learn the subject from. The

author hopes that readers will gain enormous pleasure from the subject’s

beauty and that this will encourage them to undertake further study of

Mathematics!

Once a student has grasped the principal notions of limit and continuous

function in terms of inequalities involving the three symbols ", X and �, they

will quickly understand the unity of areas of Analysis such as limits, continu-

ity, differentiability and integrability. Then they will thoroughly enjoy the

beauty of some of the arguments used to prove key theorems – whether their

proofs are short or long.

Calculus is the initial study of limits, continuity, differentiation and integra-

tion, where functions are assumed to be well-behaved. Thus all functions

continuous on an interval are assumed to be differentiable at most points in

the interval, and so on. However, Mathematics is not that simple! For example,

there exist functions that are continuous everywhere on R , but differentiable

Johann Wolfgang von Goethe(1749–1832) is said to havestudied all areas of science ofhis day except mathematics –for which he had no aptitude.

ix

nowhere on R ; this discovery by Karl Weierstrass in 1872 caused a sensation in

the mathematical community. In Analysis (sometimes called Advanced

Calculus) we make no assumptions about the behaviour of functions – and

the result is that we sometimes come across real surprises!

The book has two principal features in its approach that make it stand out

from among other Analysis texts.

Firstly, this book uses the ‘sequential approach’ to Analysis. All too often

students starting on the subject find that they cannot grasp the significance of

both " and � simultaneously. This means that the whole underlying idea about

what is happening is lost, and the student takes a very long time to master the

topic – or, in many cases in fact, never masters the topic and acquires a strong

dislike of it. In the sequential approach they proceed at a more leisurely pace to

understand the notion of limit using " and X – to handle convergent sequences –

before coming across the other symbol �, used in conjunction with " to handle

continuous functions. This approach avoids the conventional student horror at

the perceived ‘difficulty of Analysis’. Also, it avoids the necessity to re-prove

broadly similar results in a range of settings – for example, results on the sum

of two sequences, of two series, of two continuous functions and of two

differentiable functions.

Secondly, this book makes great efforts to teach the "–� approach too. After

students have had a first pass at convergence of sequences and series and at

continuity using ‘the sequential approach’, they then meet ‘the "–� approach’,

explained carefully and motivated by a clear ‘"–� game’ discussion. This

makes the new approach seem very natural, and this is motivated by using

each approach in later work in the appropriate situation. By the end of the book,

students should have a good facility at using both the sequential approach and

the "–� approach to proofs in Analysis, and should be better prepared for later

study of Analysis than students who have acquired only a weak understanding

of the conventional approach.

Outline of the content of the book

In Chapter 1, we define real numbers to be decimals. Rather than give a heavy

discussion of least upper bound and greatest lower bound, we give an intro-

duction to these matters sufficient for our purposes, and the full discussion is

postponed to Chapter 7, where it is more timely. We also study inequalities,

and their properties and proofs.

In Chapter 2, we define convergent sequences and examine their proper-

ties, basing the discussion on the notion of null sequence, which simplifies

matters considerably. We also look at divergent sequences, sequences

defined by recurrence formulas and particular sequences which converge

to p and e.

In Chapter 3, we define convergent infinite series, and establish a number of

tests for determining whether a given series is convergent or divergent. We

demonstrate the equivalence of the two definitions of the exponential function

x 7! ex, and prove that the number e is irrational.

In Chapter 4, we define carefully what we mean by a continuous function, in

terms of sequences, and establish the key properties of continuous functions.

We also give a rigorous definition of the exponential function x 7! ax.

We define ex as limn!1

1þ xn

� �n

and asP

1

n¼0

xn

n!.

x Preface

In Chapter 5, we define the limit of a function as x tends to c or as x tends

to 1 in terms of the convergence of sequences. Then we introduce the "–�definitions of limit and continuity, and check that these are equivalent to the

earlier definitions in terms of sequences. We also look briefly at uniform

continuity.

In Chapter 6, we define what we mean by a differentiable function, using

difference quotients Q(h); this enables us to use our earlier results on limits to

prove corresponding results for differentiable functions. We establish some

interesting properties of differentiable functions. Finally, we construct the

Blancmange function that is continuous everywhere on R , but differentiable

nowhere on R .

In Chapter 7, we give a careful definition of what we mean by an integrable

function, and establish a number of related criteria for establishing whether

a given function is integrable or not. Our integral is the so-called Riemann

integral, defined in terms of upper and lower Riemann sums. We check the

standard properties of integrals and verify a number of standard approaches for

calculating definite integrals. Then we give a number of applications of

integrals to limits of certain sequences and series and prove Stirling’s Formula.

Finally, in Chapter 8, we study the convergence and properties of power

series. The chapter ends with a marvellous proof of the irrationality of the

number p that uses a whole range of the techniques that have been met in the

previous chapters.

For completeness and for students’ convenience, we give a brief guide to our

notation for sets and functions, together with a brief indication of the logic

involved in proofs in Mathematics (in particular, the Principle of Mathematical

Induction) in Appendix 1. Appendix 2 contains a list of standard derivatives

and primitives and Appendix 3 the first 1000 decimal places in the values of the

numbersffiffiffi

2p

, p and e. Appendix 4 contains full solutions to all the problems set

during each chapter.

Solutions are not given to the exercises at the end of each chapter, however.

Lecturers/instructors may wish to use these exercises in homework assignments.

Study guide

This book assumes that students have a fair understanding of Calculus. The

assumptions on technical background are deliberately kept slight, however, so

that students can concentrate on the newer aspects of the subject ‘Analysis’.

Most students will have met some of the material in the early chapters

previously. Although this means that they can therefore proceed fairly quickly

through some sections, it does NOT mean that those sections can be ignored –

each section contains important ideas that are used later on and most include

something new or have a different emphasis.

Most chapters are divided into five or six sections (each often further divided

into sub-sections); sections are numbered using two digits (such as ‘Section

3.2 ’) and sub-sect ions using thr ee digits (such as ‘Sub- section 3.2.4 ’).

Generally a section is considered to be about one evening’s work for an

average student.

Chapter 7 on Integration is arguably the highlight of the book. However,

it contains some rather complicated mathematical arguments and proofs.

Stirling’s Formula says that,for large n, n! is ‘roughly’ffiffiffiffiffiffiffiffi

2pnp

ne

� �n, in a sense that we

explain.

Preface xi

Therefore, when reading Chapter 7, it is important not to get bogged down in

details, but to keep progressing through the key ideas, and to return later on to

reading the things that were left out at the earlier reading. Most students will

require three or four passes at this chapter before having a good idea of most of it.

We use wide pages with a large number of margin notes in which we place

teaching comments and some diagrams to aid in the understanding of particular

points in arguments. We also provide advice on which proofs to omit on a first

study of the topic; it is important for the student NOT to get bogged down in

a technical discussion or a proof until they have a good idea of the message

contained in the result and the situations in which it can be used. Therefore

clear encouragement is given on which portions of the text to leave till later, or

to simply skim on a first reading.

The end of the proof of a Theorem is indicated by a solid symbol ‘&’ and the

end of the solution of a worked Example by a hollow symbol ‘&’. There are

many worked examples within the text to explain the concepts being taught,

together with a good stock of problems to reinforce the teaching. The solutions

are a key part of the teaching, and tackling them on your own and then reading

our version of the solution is a key part of the learning process.

No one can learn Mathematics by simply reading – it is a ‘hands on’ activity.

The reader should not be afraid to draw pictures to illustrate what seems to be

happening to a sequence or a function, to get a feeling for their behaviour. A

wise old man once said that ‘A picture is worth a thousand words!’. A good

picture may even suggest a method of proof. However, at the same time it is

important not to regard a picture on its own as a proof of anything; it may

illustrate just one situation that can arise and miss many other possibilities!

It is important NOT to become discouraged if a topic seems difficult. It took

mathematicians hundreds of years to develop Analysis to its current polished

state, so it may take the reader a few hours at several sittings to really grasp the

more complex or subtle ideas.

Acknowledgements

The material in the Open University course on which this book is based was

contributed to in some way by many colleagues, including Phil Rippon, Robin

Wilson, Andrew Brown, Hossein Zand, Joan Aldous, Ian Harrison, Alan Best,

Alison Cadle and Roberta Cheriyan. Its eventual appearance in book form

owes much to Lynne Barber.

Without the forebearance of my family, the writing of the book would have

been impossible.

It is important to read themargin notes!

This signposting benefitsstudents greatly in theauthor’s experience.

Tackling the problems is agood use of your time, notsomething to skimp.

xii Preface

1 Numbers

In this book we study the properties of real functions defined on intervals of the

real line (possibly the whole real line) and whose image also lies on the real

line. In other words, they map R into R . Our work will be from a very precise

point of view in order to establish many of the properties of such functions

which seem intuitively obvious; in the process we will discover that some

apparently true properties are in fact not necessarily true!

The types of functions that we shall examine include:

exponential functions, such as x 7! ax;where a; x 2 R ;

trigonometric functions, such as x 7! sin x;where x 2 R ;

root functions, such as x 7!ffiffiffi

xp;where x � 0:

The types of behaviour that we shall examine include continuity, differ-

entiability and integrability – and we shall discover that functions with these

properties can be used in a number of surprising applications.

However, to put our study of such functions on a secure foundation, we need

first to clarify our ideas of the real numbers themselves and their properties. In

particular, we need to devote some time to the manipulation of inequalities,

which play a key role throughout the book.

In Section 1.1, we start by revising the properties of rational numbers and

their decimal representation. Then we introduce the real numbers as infinite

decimals, and describe the difficulties involved in doing arithmetic with such

decimals.

In Section 1.2, we revise the rules for manipulating inequalities and show

how to find the solution set of an inequality involving a real number, x, by

applying the rules. We also explain how to deal with inequalities which involve

modulus signs.

In Section 1.3, we describe various techniques for proving inequalities,

including the very important technique of Mathematical Induction.

The concept of a least upper bound, which is of great importance in

Analysis, is introduced in Section 1.4, and we discuss the Least Upper

Bound Property of R .

Finally, in Section 1.5, we describe how least upper bounds can be used to

define arithmetical operations in R .

Even though you may be familiar with much of this material we recommend

that you read through it, as we give the system of real numbers a more careful

treatment than you may have met before. The material on inequalities and least

upper bounds is particularly important for later on.

In later chapters we shall define exactly what the numbers p and e are, and

find various ways of calculating them. But, first, we examine numbers in

general.

For example, what exactly isthe number

ffiffiffi

2p

?

You may omit this sectionat a first reading.

1

1.1 Real numbers

We start our study of the real numbers with the rational numbers, and investi-

gate their decimal representations, then we proceed to the irrational numbers.

1.1.1 Rational numbers

We assume that you are familiar with the set of natural numbers

N ¼ f1; 2; 3; . . .g;and with the set of integers

Z ¼ f. . .; �2; �1; 0; 1; 2; 3; . . .g:The set of rational numbers consists of all fractions (or ratios of integers)

Q ¼ p

q: p 2 Z; q 2 N

� �

:

Remember that each rational number has many different representations as a

ratio of integers; for example

1

3¼ 2

6¼ 10

30¼ . . .:

We also assume that you are familiar with the usual arithmetical operations of

addition, subtraction, multiplication and division of rational numbers.

It is often convenient to represent rational numbers geometrically as points

on a number line. We begin by drawing a line and marking on it points

corresponding to the integers 0 and 1. If the distance between 0 and 1 is

taken as a unit of length, then the rationals can be arranged on the line with

positive rationals to the right of 0 and negative rationals to the left.

–3 –2 –1 0 1 2 332

12

52

–

negative positive

For example, the rational 32

is placed at the point which is one-half of the way

from 0 to 3.

This means that rationals have a natural order on the number-line. For

example, 1922

lies to the left of 78

because

19

22¼ 76

88and

7

8¼ 77

88:

If a lies to the left of b on the number-line, then

a is less than b or b is greater than a;

and we write

a a:

For example

19

22<

7

8or

7

8>

19

22:

We write a� b, or b� a, if either a< b or a¼ b.

Note that 0 is not a naturalnumber.

2 1: Numbers

Problem 1 Arrange the following rational numbers in order:

0; 1;�1; 1720;�17

20; 45

53;�45

53:

Problem 2 Show that between any two distinct rational numbers there

is another rational number.

1.1.2 Decimal representation of rational numbers

The decimal system enables us to represent all the natural numbers using only

the ten integers

0; 1; 2; 3; 4; 5; 6; 7; 8 and 9;

which are called digits. We now remind you of the basic facts about the

representation of rational numbers by decimals.

Definition A decimal is an expression of the form

�a0 � a1a2a3 . . .;

where a0 is a non-negative integer and a1, a2, a3, . . . are digits.

If only a finite number of the digits a1, a2, a3, . . . are non-zero, then the

decimal is called terminating or finite, and we usually omit the tail of 0s.

Terminating decimals are used to represent rational numbers in the follow-

ing way

�a0 � a1a2a3 . . . an ¼ � a0 þa1

101þ a2

102þ � � � þ an

10n

� �

:

It can be shown that any fraction whose denominator contains only powers

of 2 and/or 5 (such as 20¼ 22� 5) can be represented by such a terminating

decimal, which can be found by long division.

However, if we apply long division to many other rationals, then the process

of long division never terminates and we obtain a non-terminating or infinite

decimal. For example, applying long division to 13

gives 0.333 . . ., and for 1922

we obtain 0.86363 . . ..

Problem 3 Apply long division to 17

and 213

to find the corresponding

decimals.

These non-terminating decimals, which are obtained by applying the long

division process, have a certain common property. All of them are recurring;

that is, they have a recurring block of digits, and so can be written in shorthand

form, as follows:

0:333 . . . ¼ 0:3;

0:142857142857 . . . ¼ 0:142857 . . .;

0:86363 . . . ¼ 0:863:

It is not hard to show, whenever we apply the long division process to a

fraction pq, that the resulting decimal is recurring. To see why we notice that

there are only q possible remainders at each stage of the division, and so one

of these remainders must eventually recur. If the remainder 0 occurs, then the

resulting decimal is, of course, terminating; that is, it ends in recurring 0s.

For example

0:8500 . . .;

13:1212 . . .;

�1:111 . . .:

For example,

0:8500 . . . ¼ 0:85:

For example

0:85 ¼ 0þ 8

101þ 5

102

¼ 85

100¼ 17

20:

Another commonly usednotation is

0:3:

or 0:1:

42857:

:

1.1 Real numbers 3

Non-terminating recurring decimals which arise from the long division of

fractions are used to represent the corresponding rational numbers. This

representation is not quite so straight-forward as for terminating decimals,

however. For example, the statement

1

3¼ 0:3 ¼ 3

101þ 3

102þ 3

103þ � � �

can be made precise only when we have introduced the idea of the sum of a

convergent infinite series. For the moment, when we write the statement 13¼ 0:3

we mean simply that the decimal 0:3 arises from 13

by the long division process.

The following example illustrates one way of finding the rational number

with a given decimal representation.

Example 1 Find the rational number (expressed as a fraction) whose decimal

representation is 0:863:

Solution First we find the fraction x such that x ¼ 0:63:If we multiply both sides of this equation by 102 (because the recurring block

has length 2), then we obtain

100x ¼ 63:63 ¼ 63þ x:

Hence

99x ¼ 63) x ¼ 63

99¼ 7

11:

Thus

0:863 ¼ 8

10þ x

10¼ 8

10þ 7

110¼ 95

110¼ 19

22: &

The key idea in the above solution is that multiplication of a decimal by 10k

moves the decimal point k places to the right.

Problem 4 Using the above method, find the fractions whose decimal

representations are:

(a) 0:231; (b) 2:281.

The decimal representation of rational numbers has the advantage that it

enables us to decide immediately which of two distinct positive rational numbers

is the greater. We need only examine their decimal representations and notice

the first place at which the digits differ. For example, to order 78

and 1922

we write

7

8¼ 0:875 . . . and

19

22¼ 0:86363 . . .;

and so

0:86#363 . . . < 0:87

#5) 19

22<

7

8:

Problem 5 Find the first two digits after the decimal point in the deci-

mal representations of 1720

and 4553

, and hence determine which of these

two rational numbers is the greater.

Warning Decimals which end in recurring 9s sometimes arise as alternative

representations for terminating decimals. For example

1 ¼ 0:9 ¼ 0:999 . . . and 1:35 ¼ 1:349 ¼ 1:34999 . . .:

We return to this topic inChapter 3.

Whenever possible, we avoidusing the form of a decimalwhich ends in recurring 9s.

4 1: Numbers

You may find this rather alarming, but it is important to realise that this is

a matter of convention. We wish to allow the decimal 0.999 . . . to represent

a number x, so x must be less than or equal to 1 and greater than each of the

numbers

0:9; 0:99; 0:999; . . .:

The only rational with these properties is 1.

1.1.3 Irrational numbers

One of the surprising mathematical discoveries made by the Ancient Greeks

was that the system of rational numbers is not adequate to describe all the

magnitudes that occur in geometry. For example, consider the diagonal of a

square of side 1. What is its length? If the length is x, then, by Pythagoras’

Theorem, x must satisfy the equation x2¼ 2. However, there is no rational

number which satisfies this equation.

Theorem 1 There is no rational number x such that x2¼ 2.

Proof Suppose that such a rational number x exists. Then we can write x¼ pq.

By cancelling, if necessary, we may assume that p and q have no common

factor. The equation x2¼ 2 now becomes

p2

q2¼ 2; so p2 ¼ 2q2:

Now, the square of an odd number is odd, and so p cannot be odd. Hence p is

even, and so we can write p¼ 2r, say. Our equation now becomes

ð2rÞ2 ¼ 2q2; so q2 ¼ 2r2:

Reasoning as before, we find that q is also even.

Since p and q are both even, they have a common factor 2, which contradicts

our earlier statement that p and q have no common factors.

Arguing from our original assumption that x exists, we have obtained two

contradictory statements. Thus, our original assumption must be false. In other

words, no such x exists. &

Problem 6 By imitating the above proof, show that there is no rational

number x such that x3¼ 2.

Since we want equations such as x2¼ 2 and x3¼ 2 to have solutions, we

must introduce new numbers which are not rational numbers. We denote these

new numbers byffiffiffi

2p

andffiffiffi

23p

, respectively; thusffiffiffi

2p� �2¼ 2 and

ffiffiffi

23p� �3¼ 2. Of

course, we must introduce many other new numbers, such asffiffiffi

3p

,ffiffiffiffiffi

115p

, and

so on. Indeed, it can be shown that, if m, n are natural numbers and xm¼ n has

no integer solution, thenffiffiffi

nmp

cannot be rational. A number which is not rational

is called irrational.

There are many other mathematical quantities which cannot be described

exactly by rational numbers. For example, the number pwhich denotes the area

of a disc of radius 1 (or half the length of the perimeter of such a disc) is

irrational, as is the number e.

x

1

1

x2 ¼ 12 þ 12 ¼ 2

This is a proof bycontradiction.

For

ð2k þ 1Þ2 ¼ 4k2 þ 4k þ 1

¼ 4ðk2 þ kÞ þ 1:

The case m¼ 2 is treated inExercise 5 for this sectionin Section 1.6.

Lambert proved that p isirrational in 1770.

1.1 Real numbers 5

It is natural to ask whether irrational numbers, such asffiffiffi

2p

and p, can

be represented as decimals. Using your calculator, you can check that

(1.41421356)2 is very close to 2, and so 1.41421356 is a very good appro-

ximate value forffiffiffi

2p

. But is there a decimal which representsffiffiffi

2p

exactly? If

such a decimal exists, then it cannot be recurring, because all the recurring

decimals correspond to rational numbers.

In fact, it is possible to represent all the irrational numbers mentioned so

far by non-recurring decimals. For example, there are non-recurring decimals

such thatffiffiffi

2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .:

It is also natural to ask whether non-recurring decimals, such as

0:101001000100001 . . . and 0:123456789101112 . . .;

represent irrational numbers. In fact, a decimal corresponds to a rational

number if and only if it is recurring; so a non-recurring decimal must corres-

pond to an irrational number.

We may summarise this as:

recurring decimal, rational number

non-recurring decimal, irrational number

1.1.4 The real number system

Taken together, the rational numbers (recurring decimals) and irrational num-

bers (non-recurring decimals) form the set of real numbers, denoted by R .

As with rational numbers, we can determine which of two real numbers is

greater by comparing their decimals and noticing the first pair of correspond-

ing digits which differ. For example

0:10#100100010000 . . . < 0:12

#3456789101112 . . .:

We now associate with each irrational number a point on the number-line.

For example, the irrational number x¼ 0.123456789101112 . . . satisfies each

of the inequalities

0:1 < x < 0:2

0:12 < x < 0:13

0:123 < x < 0:124

..

.

We assume that there is a point on the number-line corresponding to x,

which lies to the right of each of the (rational) numbers 0.1, 0.12, 0.123 . . ., and

to the left of each of the (rational) numbers 0.2, 0.13, 0.124, . . ..

As usual, negative real numbers correspond to points lying to the left of 0;

and the ‘number-line’, complete with both rational and irrational points, is

called the real line.

In fact

ð1:41421356Þ2

¼1:9999999932878736:

We prove thatffiffiffi

2p

has adecimal representation inSection 1.5.

When comparing decimals inthis way, we do not alloweither decimal to end inrecurring 9s.

6 1: Numbers

There is thus a one–one correspondence between the points on the real line

and the set R of real numbers (or decimals).

We now state several properties of R , with which you will be already

familiar, although you may not have met their names before. These properties

are used frequently in Analysis, and we do not always refer to them explicitly

by name.

Order Properties of R

1. Trichotomy Property If a, b2R , then exactly one of the following

inequalities holds

a b:

2. Transitive Property If a, b, c2R , then

a a:

4. Density Property If a, b2R and a< b, then there is a rational number

x and an irrational number y such that

a < x < b and a < y < b:

Remark

The Archimedean Property is sometimes expressed in the following equiva-

lent way: for any positive real number a, there is a positive integer n such

that 1n< a.

The following example illustrates how we can prove the Density Property.

Example 2 Find a rational number x and an irrational number y satisfying

a < x < b and a < y < b;

where a ¼ 0:123 and b ¼ 0:12345 . . ..

Solution The two decimals

a ¼ 0:1233#

. . . and b ¼ 0:1234#5 . . .

differ first at the fourth digit. If we truncate b after this digit, we obtain the

rational number x¼ 0.1234, which satisfies the requirement that a< x< b.

To find an irrational number y between a and b, we attach to x a (sufficiently

small) non-recurring tail such as 010010001 . . . to give y¼ 0.1234j010010001 . . ..It is then clear that y is irrational (because its decimal is non-recurring) and that

a< y< b. &

Problem 7 Find a rational number x and an irrational number y such

that a< x< b and a< y< b, where a ¼ 0:3 and b ¼ 0:3401:

The first three of theseproperties are almost self-evident, but the DensityProperty is not so obvious.

1.1 Real numbers 7

Theorem 2 Density Property of R

If a, b2R and a< b, then there is a rational number x and an irrational

number y such that

a < x < b and a < y < b:

Proof For simplicity, we assume that a, b� 0. So, let a and b have decimal

representations

a ¼ a0 � a1a2a3 . . . and b ¼ b0 � b1b2b3 . . .;

where we arrange that a does not end in recurring 9s, whereas b does not

terminate (this latter can be arranged by replacing a terminating representation

by an equivalent representation that ends in recurring 9s).

Since a< b, there must be some integer n such that

a0 ¼ b0; a1 ¼ b1; . . .; an�1 ¼ bn�1; but an < bn:

Then x¼ a0 � a1a2a3 . . . an� 1bn is rational, and a< x< b as required.

Finally, since x< b, it follows that we can attach a sufficiently small non-

recurring tail to x to obtain an irrational number y for which a< y< b. &

Remark

One consequence of the Density Property is that between any two real numbers

there are infinitely many rational numbers and infinitely many irrational

numbers.

Problem 8 Prove that between any two real numbers a and b there are

at least two distinct rational numbers.

1.1.5 Arithmetic in R

We can do arithmetic with recurring decimals by first converting the decimals

to fractions. However, it is not obvious how to perform arithmetical operations

with non-recurring decimals.

Assuming that we can representffiffiffi

2p

and p by the non-recurring decimals

ffiffiffi

2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;

can we also represent the sumffiffiffi

2pþ p and the product

ffiffiffi

2p� p as deci-

mals? Indeed, what do we mean by the operations of addition and multi-

plication when non-recurring decimals (irrationals) are involved, and do

these operations satisfy the same properties as addition and multiplication

of rationals?

It would take many pages to answer these questions fully. Therefore, we

shall assume that it is possible to define all the usual arithmetical operations

with decimals, and that they do satisfy the usual properties. For definiteness,

we now list these properties.

You may omit the followingproof on a first reading.

Here a0, b0 are non-negativeintegers, and a1, b1, a2, b2, . . .are digits.

A proof of the previousremark would involve ideassimilar to those involved intackling Problem 8.

8 1: Numbers

Arithmetic in R

Addition Multiplication

A1 If a, b 2 R , then

aþ b 2 R .

M1 If a, b 2 R , then

a� b2 R .

A2 If a 2 R , then

aþ 0¼ 0þ a¼ a.

M2 If a 2 R , then

a� 1¼ 1� a¼ a.

A3 If a 2 R , then there is a

number �a 2 R such that

aþ (�a)¼ (� a)þ a¼ 0.

M3 If a 2 R � {0}, then there is a

number a�1 2 R such that

a� a�1¼ a�1� a¼ 1.

A4 If a, b, c 2 R , then

(aþ b)þ c¼ aþ (bþ c).

M4 If a, b, c 2 R , then

(a� b)� c¼ a� (b� c).

A5 If a, b 2 R , then

aþ b¼ bþ a.

M5 If a, b 2 R , then

a� b¼ b� a.

D If a, b, c 2 R , then a� (bþ c)¼ a� bþ a� c.

To summarise the contents of this table:

� R is an Abelian group under the operation of additionþ;

� R � {0} is an Abelian group under the operation of multiplication�;

� These two group structures are linked by the Distributive Property.

It follows from the above properties that we can perform addition, subtraction

(where a� b¼ aþ (�b)), multiplication and division (where ab¼ a� b�1) in

R , and that these operations satisfy all the usual properties.

Furthermore, we shall assume that the set R contains the nth roots and

rational powers of positive real numbers, with their usual properties. In

Section 1.5 we describe one way of justifying the existence of nth roots.

1.2 Inequalities

Much of Analysis is concerned with inequalities of various kinds; the aim of

this section and the next section is to provide practice in the manipulation of

inequalities.

1.2.1 Rearranging inequalities

The fundamental rule, upon which much manipulation of inequalities is based,

is that the statement a 0.

This fact can be stated concisely in the following way:

Rule 1 For any a, b2R , a< b, b� a> 0.

Put another way, the inequalities a 0 are equivalent.

There are several other standard rules for rearranging a given inequality into

an equivalent form. Each of these can be deduced from our first rule above. For

CLOSURE

IDENTITY

INVERSES

ASSOCIATIVITY

COMMUTATIVITY

DISTRIBUTATIVITY

Properties A1–A5

Properties M1–M5

Property D

Any system satisfying theproperties listed in the table iscalled a field. Both Q and R

are fields.

Recall that the symbol ‘,’means ‘if and only if’ or‘implies and is implied by’.

1.2 Inequalities 9

example, we obtain an equivalent inequality by adding the same expression to

both sides.

Rule 2 For any a, b, c2R , a< b, aþ c< bþ c.

Another way to rearrange an inequality is to multiply both sides by a non-zero

expression, making sure to reverse the inequality if the expression is

negative.

Rule 3

� For any a, b2R and any c > 0, a < b, ac < bc;

� For any a, b2R and any c < 0, a < b, ac > bc.

Sometimes the most effective way to rearrange an inequality is to take reci-

procals. However, in this case, both sides of the inequality should be positive,

and the direction of the inequality has to be reversed.

Rule 4 (Reciprocal Rule)

For any positive a, b2R , a5b, 1a> 1

b:

Some inequalities can be simplified only by taking powers. However, in order

to do this, both sides must be non-negative and must be raised to a positive

power.

Rule 5 (Power Rule)

For any non-negative a, b2R , and any p > 0, a < b, ap < bp.

For positive integers p, Rule 5 follows from the identity

bp� ap ¼ b� að Þ bp�1 þ bp�2aþ � � � þ bap�2 þ ap�1� �

;

thus, since the right-hand bracket is positive, we have

b� a > 0, bp � ap > 0;

which is equivalent to our desired result.

Remark

There are corresponding versions of Rules 1–5 in which the strict inequality

a < b is replaced by the weak inequality a � b.

Problem 1 State (without proof) the versions of Rules 1–5 for weak

inequalities.

We shal l give one more rule for rearra nging inequal ities in Sub-se ction 1.2.3.

1.2.2 Solving inequalities

Solving an inequality involving an unknown real number x means determin-

ing those values of x for which the given inequality holds; that is, finding

the solution set of the inequality. We can often do this by rewriting the

inequality in an equivalent, but simpler form, using the rules given in the

last sub-section.

For example

253, 20530 ðc ¼ 10Þ;

253, �20 > �30

ðc ¼ �10Þ:

For example

2 < 4, 1

2ð¼0:5Þ

>1

4¼0:25ð Þ:

For example

4 < 9, 412ð¼2Þ< 9

12ð¼3Þ:

We shall discuss the meaningof non-integer powers inSection 1.5.

For example,

b3� a3 ¼ b� að Þ�b2 þ baþ a2� �

:

The solution set is the set ofthose values of x for which theinequality holds.

10 1: Numbers

In this activity we frequently use the usual rules for the sign of a product, and

the fact that the square of any real number is non-negative. Also, we need to

remember the difference between the logical statements: ‘implies’, ‘is implied

by’ and ‘implies and is implied by’.

Example 1 Solve the inequality xþ2xþ4

> x�32x�1

:

Solution We rearrange this inequality to give a somewhat simpler inequality,

using Rule 1

xþ 2

xþ 4>

x� 3

2x� 1, xþ 2

xþ 4� x� 3

2x� 1> 0

, x2 þ 2xþ 10

xþ 4ð Þ 2x� 1ð Þ > 0

, xþ 1ð Þ2þ 9

xþ 4ð Þ 2x� 1ð Þ > 0:

Now, the numerator is always positive. The denominator vanishes when

x¼�4 or x ¼ 12. By examining separately the sign of the denominator when

x < �4,�4 < x < 12

and x > 12, we can deduce that the last fraction is positive

precisely when x < �4 or x > 12. Hence the solution set of the original

inequality is

x :xþ 2

xþ 4>

x� 3

2x� 1

� �

¼ �1;�4ð Þ [ 12;1

� �

: &

Example 2 Solve the inequality 12x2 þ 2

< 14:

Solution Since 2x2þ 2> 0, we have

1

2x2 þ 2<

1

4, 2x2 þ 2 > 4 ðby Rule 4Þ

, x2 þ 1 > 2 ðby Rule 3Þ, x2 � 1 > 0 ðby Rule 1Þ, x� 1ð Þ xþ 1ð Þ > 0:

This last inequality holds precisely when x<�1 or x> 1. It follows that the

solution set of the original inequality is

x :1

2x2 þ 2<

1

4

� �

¼ �1;�1ð Þ [ 1;1ð Þ: &

Problem 2 Use each of the following expressions to write down an

inequality with the given expression on its left-hand side which is equi-

valent to the inequality x> 2:

(a) xþ 3; (b) 2� x; (c) 5xþ 2; (d) �15xþ2

.

Problem 3 Solve the following inequalities:

(a) 4x�x2�7x2�1

� 3; (b) 2x2 � xþ 1ð Þ2:

� þ �

þ þ �� þ

It is a common strategy tobring all terms to one side.

We bring everything to acommon denominator.

Here we complete the squarein the numerator, since wecannot factorise it.

This is because the finaldisplayed inequality isequivalent to the inequalitywe are solving. The logicalimplication symbols betweenthe displayed inequalitieswere all ‘implies and isimplied by’.

Here we factorise the left-hand side of the inequalityto examine the signs of itsfactors.

1.2 Inequalities 11

Note that, for the moment, weare examining only those x forwhich x� 0.

Notice the use of theTransitive Property here.

y

x

y = x

For example

j3j ¼ j�3j ¼ 3:

We sometimes write

ja� bj ¼ dða; bÞ:

For example, the distancefrom �2 to 3 is

�2ð Þ � 3j j ¼ �5j j ¼ 5:

Warning Great care is needed when solving inequalities which involve

rational powers. In particular, when applying Rule 5 both sides of the inequal-

ity must be non-negative.

Example 3 Solve the inequalityffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

> x.

Solution The expressionffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

is defined only when 2xþ 3� 0; that is,

when x � � 32. Hence we need only consider those x in �3

2,1

�

.

We can obtain an equivalent inequality by squaring, provided that bothffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

and x are non-negative. Thus, for x� 0, we obtain

ffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

> x, 2xþ 3 > x2 ðby Rule 5; with p ¼ 2Þ, x2 � 2x� 3 < 0

, x� 3ð Þ xþ 1ð Þ < 0:

So the part of the solution set in [0,1) is [0, 3).

We now examine those x for which �32� x < 0. For such x,


2xþ 3p

� 0

and x< 0, so thatffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

ð� 0Þ > x, for all such x. It follows that all these x,

namely the set ½� 32

, 0Þ, belong to the solution set too.

Combining these results, the solution set of the original inequality is

x :ffiffiffiffiffiffiffiffiffiffiffiffiffi

2xþ 3p

> x �

¼ � 3

2; 0

�

[ 0; 3½ Þ

¼ � 3

2; 3

�

: &

Problem 4 Solve the inequalityffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

> x:

1.2.3 Inequalities involving modulus signs

We now turn our attention to inequalities involving the modulus, or abso-

lute value, of a real number. Recall that, if a2R , then its modulus jaj is

defined by

aj j ¼ a; if a � 0,

�a; if a < 0.

�

It is often useful to think of jaj as the distance along the real line from 0 to a.

In the same way, ja� bj is the distance along the real line from 0 to a� b,

which is the same as the distance from a to b.

Notice also that jaþ bj ¼ ja� (�b)j is the distance from a to �b.

12 1: Numbers

For example, if a¼�2, b¼ 1,then:

1. j�2j> 0;

2. �j�2j ��2� j�2j;3. j� 2j2¼ (� 2)2;

4. j(�2) � 1j ¼ j1 � (�2)j;5. j(�2)� 1j ¼ j� 2j � j1j.

Note that, in a similar way

aj j � b, �b � a � b:

We take a¼ x� 2, b¼ 1 inRule 6.

For jx� 2j2¼ (x� 2)2.

We now list some basic properties of the modulus, which follow immediately

from the definition:

Properties of the modulus For any real numbers a and b:

1. jaj � 0, with equality if and only if a¼ 0;

2. –jaj � a� jaj;3. jaj2¼ a2;

4. ja� bj ¼ jb� aj;5. jabj ¼ jaj � jbj.

There is a basic rule for rearranging inequalities involving modulus signs:

Rule 6 For any real numbers a and b, where b> 0: aj j < b,�b < a < b.

Also, it is often possible, and sometimes easier, to use Rule 5 with p¼ 2 than to

use Rule 6. The following example illustrates the use of both rules.

Example 4 Solve the inequality jx� 2j < 1.

Solution Using Rule 6, we obtain

x� 2j j < 1, �1 < x� 2 < 1

, 1 < x < 3:

So the solution set of the original inequality is

x: x� 2j j < 1f g ¼ 1; 3ð Þ:

Alternatively, using Rule 5 (with p¼ 2), we obtain

x� 2j j < 1, x� 2ð Þ2< 1

, x2 � 4xþ 3 < 0

, x� 1ð Þ x� 3ð Þ < 0:

Again, this shows that the required solution set is (1, 3). &

Example 5 Solve the inequality jx� 2j � jxþ 1j.

Solution Using Rule 5 (with p¼ 2), we obtain

x� 2j j � xþ 1j j , x� 2ð Þ2� xþ 1ð Þ2

, x2 � 4xþ 4 � x2 þ 2xþ 1

, 3 � 6x

, 1

2� x:

So the solution set of the original inequality is

x: x� 2j j � xþ 1j jf g ¼ 1

2;1

�

: &

The inequalities in Examples 4 and 5 can easily be interpreted geometrically.

In Example 4, the inequality jx� 2j < 1 holds when the distance from x to 2

is strictly less than 1. So it holds for all points on either side of 2 at a distance

less than 1 from 2� namely, in the open interval (1, 3).

1.2 Inequalities 13

In Example 5, the inequality jx� 2j � jxþ 1j holds when the distance from x

to 2 is less than or equal to the distance from x to�1, since jxþ 1j ¼ jx� (�1)j.The mid-point of 2 and �1 (that is, the point x where the distance from x to 2

equals the distance from x to �1) is 12. So the inequality holds when x lies

in [12

, 1).

Some good ideas when tackling problems involving inequalities of these

types are:

� use your geometrical intuition, where possible, to give yourself an idea of

the sets involved;

� test one or two values of x in your final solution set to see if they are valid –

this often detects errors in manipulating inequality signs!

Problem 5 Solve the following inequalities:

(a) j2x2 � 13j < 5; (b) jx� 1j � 2jxþ 1j.

1.3 Proving inequalities

In this section we show you how to prove inequalities of various types. We

shall use the rules for rearranging inequalities given in Section 1.2, and also use

other rules which enable us to deduce new inequalities from old. We have

already met the first rule in Section 1.1, where it was called the Transitive

Property of R .

Transitive Rule a < b and b < c) a < c.

We use the Transitive Rule when we want to prove that a < c, and we know

that a < b and b < c.

The following rules are also useful:

Combination Rules

If a < b and c < d, then:

Sum Rule aþ c< bþ d;

Product Rule ac< bd (provided that a,c� 0).

There are also weak and weak/strict versions of the Transitive Rule and

Combination Rules, which we will ask you to work out and use as they arise.

Remark

It is important to appreciate that the Transitive Rule and the Combination

Rules have a different nature from Rules 1–6 in Section 1.2 . Rules 1–6 tell you

how to rearrange inequalities into equivalent forms, whereas the Transitive

Rule and the Combination Rules enable you to deduce new inequalities which

are not equivalent to the old ones.

For example, since 2< 3 and4< 5, then

2þ 4 < 3þ 5;

2� 4 < 3� 5:

For example, if a 0:

14 1: Numbers

1.3.1 The Triangle Inequality

If a and b are both vectors in R2, then the vector aþ b is obtained from the

‘parallelogram construction’:

b

a

a + b

By elementary geometry, the length of any side of a triangle is less than or

equal to the sum of the lengths of the other two sides. In the special case of R ,

when all the vectors lie on a line, this can be interpreted as the Triangle

Inequality, which involves the absolute value of the real numbers a, b and aþ b.

Triangle Inequality If a, b 2 R , then:

1. aþ bj j � aj j þ bj j;2. a� bj j �

�

� aj j � bj j�

� (the ‘reverse form’ of the Triangle Inequality).

Proof In order to prove part 1, we use Rule 5, with p¼ 2

aþ bj j � aj j þ bj j , aþ bð Þ2� aj j þ bj jð Þ2

, a2 þ 2abþ b2 � a2 þ 2 aj j bj j þ b2

, 2ab � 2 aj j bj j:The final inequality is certainly true for all a, b 2 R , and so the first

inequality must also be true for all a, b 2 R . Hence we have proved part 1.

We prove part 2 by using the same method

a� bj j ��

� aj j � bj j�

�, a� bð Þ2� aj j � bj jð Þ2

, a2 � 2abþ b2 � a2 � 2 aj j bj j þ b2

, �2ab � �2 aj j bj j, 2ab � 2 aj j bj j;

which is again true for all a, b 2 R . &

Remarks

1. Although we have used double-headed arrows here, the actual proof

requires only the arrows going from right to left. For example, in the

proof of part 1 the important implication is

aþ bj j � aj j þ bj j ( 2ab � 2 aj j bj j:2. Part 1 of the Triangle Inequality can also be proved by using Rule 6, as

follows. By Rule 6

aþ bj j � aj j þ bj j , � aj j þ bj jð Þ � aþ b � aj j þ bj jð Þ: (1)

Here we discuss R2, rather

than R , simply because theargument is thengeometrically clearer.

In the ‘parallelogramconstruction’, you draw thevector a from the origin tosome point, then the vector bfrom that point to a final point.The vector aþ b is then thevector from the origin to thatfinal point.

For example, with a¼�1 andb¼ 3:

1. j�1þ 3j � j�1j þ j3j;2. j(�1)� 3j � j j�1 j� j3j j.

Remember that aj j2¼ a2:

Part 2 is sometimes called the‘cunning form’ or ‘backwardsform’ of the TriangleInequality.


Now, we know that�|a|� a� |a| and�|b|� b� |b|; so, by the Sum Rule

� aj j þ bj jð Þ � aþ b � aj j þ bj jð Þ:It follows that the left-hand inequality in (1) must also hold, as required.

3. An obvious modification of the proof in Remark 2 shows that the following

more general form of the Triangle Inequality also holds:

Triangle Inequality for n terms For any real numbers a1, a2, . . ., an

a1 þ a2 þ � � � þ anj j � a1j j þ a2j j þ � � � þ anj j:

The following example is a typical application of the Triangle Inequality.

Example 1 Use the Triangle Inequality to prove that:

(a) jaj � 1)j3þ a3j � 4; (b) jbj< 1) j3� bj> 2.

Solution

(a) Suppose that aj j � 1. The Triangle Inequality then gives

3þ a3�

�

�

� � 3j j þ a3�

�

�

�

¼ 3þ aj j3

� 3þ 1 ðsince aj j � 1Þ¼ 4:

(b) Suppose that jbj< 1. The ‘reverse form’ of the Triangle Inequality then

gives

3� bj j ��

� 3j j � bj j�

�

¼�

�3� bj j�

�

� 3� bj j:

Now jbj< 1, so that � jbj>�1. Thus

3� bj j > 3� 1

¼ 2;

and we can then deduce from the previous chain of inequalities that

j3� bj> 2, as desired. &

Remarks

1. The results of Example 1 can also be stated in the form:

(a) j3þ a3j � 4, for jaj � 1;

(b) j3� bj> 2, for jbj< 1.

2. The reverse implications

3þ a3�

�

�

� � 4) aj j � 1 and 3� bj j > 2) bj j < 1

are FALSE. For example, try putting a ¼ �32

and b¼�2!

Problem 1 Use the Triangle Inequality to prove that:

(a) aj j � 12) aþ 1j j � 3

2; (b) bj j < 1

2) b3 � 1�

�

�

� > 78.

Note the use of the TransitiveRule here.

Again, we use the TransitiveRule.

16 1: Numbers

1.3.2 Inequalities involving n

In Analysis we often need to prove inequalities involving an integer n. It is a

common convention in mathematics that the symbol n is used to denote an

integer (frequently a natural number).

It is often possible to deal with inequalities involving n by using the

rearrangement rules given in Section 1.2. Here is such an example.

Example 2 Prove that 2n2 � nþ 1ð Þ2; for n � 3:

Solution Rearranging this inequality into an equivalent form, we obtain

2n2 � nþ 1ð Þ2 , 2n2 � nþ 1ð Þ2� 0

, n2 � 2n� 1 � 0

, n� 1ð Þ2� 2 � 0 ðby ‘completing the square’Þ, n� 1ð Þ2� 2:

This final inequality is clearly true for n� 3, and so the original inequality

2n2� (nþ 1)2 is true for n� 3. &

Remarks

1. In Problem 3 of Section 1.2, we asked you to solve the inequality 2x2� (xþ 1)2;

its solution set was �1; 1�ffiffiffi

2p

� �

[ 1þffiffiffi

2p

;1 �

. In Example 2, above,

we found those natural numbers n lying in this solution set.

2. An alternative solution to Example 2 is as follows

2n2 � nþ 1ð Þ2 , 2 � nþ 1

n

� 2

ðby Rule 3Þ

,ffiffiffi

2p� 1þ 1

nðby Rule 5, with p ¼ 1

2Þ;

and this final inequality certainly holds for n� 3.

Problem 2 Prove that 3nn2þ2

< 1; for n > 2.

1.3.3 More on inequalities

We now look at a number of inequalities and methods for proving inequalities

that will be useful later on.

Example 3 Prove that ab � aþ b2

� �2, for a, b 2R .

Solution We tackle this inequality using the various rearrangement rules and

a chain of equivalent inequalities until we obtain an inequality that we know

must be true

ab � aþ b

2

� 2

, ab � a2 þ 2abþ b2

4

, 4ab � a2 þ 2abþ b2

, 0 � a2 � 2abþ b2

, 0 � a� bð Þ2:

n 1 2 3 4

2n2 2 8 18 32

(nþ 1)2 4 9 16 25

This has the followinggeometric interpretation: Thearea of a rectangle withsides of length a and b isless than or equal to the areaof a square with sides oflength aþb

2.

a

b

This final inequality is certainly true, since all squares are non-negative. It

follows that the original inequality ab � aþb2

� �2is also true, for a, b 2 R . &

Remark

A close examination of the above chain of equivalent statements shows that in

fact ab ¼ aþb2

� �2if and only if a¼ b.

Problem 3 Prove that aþbffiffi

2p �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

; for a, b 2 R .

Problem 4 Suppose that a >ffiffiffi

2p

. Prove the following inequalities:

(a) 12

aþ 2a

� �

< a; (b) 12

aþ 2a

� �� 2> 2:

Hint: In part (b), use the result of Example 3 and the subsequent remark.

Example 4 Prove thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

� aþ b; for a; b � 0:

Solution We tackle this inequality using the various rearrangement rules and

a chain of equivalent inequalities until we obtain an inequality that we know

must be trueffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

� aþ b, a2 þ b2 � aþ bð Þ2

, a2 þ b2 � a2 þ 2abþ b2

, 0 � 2ab:

This final inequality is certainly true, since a, b� 0. It follows that the

original inequalityffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

� aþ b is also true, for a, b� 0. &

Problem 5 Use the result of Example 4 to prove thatffiffiffiffiffiffiffiffiffiffiffi

cþ dp

�ffiffiffi

cpþ

ffiffiffi

dp

; for c; d � 0:

Example 5 Prove thatffiffiffi

ap�

ffiffiffi

bp�

�

�

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a� bj jp

, for a, b� 0.

Solution Notice first that interchanging the roles of a and b leaves the

inequality unaltered. It follows that it is sufficient to prove the inequality

under the assumption that a� b.

So, assume that a� b. Then we know thatffiffiffi

ap�

ffiffiffi

bp

and a� bj j ¼ a� b:Hence

ffiffiffi

ap�

ffiffiffi

bp�

�

�

�

�

��

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a� bj jp

,ffiffiffi

ap�

ffiffiffi

bp�

ffiffiffiffiffiffiffiffiffiffiffi

a� bp

,ffiffiffi

ap�


a� bp

þffiffiffi

bp

:

This final inequality is certainly true, and is obtained from the result of

Problem 5 by simply substituting a� b in place of c and b in place of d. It

follows that the original inequalityffiffiffi

ap�

ffiffiffi

bp�

�

�

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a� bj jp

is also true, for

a, b� 0. &

Problem 6 Prove thatffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

aþ bþ cp

�ffiffiffi

apþ

ffiffiffi

bpþ

ffiffiffi

cp; for a; b; c � 0:

We often use the Binomial Theorem and the Principle of Mathematical

Induction (see Appendix 1) to prove inequalities.

Example 6 Prove the following inequalities, for n� 1:

(a) 2n� 1þ n; (b) 21n � 1þ 1

n:

In the formffiffiffiffiffi

abp

� aþb2

thisinequality is sometimes calledthe Arithmetic–GeometricMean Inequality for a, b.

This has the followinggeometric interpretation: Thelength of the hypotenuse of aright-angled triangle whoseother sides are of lengthsa and b is less than or equal tothe sum of the lengths of thosetwo sides.

ba2 + b2

a

This will simplify the detailsof our chain of inequalities.

We avoid one modulus as aresult of our simplifyingassumption!

Never be ashamed to utiliseevery tool at your disposal!(Why do the same worktwice?)

n 1 2 3 4

2n 2 4 8 16

1þ n 2 3 4 5

18 1: Numbers

Solution

(a) By the Binomial Theorem for n� 1

1þ xð Þn ¼ 1þ nxþ n n� 1ð Þ2!

x2 þ � � � þ xn

� 1þ nx; for x � 0:

Then, if we substitute x¼ 1 in this last inequality, we get

2n � 1þ n; for n � 1:

(b) We start by rewriting the required result in an equivalent form

21n � 1þ 1

n, 2 � 1þ 1

n

� n

ðby the Power RuleÞ:

Now, if we substitute x ¼ 1n

in the Binomial Theorem for (1þ x)n, we get

1þ 1

n

� n

¼ 1þ n1

n

�

þ n n� 1ð Þ2!

1

n

� 2

þ � � � þ 1

n

� n

� 1þ 1 ¼ 2:

Since the inequality 2 � 1þ 1n

� �nis true, it follows that the original

inequality 21n � 1þ 1

n, for n� 1, is also true, as required. &

Problem 7 Prove the inequality 1þ 1n

� �n� 52� 1

2n; for n� 1.

Hint: consider the first three terms in the binomial expansion.

Example 7 Prove that 2n� n2, for n� 4.

Solution Let P(n) be the statement

PðnÞ : 2n � n2:

First we show that P(4) is true: 24� 42.

STEP 1 Since 24¼ 16 and 42¼ 16, P(4) is certainly true.

STEP 2 We now assume that P(k) holds for some k� 4, and deduce that

P(kþ 1) is then true.

So, we are assuming that 2k� k2. Multiplying this inequality by 2

we get

2kþ1 � 2k2;

so it is therefore sufficient for our purposes to prove that 2k2� (kþ 1)2.

Now

2k2 � k þ 1ð Þ2 , 2k2 � k2 þ 2k þ 1

, k2 � 2k � 1 � 0 ðby ‘completing the square’Þ

, k � 1ð Þ2�2 � 0:

This last inequality certainly holds for k� 4, and so 2kþ1� (kþ 1)2

also holds for k� 4.

In other words: P(k) true for some k� 4)P(kþ 1) true.

It follows, by the Principle of Mathematical Induction, that 2n � n2, for

n� 4. &

Problem 8 Prove that 4n> n4, for n� 5.

n 1 2 3 4

21n 2 1.41 1.26 1.19

1þ 1n

2 1.5 1.33 1.25

We decrease the sum byomitting subsequentnon-negative terms.

We decrease the sum byomitting all but the first twoterms.

n 1 2 3 4 5

2n 2 4 8 16 32

n2 1 4 9 16 25

This assumption is just P(k).

Since P(kþ 1) is:2kþ1� (kþ 1)2.


The value of this result willcome from making suitablechoices of x and n forparticular purposes.

We prove the result usingMathematical Induction.

This assumption is P(k).This multiplication is validsince 1þ xð Þ � 0.

We decrease the expressionif we omit the final non-negative term.

You saw in part (b) of

Example 6 that 21n � 1þ 1

n:

We give the proof ofTheorem 2 at the end of thesub-section.

Three important inequalities in Analysis

Our first inequality, called Bernoulli’s Inequality, will be of regular use in later

chapters.

Theorem 1 Bernoulli’s Inequality

For any real number x��1 and any natural number n, (1þ x)n� 1þ nx.

Remark

In part (a) of Example 6, you saw that (1þ x)n� 1þ nx, for x> 0 and n a

natural number. Theorem 1 asserts that the same result holds under the weaker

assumption that x��1.

Proof Let P(n) be the statement

PðnÞ : 1þ xð Þn� 1þ nx; for x � �1:

STEP 1 First we show that P(1) is true: (1þ x)1� 1þ x. This is obviously

true.

STEP 2 We now assume that P(k) holds for some k� 1, and prove that P(kþ 1)

is then true.

So, we are assuming that 1þ xð Þk � 1þ kx; for x��1. Multiplying

this inequality by (1þ x), we get

1þ xð Þkþ1 � 1þ xð Þ 1þ kxð Þ¼ 1þ k þ 1ð Þxþ kx2

� 1þ k þ 1ð Þx:

Thus, we have 1þ xð Þkþ1� 1þ k þ 1ð Þx; in other words the statement

P(kþ 1) holds.

So, P(k) true for some k� 1)P(kþ 1) true.

It follows, by the Principle of Mathematical Induction, that 1þ xð Þn�1þ nx, for x � �1, n � 1: &

Problem 9 By applying Bernoulli’s Inequality with x ¼ � 1ð2nÞ, prove

that 21n � 1þ 1

2n�1, for any natural number n.

Our second inequality is of considerable use in various branches of

Analysis. In Problem 3 you proved that aþbffiffi

2p �


a2 þ b2p

, for a, b2R . We

can rewrite this inequality in the equivalent form aþ bð Þ2� 2 a2 þ b2ð Þ or

aþ bð Þ2� a2 þ b2ð Þ 12 þ 12ð Þ. The Cauchy–Schwarz Inequality is a general-

isation of this result to 2n real numbers.

Theorem 2 Cauchy–Schwarz Inequality

For any real numbers a1; a2; . . .; an and b1; b2; . . .; bn; we have

a1b1þ a2b2 þ � � � þ anbnð Þ2

� a21 þ a2

2 þ � � � þ a2n

� �

b21 þ b2

2 þ � � � þ b2n

� �

:

20 1: Numbers

For example, with n¼ 3

1þ 2þ 3ð Þ 1

1þ 1

2þ 1

3

�

¼ 6� 11

6

¼ 11 � 32:

We give the proof ofTheorem 3 at the end of thesub-section.

For example, with n¼ 3

1þ 1

3

� 3

¼ 64

27¼ 2:37 . . .

5 1þ 1

4

� 4

¼ 625

256¼ 2:44 . . .

You may omit these proofs ata first reading.

Note that we will use the �notation to keep the argumentbrief.

A is non-zero, by assumption,so that 1/A makes sense.

Problem 10 Use Theorem 2 to prove that for any positive real numbers

a1, a2, . . ., an, then a1 þ a2 þ � � � þ anð Þ 1a1þ 1

a2þ � � � þ 1

an

� �

� n2:

Our final result also has many useful applications. In Example 3 you proved

that ab � aþ b2

� �2, for a, b2R ; it follows that, if a and b are positive, then

abð Þ12� aþb

2. The Arithmetic Mean–Geometric Mean Inequality is a general-

isation of this result for two real numbers to n real numbers.

Theorem 3 Arithmetic Mean–Geometric Mean Inequality

For any positive real numbers a1, a2, . . ., an, we have

a1a2 . . . anð Þ1n� a1 þ a2þ � � �þ an

n:

Problem 11 Use Theorem 3 with the nþ 1 positive numbers 1, 1þ 1n,

1þ 1n, . . ., 1þ 1

nto prove that, for any positive integer n

1þ 1

n

� n

� 1þ 1

nþ 1

� nþ1

:

Proofs of Theorems 2 and 3

Theorem 2 Cauchy–Schwarz Inequality

For any real numbers a1, a2, . . ., an and b1, b2, . . ., bn, we have

a1b1þ a2b2 þ � � � þanbnð Þ2� a21 þ a2

2 þ � � � þ a2n

� �

b21 þ b2

2 þ � � � þ b2n

� �

:

Proof If all the as are zero, the result is obvious; so we need only examine

the case when not all the as are zero. It follows that, if we denote the

sumPn

k¼1a2k ¼ a2

1 þ a22 þ � � � þ a2

n by A, then A> 0. Also, denotePn

k¼1b2k¼b2

1þb22þ��þb2

n by B andPn

k¼1akbk¼ a1b1þ a2b2þ��þanbn by C.

Now, for any real number l, we have that lak þ bkð Þ2� 0, so that

X

n

k¼1

l2a2k þ 2lakbk þ b2

k

� �

¼X

n

k¼1

lak þ bkð Þ2 � 0;

which we may rewrite in the form

l2Aþ 2lC þ B � 0:

But this inequality is equivalent to the inequality

lAþ Cð Þ2þAB � C2; for any real number l:

Since A is non-zero, we may now choose l ¼ � CA. It follows from the last

inequality that AB�C2, which is exactly what we had to prove. &

Remark

If not all the as are zero, equality can only occur ifP

n

k¼1

lak þ bkð Þ2¼ 0; that is,

if all the numbers ak are proportional to all the numbers bk, 1 � k � n:


Theorem 3 Arithmetic Mean–Geometric Mean Inequality

For any positive real numbers a1, a2, . . ., an, we have

a1a2 . . . anð Þ1n� a1 þ a2þ . . .þan

n: (2)

Proof Since the ai are positive, we can rewrite (2) in the equivalent form

a1a2 . . . anð Þ1n

a1 þ a2þ � � � þanð Þ=n� 1: (3)

Now, replacing each term ai by lai for any non-zero number l does not alter

the left-hand side of the inequality (3). It follows that it is sufficient to prove the

inequality (2) in the special case when the product of the terms ai is 1. Hence it

is sufficient to prove the following statement P(n) for each natural number n:

P(n): For any positive real numbers ai with a1a2 . . . an¼ 1, then

a1þ a2þ � � � þ an� n.

First, the statement P(1) is obviously true.

Next, we assume that P(k) holds for some k� 1, and prove that P(kþ 1) is

then true.

Now, if all the terms a1, a2, . . ., akþ1 are equal to 1, the result P(kþ 1)

certainly holds. Otherwise, at least two of the terms differ from 1, say a1 and a2,

such that a1> 1 and a2< 1. Hence

a1 � 1ð Þ � a2 � 1ð Þ � 0;

which after some manipulation we may rewrite as

a1 þ a2 � 1þ a1a2: (4)

We are now ready to tackle P(kþ 1). Then

a1 þ a2þ � � � þ akþ1 � 1þ a1a2 þ a3 þ a4 þ � � � þ akþ1

� k þ 1;

since we may apply the assumption that P(k) holds to the k quantities a1a2,

a3, a4, . . ., akþ1. This last inequality is simply the statement that P(kþ 1) is

indeed true.

It follows by the Principle of Mathematical Induction that P(n) holds for all

natural numbers n, and so the inequality (2) must also hold. &

Remark

A careful examination of the proof of Theorem 3 shows that equality can only

occur if all the terms ai are equal.

1.4 Least upper bounds and greatest lower bounds

1.4.1 Upper and lower bounds

Any finite set {x1, x2, . . ., xn} of real numbers obviously has a greatest element

and a least element, but this property does not necessarily hold for infinite sets.

We denote the typical term byai rather than ak to avoidconfusion with a different useof the letter k in theMathematical Inductionargument below.

We will prove this byMathematical Induction.

The argument is exactly thesame whichever two termsactually differ from 1.

You should check thisyourself.

By (4).

That isa1a2ð Þ a3a4 . . . akþ1 ¼ 1

) a1a2ð Þ þ a3 þ a4 þ � � �þ akþ1 � k:

22 1: Numbers

For example, the interval (0, 2] has greatest element 2, but neither of the sets

N ¼ {1, 2, 3, . . .} nor [0, 2) has a greatest element. However the set [0, 2) is

bounded above by 2, since all points of [0, 2) are less than or equal to 2.

Definitions A set ER is bounded above if there is a real number, M

say, called an upper bound of E, such that

x � M; for all x 2 E:

If the upper bound M belongs to E, then M is called the maximum element

of E, denoted by max E.

Geometrically, the set E is bounded above by M if no point of E lies to the

right of M on the real line.

For example, if E ¼ [0, 2), then the numbers 2, 3, 3.5 and 157.1 are all upper

bounds of E, whereas the numbers 1.995, 1.5, 0 and �157.1 are not upper

bounds of E. Although it seems obvious that [0, 2) has no maximum element,

you may find it difficult to write down a formal proof. The following example

shows you how to do this:

Example 1 Determine which of the following sets are bounded above, and

which have a maximum element:

(a) E1 ¼ [0, 2); (b) E2 ¼ f1n

: n ¼ 1; 2; . . .g; (c) E3 ¼ N .

Solution

(a) The set E1 is bounded above. For example, M ¼ 2 is an upper bound

of E1, since

x � 2; for all x 2 E1:

However, E1 has no maximum element. For each x in E1, we have x< 2,

and so there is some real number y such that

x < y < 2;

by the Density Property of R .

Hence y2E1, and so x cannot be a maximum element.

(b) The set E2 is bounded above. For example, M¼ 1 is an upper bound of E2,

since

1

n� 1; for all n ¼ 1; 2; . . .:

Also, since 12E2

max E2 ¼ 1:

(c) The set E3 is not bounded above. For each real number M, there is a

positive integer n such that n>M, by the Archimedean Property of R .

Hence M cannot be an upper bound of E3.

This also means that E3 cannot have a maximum element. &

Problem 1 Sketch the following sets, and determine which are bounded

above, and which have a maximum element:

(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n

: n ¼ 1; 2; . . .g;(c) E3¼ {n2: n¼ 1, 2, . . .}.

For example, y can be of theform 1.99. . . 9 or y¼ 1

2(xþ 2).

2 is not a maximum element,since 2 =2E1.

Similarly, we define lower bounds. For example, the interval (0, 2) is

bounded below by 0, since

0 � x; for all x 2 0; 2ð Þ:However, 0 does not belong to (0, 2), and so 0 is not a minimum element of

(0, 2). In fact, (0, 2) has no minimum element.

Definitions A set ER is bounded below if there is a real number, m say,

called a lower bound of E, such that

m � x; for all x 2 E:

If the lower bound m belongs to E, then m is called the minimum element of

E, denoted by min E.

Geometrically, the set E is bounded below by m if no point of E lies to the

left of m on the real line.

Problem 2 Determine which of the following sets are bounded below,

and which have a minimum element:

(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n: n ¼ 1; 2; . . .g;

(c) E3¼ {n2: n¼ 1, 2, . . .}.

The following terminology is also useful:

Definition A set ER is bounded if it is bounded above and bounded

below.

For example, the set E2 ¼ f1� 1n: n ¼ 1; 2; . . .g is bounded, but the sets

E1¼ (�1, 1] and E3¼ {n2: n¼ 1, 2, . . .} are not bounded.

Similar terminology applies to functions.

Definitions A function ƒ defined on an interval IR is said to:

� be bounded above by M if f(x)�M, for all x2 I; M is an upper bound of ƒ;

� be bounded below by m if f(x)�m, for all x2 I; m is a lower bound of ƒ;

� have a maximum (or maximum value) M if M is an upper bound of ƒ and

f(x)¼M, for at least one x2 I;

� have a minimum (or minimum value) m if m is a lower bound of ƒ and

f(x)¼m, for at least one x2 I.

Example 2 Let ƒ be the function defined by f (x)¼ x2, x 2 12

, 3 �

. Determine

whether ƒ is bounded above or below, and any maximum or minimum value

of ƒ.

Solution First, ƒ is increasing on the interval 12

, 3Þ

, so that since 12� x < 3 it

follows that 14� f ðxÞ < 9. Hence ƒ is bounded above and bounded below.

Next, since f 12

� �

¼ 14

and 14

is a lower bound for ƒ on the interval 12

, 3Þ

, it

follows that ƒ has a minimum value of 14

on this interval.

Finally, 9 is an upper bound for ƒ on the interval 12

, 3Þ

but there is no point

x in 12

, 3Þ

for which ƒ(x)¼ 9. So 9 cannot be a maximum of ƒ on the interval.

However, if y is any number in (8, 9) there is a number x>ffiffiffi

yp

inffiffiffi

yp

, 3� �

2ffiffiffi

2p

, 3� �

12

, 3 �

such that ƒ(x)¼ x2> y, so that no number in (8, 9) will serve

Strictly speaking, M and m arethe upper bound and lowerbound of the image set{ƒ(x): x 2 I}.

2√2 y 312

24 1: Numbers

as a maximum of ƒ on the interval. It follows that ƒ has no maximum value

on 12

, 3Þ

. &

Problem 3 Let ƒ be the function defined by f xð Þ ¼ 1x2 ; x 2 �3;�2½ Þ.

Determine whether ƒ is bounded above or below, and any maximum or

minimum value of ƒ.

1.4.2 Least upper bounds and greatest lower bounds

We have seen that the interval [0, 2] has a maximum element 2, but [0, 2) has

no maximum element. However, the number 2 is ‘rather like’ a maximum

element of [0, 2), because 2 is an upper bound of [0, 2) and any number less

than 2 is not an upper bound of [0, 2). In other words, 2 is the least upper

bound of [0, 2).

Definition A real number M is the least upper bound, or supremum, of a

set ER if:

1. M is an upper bound of E;

2. if M0<M, then M0 is not an upper bound of E.

In this case, we write M¼ sup E.

If E has a maximum element, max E, then sup E¼max E. For example,

the closed interval [0, 2] has least upper bound 2. We can think of the least

upper bound of a set, when it exists, as a kind of ‘generalised maximum

element’.

If a set does not have a maximum element, but is bounded above, then we may

be able to guess the value of its least upper bound. As in the case E¼ [0, 2), there

may be an obvious ‘missing point’ at the upper end of the set. However it is

important to prove that your guess is correct. We now show you how to do this.

Example 3 Prove that the least upper bound of [0, 2) is 2.

Solution We know that M¼ 2 is an upper bound of [0, 2), because

x � 2; for all x 2 ½0; 2Þ:To show that 2 is the least upper bound, we must prove that each number

M0< 2 is not an upper bound of [0, 2). To do this, we must find an element

x in [0, 2) which is greater than M0. But, if M0< 2, then there is a real number

x such that

M0 < x < 2

and also

0 < x < 2:

Since x2 [0, 2), the number M0 cannot be an upper bound of [0, 2). Hence

M¼ 2 is the least upper bound, or supremum, of [0, 2). &

Although the conclusion of Example 3 may seem painfully obvious, we have

written out the solution in detail because it illustrates the strategy for determin-

ing the least upper bound of a set, if it has one.

Part 1 says that M is an upperbound.Part 2 says that no smallernumber can be an upperbound.

For example, x can be of theform 1.99 . . . 9 for a suitablylarge number of digits, or it

can be 12

M0 þ 2ð Þ since

M0< 12

M0 þ 2ð Þ < 2:

Strategy Given a subset E of R , to show that M is the least upper bound, or

supremum, of E, check that:

1. x�M, for all x2E;

2. if M0<M, then there is some x2E such that x>M0.

Notice that, if M is an upper bound of E and M2E, then part 2 is auto-

matically satisfied, and so M¼ sup E¼max E.

Example 4 Determine the least upper bound of E ¼ f1� 1n2 : n ¼ 1; 2; . . .g.

Solution We guess that the least upper bound of E is M¼ 1. Certainly, 1 is

an upper bound of E, since

1� 1

n2� 1; for n ¼ 1; 2; . . .:

To check part 2 of the strategy, we need to show that, if M0< 1, then there

is some natural number n such that

1� 1

n2> M0: (1)

However

1� 1

n2> M0 , 1�M0 >

1

n2

, 1

1�M0< n2 ðsince 1�M0 > 0Þ

,ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

1�M0

r

< n ðsince1

1�M0> 0

and n > 0Þ:We can certainly choose n so that this final inequality holds, by the

Archimedean Property of R , and so we can choose n so that inequality (1) holds.

Hence 1 is the least upper bound of E. &

Remark

Although we used double-headed arrows in this solution, the actual proof

required only the implications going from right to left. In other words, the

proof uses only the fact that

1� 1

n2> M0 (


1

1�M0

r

< n:

Problem 4 Determine sup E, if it exists, for each of the following sets:

(a) E1¼ (�1, 1]; (b) E2 ¼ f1� 1n

: n ¼ 1, 2, . . .g;(c) E3¼ {n2: n¼ 1, 2, . . .}.

Similarly, we define the notion of a greatest lower bound.

Definition A real number m is the greatest lower bound, or infimum, of

a set ER if:

1. m is a lower bound of E;

2. if m0>m, then m0 is not a lower bound of E.

In this case, we write m¼ inf E.

GUESS the value of M, thenCHECK parts 1 and 2.

That is, 1¼ sup E.

Part 1 says that m is a lowerbound.Part 2 says that no largernumber can be a lower bound.

26 1: Numbers

If E has a minimum element, min E, then inf E¼min E. For example, the

closed interval [0, 2] has greatest lower bound 0. We can think of the greatest

lower bound of a set, when it exists, as a kind of ‘generalised minimum

element’.

The strategy for establishing that a number is the greatest lower bound of

a set is very similar to that for proving that a number is the least upper bound

of a set.

Strategy Given a subset E of R , to show that m is the greatest lower

bound, or infimum, of E, check that:

1. x�m, for all x2E;

2. if m0>m, then there is some x2E such that x<m0.

Notice that, if m is a lower bound of E and m2 E, then part 2 is automatically

satisfied, and so m¼ inf E¼min E.

Problem 5 Determine inf E, if it exists, for each of the following sets:

(a) E1¼ (1, 5]; (b) E2 ¼ f 1n2 : n ¼ 1; 2; . . .g.

Remarks

1. For any subset E of R , inf E� sup E. This follows from the fact that, for

any x2E, we have inf E� x� sup E.

2. For any bounded interval I of R , let a be its left end-point and b its right

end-point. Then inf I¼ a and sup I¼ b.

Least upper bounds and greatest lower boundsof functions

Similar terminology applies to bounds for functions.

Definitions Let ƒ be a function defined on an interval IR . Then:

� A real number M is the least upper bound, or supremum, of ƒ on I if:

1. M is an upper bound of ƒ(I);

2. if M0<M, then M0 is not an upper bound of ƒ(I).

In this case, we write M ¼ sup f or supI

f or sup f xð Þ : x 2 If g or supx2I

f xð Þ:� A real number m is the greatest lower bound, or infimum, of ƒ on I if:

1. m is a lower bound of ƒ(I);

2. if m0>m, then m0 is not a lower bound of ƒ(I).

In this case, we write m ¼ inf f or infI

f or inf f xð Þ : x 2 If g or infx2I

f xð Þ.

Notice, for instance, that:

� if M is an upper bound for f on I, then supI

f � M,

� if m is a lower bound for f on I, then infI

f � m.

GUESS the value of m, thenCHECK parts 1 and 2.

There are similar definitionsfor the least upper bound andthe greatest lower bound of fon a general set S in R .

These are really thedefinitions for the least upperbound or the greatest lowerbound of the set {ƒ(x): x 2 I}.

Forsup

I

f is the least upper bound,

and infI

f is the greatest lower

bound, for f on I.

The strategies for proving that M is the least upper bound or m the greatest

lower bound of ƒ on I are similar to the corresponding strategies for the least

upper bound or the greatest lower bound of a set E.

Strategies Let ƒ be a function defined on an interval IR . Then:

� To show that m is the greatest lower bound, or infimum, of f on I, check

that:

1. f(x)�m, for all x2 I;

2. if m0>m, then there is some x2 I such that f(x)<m0.

� To show that M is the least upper bound, or supremum, of f on I, check that:

1. f(x)�M, for all x2 I;

2. if M0<M, then there is some x2 I such that f(x)>M0.

Example 5 Let ƒ be the function defined by f xð Þ ¼ x2; x 2 12; 3

�

. Determine

the least upper bound and the greatest lower bound of ƒ on 12; 3

�

.

Solution We have already seen that 9 is an upper bound for ƒ on 12; 3

�

, and

that no smaller number will serve as an upper bound. It follows that 9 must be

the least upper bound of ƒ on 12; 3

�

.

Similarly, we have already seen that 14

is the minimum value of ƒ on 12; 3

�

;

it follows that 14

is the greatest lower bound of ƒ on 12; 3

�

, and this is actually

attained ðat the point 12Þ. &

Problem 6 Let ƒ be the function defined by f xð Þ ¼ 1x2 ; x 2 ½1; 4Þ:

Determine the least upper bound and the greatest lower bound of ƒ on ½1, 4).

Remark

For any interval I of R , infI

f � supI

f . This follows from the fact that, for

any x2 I, we have infI

f � f xð Þ � supI

f .

The least upper bound and the greatest lower bound of a function on an

interval will be particularly significant in our later work on continuity and

integrability of functions.

1.4.3 The Least Upper Bound Property

In the examples in the previous sub-section, it was easy to guess the values of

sup E and inf E. At times, however, we shall meet sets for which these values

are not so easy to determine. For example, if

E ¼ 1þ 1

n

� n

: n ¼ 1; 2; . . .

� �

;

then it can be shown that E is bounded above by 3, but it is not easy to guess the

least upper bound of E.

In such circumstances, it is reassuring to know that sup E does exist, even

though it may be difficult to find. This existence is guaranteed by the following

fundamental result.

You saw this in Example 2.

Example 2.

Chapters 4 and 7.

We will study this set closelyin Section 2.5.

28 1: Numbers

The Least Upper Bound Property of R Let E be a non-empty subset

of R . If E is bounded above, then E has a least upper bound.

We leave the proof of the Least Upper Bound Property of R to the next sub-

section. However, the Property itself is intuitively obvious. If the set E lies

entirely to the left of some number M, then you can imagine moving M steadily

to the left until you meet E. At this point, sup E has been reached.

The Least Upper Bound Property of R can be used to show that R does

include decimals which represent irrational numbers such asffiffiffi

2p

, as we

claimed in Section 1.1.

In Sections 1.2 and 1.3 we have taken for granted the existence of rational

powers and their properties, without giving formal definitions. How can we

supply these definitions? For example, how can we defineffiffiffi

2p

as a decimal?

Consider the set

E ¼ x 2 Q : x > 0; x2< 2 �

:

This is the set of positive rational numbers whose squares are less than 2.

Intuitively,ffiffiffi

2p

lies on the number line to the right of the numbers in E, but

‘only just’! In fact, we should expectffiffiffi

2p

to be the least upper bound of E.

Certainly E has a least upper bound, by the Least Upper Bound Property,

because E is bounded above, by 1.5 for example. Thus it seems likely that sup E

is the decimal representation offfiffiffi

2p

. But how can we prove that (sup E)2¼ 2?

We shall prove this in Section 1.5, once we have described how to do

arithmetic with real numbers (decimals).

Finally, note that there is a corresponding result about lower bounds.

The Greatest Lower Bound Property of R Let E be a non-empty subset

of R . If E is bounded below, then E has a greatest lower bound.

1.4.4 Proof of the Least Upper Bound Property

We know that E is a non-empty set, and we shall assume for simplicity that E

contains at least one positive number. We also know that E is bounded above.

The following procedure gives us the successive digits in a particular decimal,

which we then prove to be the least upper bound of E.

Procedure to find a¼ a0 � a1a2 . . .¼ sup E

Choose in succession:

� the greatest integer a0 such that a0 is not an upper bound of E;

� the greatest digit a1 such that a0 � a1 is not an upper bound of E;

� the greatest digit a2 such that a0 � a1a2 is not an upper bound of E;

..

.

� the greatest digit an such that a0 � a1a2 . . . an is not an upper bound of E;

..

.

The Least Upper BoundProperty of R is an exampleof an existence theorem, onewhich asserts that a realnumber exists having a certainproperty. Analysis containsmany such results whichdepend on the Least UpperBound Property of R . Whilethese results are often verygeneral, and their proofselegant, they do not alwaysprovide the most efficientmethods of calculating goodapproximate values for thenumbers in question.

You may omit this proof at afirst reading.

For example, if

E ¼fx 2 Q : x > 0;

x2 < 2g;then

a0 ¼ 1; since 12 < 2 < 22;

a0 � a1 ¼ 1:4; since

1:42 < 2 < 1:52;

a0 � a1a2 ¼ 1:41; since

1:412 < 2 < 1:422;

..

.

Thus, at the nth stage we choose the digit an so that:

� a0 � a1a2 . . . an is not an upper bound of E;

� a0 � a1a2 . . . an þ 110n is an upper bound of E.

We now prove that the least upper bound of E is a¼ a0 � a1a2 . . ..First, we have to prove that a is an upper bound of E. To do this, we prove

that, if x> a, then x =2E (this is equivalent to proving, that, if x2E, then x� a).

We begin by representing x as a non-terminating decimal x¼ x0 � x1x2 . . ..Since x> a, there is an integer n such that

a < x0 � x1x2 . . . xn:

Hence

x0 � x1x2 . . . xn � a0 � a1a2 . . . an þ1

10n;

and so, by our choice of an, x¼ x0 � x1 x2 . . . xn is an upper bound of E. Since

x> x0 � x1x2 . . . xn, we have that x =2E, as required.

Next, we have to show that, if x< a, then x is not an upper bound of E. Since

x< a, there is an integer n such that

x < a0 � a1a2 . . . an;

and so x is not an upper bound of E, by our choice of an.

Thus we have proved that a is the least upper bound of E. &

Remark

Notice that this proof does not use any arithmetical properties of the real

numbers but only their order properties, together with the arithmetical proper-

ties of rational numbers. In the next section, we use the Least Upper Bound

Property to define some of the arithmetical operations on R .

1.5 Manipulating real numbers

1.5.1 Arithmetic in R

At the end of Section 1.1 we discussed the decimalsffiffiffi

2p¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;

and asked whether it is possible to add and multiply these numbers to obtain

another real number. We now explain how this can be done, using the Least

Upper Bound Property of R .

A natural way to obtain a sequence of approximations to the sumffiffiffi

2pþ p

is to truncate each of the above decimals, and form the sums of the

truncations.

If each of the decimals is truncated at the same decimal place, this gives a

sequence of approximations which is increasing:

Here we are assuming thatffiffiffi

2p

and p can be represented asdecimals.

30 1: Numbers

ffiffiffi

2p

pffiffiffi

2pþ p

1 3 4

1.4 3.1 4.5

1.41 3.14 4.55

1.414 3.141 4.555

1.4142 3.1415 4.5557... ..

. ...

Intuitively, we should expect that the sumffiffiffi

2pþ p is greater than each of the

numbers in the right-hand column, but ‘only just’! To accord with our intuition,

therefore, we define the sumffiffiffi

2pþ p to be the least upper bound of the set of

numbers in the right-hand column; that isffiffiffi

2pþ p ¼ sup 4; 4:5; 4:55; 4:555; 4:5557; . . .f g:

To be sure that this definition makes sense, we need to show that this set

is bounded above. But all the truncations offfiffiffi

2p

are less than 1.5, and all the

truncations of p are less than, say, 4. Hence, all the sums in the right-hand

column are less than 1.5þ 4¼ 5.5. So, by the Least Upper Bound Property, the

set of numbers in the right-hand column does have a least upper bound, and we

can defineffiffiffi

2pþ p in this way.

This method can be used to define the sum of any pair of positive real

numbers.

Let us check that this method of adding decimals gives the correct answer

when we use it in a familiar case. Consider the simple calculation

1

3þ 2

3¼ 0:333 . . .þ 0:666 . . .:

Truncating each of these decimals and forming the sums, we obtain the set

0; 0:9; 0:99; 0:999; . . .f g:

The supremum of this set is, of course, the number 0.999. . .¼ 1, which is the

correct answer.

Similarly, we can define the product of any two positive real numbers. For

example, to defineffiffiffi

2p� p, we can form the sequence of products of their

truncations:

ffiffiffi

2p

pffiffiffi

2p� p

1 3 3

1.4 3.1 4.34

1.41 3.14 4.4274

1.414 3.141 4.441374

1.4142 3.1415 4.4427093... ..

. ...

We do not expect you to usethis method to add decimals!


As before, we defineffiffiffi

2p� p to be the least upper bound of the set of numbers

in the right-hand column.

Similar ideas can be used to define the operations of subtraction and

division.

Thus we can define arithmetic with real numbers in terms of the familiar

arithmetic with rationals, using the Least Upper Bound Property of R .

Moreover, it can be proved that these operations in R satisfy all the usual

properties of a field.

1.5.2 The existence of roots

Just as we usually take for granted the basic arithmetical operations with real

numbers, so we usually assume that, given any positive real number a, there is

a unique positive real number b ¼ffiffiffi

ap

such that b2¼ a. We now discuss the

justification for this assumption.

First, here is a geometrical justification. Given line segments of lengths 1

and a, we can construct a semi-circle with diameter aþ 1 as shown.

a

b

1

Using similar triangles, we see that

a

b¼ b

1;

and so

b2 ¼ a:

This shows that there should be a positive real number b such that b2¼ a,

so that the length of the vertical line segment in the figure can be described

exactly by the expressionffiffiffi

ap

. But does b ¼ffiffiffi

ap

exist exactly as a real number?

In fact it does, and a more general result is true.

Theorem 1 For each positive real number a and each integer n> 1, there

is a unique positive real number b such that

bn ¼ a:

We call this number b the nth root of a, and we write b ¼ffiffiffi

anp

. We also defineffiffiffi

0np¼ 0, since 0n¼ 0, and if n is odd we define

ffiffiffiffiffiffiffiffiffiffi

�að Þnp

¼ �ffiffiffi

anp

, since

�ffiffiffi

anp

ð Þn¼ �a if n is odd.

Let us illustrate Theorem 1 with the special case a¼ 2 and n¼ 2. In this case,

Theorem 1 asserts the existence of a real number b such that b2¼ 2. In other

words, it asserts the existence of a decimal b which can be used to defineffiffiffi

2p

precisely.

Here is a direct proof of Theorem 1 in this special case. We choose the

numbers 1, 1.4, 1.41, 1.414, . . . to satisfy the inequalities

We omit the details.

These properties were listedin Sub-section 1.1.5.

For each positive integer n,we can also construct

ffiffiffi

np

asfollows:

1

1

1 1

1

1

1

2

345

6

7

We shall prove Theorem 1 inSub-section 4.3.3.

For example,ffiffiffiffiffiffiffiffiffiffi

�8ð Þ3p

¼ �2:

32 1: Numbers

12< 2 < 22

ð1:4Þ2< 2 < ð1:5Þ2

ð1:41Þ2< 2 < ð1:42Þ2

ð1:414Þ2< 2 < ð1:415Þ2

..

.

(1)

This process gives an infinite decimal

b ¼ 1:414 . . .;

and we claim that

b2 ¼ ð1:414 . . .Þ2 ¼ 2:

This can be proved using our method of multiplying decimals:

b b b2

1 1 1

1.4 1.4 1.96

1.41 1.41 1.9881

1.414 1.414 1.999396... ..

. ...

We have to prove that the least upper bound of the set E of numbers in the right-

hand column is 2, in other words that

sup E ¼ sup f1; ð1:4Þ2; ð1:41Þ2; ð1:414Þ2; . . .g ¼ 2:

To do this, we employ the strategy given in Sub-section 1.4.2.

First, we check that M¼ 2 is an upper bound of E. This follows from the left-

hand inequalities in (1).

Next, we check that, if M0< 2, then there is a number in E which is greater

than M0. To prove this, put

x0 ¼ 1; x1 ¼ 1:4; x2 ¼ 1:41; x3 ¼ 1:414; . . .:

Then, by the right-hand inequalities in (1), we have that

xn þ1

10n

� 2

> 2:

Also

xn þ1

10n

� 2

� x2n ¼

1

10n2xn þ

1

10n

�

51

10n2� 2þ 1ð Þ ¼ 5

10n;

and so

x2n > xn þ

1

10n

� 2

� 5

10n

> 2� 5

10n

¼ 1:99 . . . 95:n digits

Notice that

b ¼ 1:414 . . .

is the decimal that weobtained as the least upperbound of the set

fx 2 Q : x > 0; x2 < 2gin Sub-section 1.5.1.

For example, if n¼ 1, then

1:4þ 1

10

� 2

¼ 1:52 > 2:

For example, if n¼ 2, then

ð1:41Þ2 > 1:95:

So, if M0< 2, then we can choose n so large that xn2>M0 (while still having

xn2E). This proves that the least upper bound of E is 2, and so (1.414. . .)2¼ 2.

Thus we can defineffiffiffi

2p¼ 1:414 . . .;

which justifies our earlier claim thatffiffiffi

2p

can be represented exactly by a

decimal.

1.5.3 Rational powers

Having discussed nth roots, we are now in a position to define the expression

ax, where a is positive and x is rational.

Definition If a> 0, m2Z and n2N , then

amn ¼

ffiffiffi

anp� �m

:

For example, for a> 0, with m¼ 1 we have a1n ¼

ffiffiffi

anp

, and with m¼ 2 and

n¼ 3 we have a23 ¼

ffiffiffi

a3pð Þ2:

This notation is particularly useful, because rational powers (or rational expo-

nents) satisfy the following exponent laws (whose proofs depend on Theorem 1):

Exponent Laws

� If a, b> 0 and x2Q , then axbx¼ (ab)x.

� If a> 0 and x, y2Q , then ax ay¼ axþ y.

� If a> 0 and x, y2Q , then axð Þy¼ axy.

If x and y are integers, these laws actually hold for all non-zero real numbers

a and b. However, if x and y are not integers, then we must have a, b> 0. For

example, �1ð Þ12 is not defined as a real number.

However, if a is a negative real number, then amn can be defined whenever

m2Z, n2N and mn

is reduced to its lowest terms with n odd, as follows

amn ¼

ffiffiffi

anp� �m

:

Finally, you may have wondered why we did not mention that each positive

number has two nth roots when n is even. For example, 22¼ (�2)2¼ 4. We

shall adopt the convention that, for a> 0,ffiffiffi

anp

and a1n always refer to the positive

nth root of a. If we wish to refer to both roots (for example, when solving

equations), we write �ffiffiffi

anp

.

1.5.4 Real powers

We conclude this section by briefly discussing the meaning of ax when a> 0

and x is an arbitrary real number. We have defined this expression when x is

rational, but the same definition does not work if x is irrational. However, it

is common practice to write down expressions such asffiffiffi

2p ffiffi

2p

, and even to apply

the Exponent Laws to give equalities such as

For example

212 � 3

12 ¼ 6

12;

212 � 2

13 ¼ 2

56;

212

� �13¼ 2

16:

This extends our abovedefinition of a

mn ; for instance,

it defines a1n whenever n2N

and n is odd. For example,�8ð Þ

23¼ 4.

34 1: Numbers

ffiffiffi

2p ffiffi

2p�

ffiffi

2p

¼ffiffiffi

2p� �

ffiffi

2p�ffiffi

2p

¼ffiffiffi

2p� �2

¼ 2:

Can such manipulations be justified?

In fact, it is possible to define ax, for a> 0 and x2R , using the Least Upper

Bound Property of R , but it is then rather tricky to check that the Exponent

Laws work. In Chapters 2 and 3 we shall explain how to define the expression

ex, and in Chapter 4 we use ex to define the real powers in general and show that

the Exponent Laws hold. For the time being, whenever the expression ax

appears, you should assume that x is rational.

1.6 Exercises

Section 1.1

1. Arrange the following numbers in increasing order:

(a) 736; 3

20; 1

6; 7

45; 11

60;

(b) 0:465, 0:465, 0:465, 0.4655, 0.4656.

2. Find the fractions whose decimal expansions are:

(a) 0:481; (b) 0:481.

3. Let x ¼ 0:21 and y ¼ 0:2. Find xþ y and xy (in decimal form).

4. Find a rational number x and an irrational number y in the interval

(0.119, 0.12).

5. Prove that, if n is a positive integer which is not a perfect square, thenffiffiffi

np

is irrational.

Hint: If p, q are positive integers such that�

pq

�2 ¼ n, and k is the positive

integer such that k < pq< k þ 1 (why does such a positive integer k exist?),

show that 0< p� kq< q and nq�kpp�kq

¼ pq, and hence obtain a contradiction.

Section 1.2

1. Solve the following inequalities:

(a) x�1x2þ4

< xþ1x2�4

; (b)ffiffiffiffiffiffiffiffiffiffiffiffiffi

4x� 3p

> x;

(c) 17� 2x4�

�

�

� � 15; (d) xþ 1j j þ x� 1j j < 4.

Section 1.3

1. Use the Triangle Inequality to prove that

aj j � 1) a� 3j j � 2:

2. Prove that (a2þ b2)(c2þ d2)� (acþ ad)2, for any a, b, c, d2R .

3. Prove the inequality 3n� 2n2þ 1, for n¼ 1, 2, . . .:

1.6 Exercises 35

(a) by using the Binomial Theorem, applied to (1þ x)n with x¼ 2;

(b) by using the Principle of Mathematical Induction.

4. Use the Principle of Mathematical Induction to prove that, for n¼ 1, 2, . . . :

(a) 12 þ 22 þ 32 þ � � � þ n2 ¼ n nþ1ð Þ 2nþ1ð Þ6

;

(b)

ffiffiffiffiffiffiffiffi

5=4

4nþ1

q

� 1�3�5� ... � 2n�1ð Þ2�4�6� ... � 2nð Þ �


3=4

2nþ1

q

.

5. Apply Bernoulli’s Inequality, first with x ¼ 2n

and then with x ¼ �2ð3nÞ to

prove that1þ 2

3n� 2� 3

1n � 1þ 2

n; for n ¼ 1; 2; . . . :

6. By applying the Arithmetic Mean–Geometric Mean Inequality to the nþ 1

positive numbers 1, 1� 1n

, 1� 1n

, 1� 1n

, . . . , 1� 1n, prove that

1� 1n

� �n� 1� 1nþ1

� �nþ1

; for n ¼ 1; 2; . . . :

7. Use the Cauchy–Schwarz Inequality to prove that, if a1, a2, . . . , an are

positive numbers with a1þ a2þ � � � þ an¼ 1, thenffiffiffi

ap

1 þffiffiffi

ap

2 þ � � � þffiffiffi

ap

n �ffiffiffi

np

:

8. Use the Cauchy–Schwarz Inequality to prove that

3ffiffiffiffiffiffiffiffiffiffi

cos x4p

þ 4

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�ffiffiffiffiffiffiffiffiffiffi

cos xpq

� 5; for x 2 0; p2

�

:

Section 1.4

In Exerc ises 1–4 , take E1¼ {x: x2Q , 0� x< 1} and E2 ¼

1þ 1n

� �2:

n ¼ 1; 2; . . .g:1. Prove that each of the sets E1 and E2 is bounded above. Which of them has

a maximum element?

2. Prove that each of the sets E1 and E2 is bounded below. Which of them has

a minimum element?

3. Determine the least upper bound of each of the sets E1 and E2.

4. Determine the greatest lower bound of each of the sets E1 and E2.

5. For each of the following functions, determine whether it has a maximum

or a minimum, and determine its supremum and infimum:

(a) f xð Þ ¼ 11þx2 ; x 2 0; 1½ Þ; (b) f xð Þ ¼ 1� xþ x2; x 2 0; 2½ Þ.

6. Prove that, for any two numbers a, b2R

minfa; bg ¼ 12

aþ b� ja� bjð Þ and

maxfa; bg ¼ 12

aþ bþ jaþ bjð Þ:

36 1: Numbers

2 Sequences

This chapter deals with sequences of real numbers, such as

1;1

2;1

3;1

4;1

5;1

6; . . .;

0; 1; 0; 1; 0; 1; . . .;

1; 2; 4; 8; 16; 32; . . .:

It describes in detail various properties that a sequence may possess, the most

important of which is convergence. Roughly speaking, a sequence is conver-

gent, or tends to a limit, if the numbers, or terms, in the sequence approach

arbitrarily close to a unique real number, which is called the limit of the

sequence. For example, we shall see that the sequence

1;1

2;1

3;1

4;1

5;1

6; . . .

is convergent with limit 0. On the other hand, the terms of the sequence

0; 1; 0; 1; 0; 1; . . .

do not approach arbitrarily close to any unique real number, and so this

sequence is not convergent. Likewise, the sequence

1; 2; 4; 8; 16; 32; . . .

is not convergent.

A sequence which is not convergent is called divergent. The sequence

0; 1; 0; 1; 0; 1; . . .;

is a bounded divergent sequence. The sequence

1; 2; 4; 8; 16; 32; . . .

is unbounded; its terms become arbitrarily large and positive, and we say that it

tends to infinity.

Intuitively, it seems plausible that some sequences are convergent, whereas

others are not. However, the above description of convergence, involving the

phrase ‘approach arbitrarily close to’, lacks the precision required in Pure

Mathematics. If we wish to work in a serious way with convergent sequences,

prove results about them and decide whether a given sequence is convergent,

then we need a rigorous definition of the concept of convergence.

Historically, such a definition emerged only in the late nineteenth century,

when mathematicians such as Cantor, Cauchy, Dedekind and Weierstrass sought

to place Analysis on a rigorous non-intuitive footing. It is not surprising, there-

fore, that the definition of convergence seems at first sight rather obscure, and it

may take you a little time to master the logic that it involves.

Three dots are used to indicatethat the sequence continuesindefinitely.

37

In Section 2.1 we show how to picture the behaviour of a sequence by

drawing a sequence diagram. We also introduce monotonic sequences; that is,

sequences which are either increasing or decreasing.

In Section 2.2 we explain the definition of a null sequence; that is, a

sequence which is convergent with limit 0. We then establish various proper-

ties of null sequences, and list some basic null sequences.

In Section 2.3 we discuss general convergent sequences (that is, sequences

which converge but whose limit is not necessarily 0), together with techniques

for calculating their limits.

In Section 2.4 we study divergent sequences, giving particular emphasis to

sequences which tend to infinity or tend to minus infinity. We also show that

convergent sequences are bounded; it follows that unbounded sequences are

necessarily divergent.

In Section 2.5 we prove the Monotone Convergence Theorem, which states

that any increasing sequence which is bounded above must be convergent, and,

similarly, that any decreasing sequence which is bounded below must be con-

vergent. We use this theorem to study simple examples of sequences defined by

recurrence formulas, and particular sequences which converge to p and e.

2.1 Introducing sequences

2.1.1 What is a sequence?

Ever since learning to count you have been familiar with the sequence of

natural numbers

1; 2; 3; 4; 5; 6; . . .:

You have also encountered many other sequences of numbers, such as

2; 4; 6; 8; 10; 12; . . .;

1

2;1

4;1

8;

1

16;

1

32;

1

64; . . .:

We begin our study of sequences with a definition and some notation.

Definition A sequence is an unending list of real numbers

a1; a2; a3; . . .:

The real number an is called the nth term of the sequence, and the sequence

is denoted by

anf g:

In each of the sequences above, we wrote down the first few terms and left

you to assume that subsequent terms were obtained by continuing the pattern in

an obvious way. It is sometimes better, however, to give a precise description

of a typical term of a sequence, and we do this by stating an explicit formula for

the nth term. Thus the expression {2n� 1} denotes the sequence

1; 3; 5; 7; 9; 11; . . .;

Alternative notations areanf g11 and anf g1n¼1:

38 2: Sequences

and the sequence {an} defined by the statement

an ¼ �1ð Þn; n ¼ 1; 2; . . .;

has terms

a1 ¼ �1; a2 ¼ 1; a3 ¼ �1; a4 ¼ 1; a5 ¼ �1; . . .:

Problem 1

(a) Calculate the first five terms of each of the following sequences:

(i) {3nþ 1}; (ii) {3�n}; (iii) {(� 1)nn}.

(b) Calculate the first five terms of each of the following sequences {an}:

(i) an ¼ n!; n ¼ 1; 2; . . .;

(ii) an ¼ 1þ 1n

� �n; n ¼ 1; 2; . . . (to 2 decimal places).

Sequences often begin with a term corresponding to n¼ 1. Sometimes,

however, it is necessary to begin a sequence with some other value of n. We

indicate this by writing, for example, anf g13 to represent the sequence

a3; a4; a5; . . .:

Sequence diagrams

It is often helpful to picture how a given sequence {an} behaves by drawing a

sequence diagram; that is, a graph of the sequence in R2. To do this, we mark

the values n¼ 1, 2, 3, . . . on the x-axis and, for each value of n, we plot the

point (n,an). For example, the sequence diagrams for the sequences {2n� 1},1n

� �

and {(� 1n)} are as follows:

Problem 2 Draw a sequence diagram, showing the first five points, for

each of the following sequences:

(a) {n2}; (b) {3}; (c) 1þ 1n

�ng��

; (d)ð�1Þn

n

n o

.

(In part (c), use the result of Problem 1, part (b).)

2.1.2 Monotonic sequences

Many of the sequences considered so far have the property that, as n increases,

their terms are either increasing or decreasing. For example, the sequence

{2n� 1} has terms 1, 3, 5, 7, . . ., which are increasing, whereas the sequence

For example, the sequence1

n!�n

� �

cannot begin withn¼ 1 or n¼ 2.

In Figure (a), the pointsplotted all lie on the straightline y¼ 2x� 1.

In Figure (b), they all lie onthe hyperbola y ¼ 1

x:


1n

� �

has terms 1; 12; 1

3; 1

4; . . .; which are decreasing. The sequence {(� 1)n} is

neither increasing nor decreasing. All this can be seen clearly on the above

sequence diagrams.

We now give a precise meaning to these words increasing and decreasing,

and introduce the term monotonic.

Definition A sequence {an} is:

� constant, if anþ1 ¼ an; for n ¼ 1; 2; . . .;

� increasing, if anþ1 � an; for n ¼ 1; 2; . . .;

� decreasing, if anþ1 � an; for n ¼ 1; 2; . . .;

� monotonic, if {an} is either increasing or decreasing.

Remarks

1. Note that, for a sequence {an} to be increasing, it is essential that anþ1� an,

for all n� 1. However, we do not require strict inequalities, because we

wish to describe sequences such as

1; 1; 2; 2; 3; 3; 4; 4; . . . and 1; 2; 2; 3; 4; 4; 5; 6; 6; . . .

as increasing. One slightly bizarre consequence of the definition is that

constant sequences are both increasing and decreasing!

2. A sequence {an} is said to be:

strictly increasing, if anþ1 > an; for n ¼ 1; 2; . . .;

strictly decreasing, if anþ1 < an; for n ¼ 1; 2; . . .;

strictly monotonic, if {an} is either strictly increasing or strictly

decreasing.

3. A diagram does NOT constitute a proof! In our first example we formally

establish the monotonicity properties of our three sequences.

Example 1 Determine which of the following sequences {an} are

monotonic:

(a) an¼ 2n�1, n¼ 1, 2, . . .;

(b) an ¼ 1n

, n ¼ 1, 2, . . .;

(c) an¼ (� 1)n, n¼ 1, 2, . . .

Solution

(a) The sequence {2n� 1} is monotonic because

an ¼ 2n� 1 and anþ1 ¼ 2 nþ 1ð Þ � 1 ¼ 2nþ 1;

so that

anþ1 � an ¼ 2nþ 1ð Þ � 2n� 1ð Þ ¼ 2 > 0; for n ¼ 1; 2; . . .:

Thus {2n� 1} is increasing.

(b) The sequence 1n

� �

is monotonic because

an ¼1

nand anþ1 ¼

1

nþ 1;

so that

In fact, strictly increasing.

40 2: Sequences

anþ1�an¼1

nþ1�1

n¼ n� nþ1ð Þ

nþ1ð Þn ¼�1

nþ1ð Þn< 0; for n¼ 1; 2; . . .:

Thus 1n

� �

is decreasing.

Alternatively, since an> 0, for all n, and

anþ1

an

¼ n

nþ 1< 1; for n ¼ 1; 2; . . .;

it follows that

anþ1 < an; for n ¼ 1; 2; . . .:

Thus {an} is decreasing.

(c) The sequence {(�1)n} is not monotonic. In fact, a1¼�1, a2¼ 1 and

a3¼�1.

Hence a3< a2, which means that {an} is not increasing.

Also, a2> a1, which means that {an} is not decreasing.

Thus {(�1)n} is neither increasing nor decreasing, and so is not mono-

tonic. &

Example 1 illustrates the use of the following strategies:

Strategy To show that a given sequence {an} is monotonic, consider the

expression anþ1� an.

� If anþ1 � an � 0; for n ¼ 1; 2; . . .; then anf g is increasing.

� If anþ1 � an � 0; for n ¼ 1; 2; . . .; then anf g is decreasing.

If an> 0 for all n, it may be more convenient to use the following version

of the strategy:

Strategy To show that a given sequence of positive terms, {an}, is mono-

tonic, consider the expression anþ1

an.

� If anþ1

an� 1; for n ¼ 1; 2; . . .; then anf g is increasing.

� If anþ1

an� 1; for n ¼ 1; 2; . . .; then anf g is decreasing.

Problem 3 Show that the following sequences {an} are monotonic:

(a) an ¼ n!; n ¼ 1; 2; . . .;

(b) an ¼ 2�n; n ¼ 1; 2; . . .;

(c) an ¼ nþ 1n; n ¼ 1; 2; . . .:

It is often possible to guess whether a sequence given by a specific formula is

monotonic by calculating the first few terms. For example, consider the

sequence {an} given by

an ¼ 1þ 1

n

� �n

; n ¼ 1; 2; . . .:

In Problem 1 you found that the first five terms of this sequence are

approximately

2; 2:25; 2:37; 2:44 and 2:49:

In fact, strictly decreasing.

A single counter-example issufficient to show that{(�1)n} is not increasing;similarly a (different!) singlecounter-example is sufficientto show that {(�1)n} is notdecreasing.

We will study this importantsequence in detail inSection 2.5.

These terms suggest that the sequence {an} is increasing, and in fact it is.

However, the first few terms of a sequence are not always a reliable guide to

the sequence’s behaviour. Consider, for example, the sequence

an ¼10n

n!; n ¼ 1; 2; . . .:

The first five terms of this sequence are approximately

10; 50; 167; 417 and 833;

this suggests that the sequence {an} is increasing. However, calculation of

more terms shows that this is not so, and the sequence diagram for {an} looks

like this:

Simplifying anþ1

an, we find that

anþ1

an

¼ 10nþ1

nþ 1ð Þ!

�

10n

n!¼ 10

nþ 1:

But 10nþ1� 1, for n¼ 9, 10, . . ., so that anþ1

an� 1, for n¼ 9, 10, . . .; it follows

that the sequence {an} is decreasing, if we ignore the first eight terms.

In a situation like this, when a given sequence has a certain property

provided that we ignore a finite number of terms, we say that the sequence

eventually has the property. Thus we have just seen that the sequence 10n

n!

� �

is

eventually decreasing.

Another example of this usage is the following statement:

the terms of the sequence {n2} are eventually greater than 100.

This statement is true because n2> 100, for all n> 10.

Problem 4 Classify each of the following statements as TRUE or

FALSE, and justify your answers (that is, if a statement is TRUE, prove it;

if a statement is FALSE, give a specific counter-example):

(a) The terms of the sequence {2n} are eventually greater than 1000.

(b) The terms of the sequence {(�1)n} are eventually positive.

(c) The terms of the sequence 1n

� �

are eventually less than 0.025.

(d) The sequence n4

4n

n o

is eventually decreasing.

The first two terms certainlyprove that the sequence isneither decreasing norconstant.

In fact

a6 ’ 1389; a7 ’ 1984;

a8 ’ 2480; a9 ’ 2756;

a10 ’ 2756; a11 ’ 2505;

a12 ’ 2088:

In particular, the fact thata12< a11 shows that thesequence {an} cannot beincreasing.

In fact, a10

a9¼1 and anþ1

an< 1;

for n ¼ 10; 11; . . .:

Note this general approach to‘TRUE’ and ‘FALSE’.

42 2: Sequences

2.2 Null sequences

2.2.1 What is a null sequence?

In this sub-section we give a precise definition of a null sequence (that is, a

sequence which converges to 0) and introduce some properties of null

sequences.

We shall frequently use the rules for rearranging inequalities which you met

in Chapter 1, so you may find it helpful to reread that section quickly before

starting here. We shall also use the following inequalities which were proved in

Sub-section 1.3.3

2n � 1þ n; for n ¼ 1; 2; . . .;

and

2n � n2; for n � 4:

Problem 1 For each of the following statements, find a number X such

that the statement is true:

(a) 1n< 1

100; for all n >X; (b) 1

n< 3

1000; for all n > X:

Problem 2 For each of the following statements, find a number X such

that the statement is true:

(a)�1ð Þnn2

< 1

100; for all n> X; (b)

�1ð Þnn2

< 3

1000; for all n> X:

The solutions of Problems 1 and 2 both suggest that the larger and larger we

choose n, the closer and closer to 0 the terms of the sequences 1n

� �

and�1ð Þnn2

n o

become.

We can express this in terms of the Greek letter " (pronounced ‘epsilon’),

which we introduce to denote a positive number that may be as small as we

please in any given particular instance. In terms of the sequence diagrams for1n

� �

and�1ð Þnn2

n o

, this means that the terms of these sequences eventually lie

inside a horizontal strip in the sequence diagram from�" up to ". However, the

smaller we choose ", the further to the right we have to go before we can be sure

that all the terms of the sequence from that point onwards lie inside the strip.

That is, the smaller we choose " the larger we have to choose X if we wish to have

1

n< "; for all n > X; or

�1ð Þn

n2

< "; for all n > X:

Definition A sequence {an} is a null sequence if:

for each positive number ", there is a number X such that

anj j < "; for all n > X: (1)

Using this terminology, it follows from our previous discussion that the

sequences 1n

� �

and�1ð Þnn2

n o

are both null.

We can interpret our finding of a suitable number X for (1) to hold as an

‘"�X game’ in which player A chooses a positive number " and challenges

player B to find some number X such that the property (1) holds.

Section 1.2

In fact, we use

2n � n; for n ¼ 1; 2; . . .:

Thus in Problems 1 and 2 wechose the particular examplesof 1

100and 3

1000in place of the

‘general’ positive number ".

ε

–εX n

Note that here X need not bean integer; any appropriatereal number will serve. X doesNOT depend on n, but does ingeneral depend on ".

This is the case because, if

n > 1ffiffi

"p ½¼ X�,

then we have n2 > 1" or " > 1

n2;

hence, for n>X, we have

anj j ¼ 1n2 <

1X2 ¼ ":

All that happens is that thedefinition (1) remains validwith a possibly different valueof X having to be chosen.

This is often expressed in thefollowing memorable way:‘a finite number of terms donot matter’.

For example, if 7þ p willserve as a suitable value for X,then so will 12 or 37; but 10might not.

The term inside the modulusis positive.

Here we use the ReciprocalRule for inequalities that youmet in Sub-section 1.2.1.

Here we use the Power Rulefor inequalities that you metin Sub-section 1.2.1.

For example, consider the sequence {an} where an ¼ �1ð Þnn2 ; n ¼ 1; 2; . . ..

This sequence is null. For

anj j ¼�1ð Þn

n2

¼ 1

n2;

so that, if we make the choice X ¼ 1ffiffi

"p , it is certainly true that

anj j < "; for all n > X:

In terms of the "�X game, this simply means that whatever choice of " is

made by player A, player B can always win by making the choice X ¼ 1ffiffi

np !

Remarks

1. In the sequence�1ð Þnn2

n o

, the signs of the terms made no difference to

whether the sequence was null, for they disappeared immediately we took

the modulus of the terms in order to examine whether the definition (1) of a

null sequence was satisfied. Indeed, in general, the signs of terms in a

sequence make no difference to whether a sequence is null. We can express

this formally as follows:

A sequence {an} is null if and only if the corresponding sequence {janj} is null.

2. A null sequence {an} remains null if we add, delete or alter a finite number

of terms in the sequence.

Similarly, a non-null sequence remains non-null if we add, delete or alter

a finite number of terms.

3. If one number serves as a suitable value of X for the inequality in (1) to hold,

then any larger number will also serve as a suitable X. Hence, for simplicity

in some proofs, we may assume if we wish that our initial choice of X in (1)

is a positive integer.

Example 1 Prove that the sequence 1n3

� �

is a null sequence.

Solution We have to prove that


1

n3

< "; for all n > X: (2)

In order to find a suitable value of X for (2) to hold, we rewrite the inequality1n3

< " in various equivalent ways until we spy a value for X that will suit

our purpose. Now

1

n3

< " , 1

n3< "

, n3 >1

"

, n >1ffiffiffi

"3p :

So, let us choose X to be 1ffiffi

"3p . With this choice of X, the above chain of

equivalent inequalities shows us that, if n>X (the last line in the chain), then1n3

< " (in the first line). Thus, with this choice of X, (2) holds; so 1n3

� �

is

indeed null. &

44 2: Sequences

However, any value of "greater than 1 providesno information!

Example 2 Prove that the following sequence is not null

an ¼1; if n is odd;0; if n is even:

�

Solution To prove that the sequence is not null, we have to show that it does

not satisfy the definition. In other words, we must show that the following

statement is not true:

for each positive number ", there is a number X such thatjanj5 ", for all n4X:

So, what we have to show is that the following is true:

for some positive number ", whatever X one choosesjanj 65", for all n4X:

which we may rephrase as:

there is some positive number ", such that whatever X one choosesjanj 65 ", for all n4X:

2

n

–2

1

Sequence diagram

1

n–1/2

1/2

Strip of half-width 12

So, we need to find some positive number " with such a property. The

sequence diagram provides the clue! If we choose " ¼ 12, then the point (n, an)

lies outside the strip � 12; 1

2

� �

for every odd n. In other words, whatever X one

then chooses, the statement

anj j < "; for all n > X;

is false. It follows that the sequence is not a null sequence. &

Note

Notice that any positive value of " less than 1 will serve for our purpose here;

there is nothing special about the number 12.

These two examples illustrate the following strategy:

Strategy for using the definition of null sequence

1. To show that {an} is null, solve the inequality janj<" to find a number X

(generally depending on ") such that janj<", for all n>X.

2. To show that {an} is not null, find ONE value of " for which there is NO

number X such that janj<", for all n>X.

Problem 3 Use the above strategy to determine which of the following

sequences are null:

(a) 12n�1

� �

; (b)�1ð Þn10

n o

; (c)�1ð Þn

n4þ1

n o

:

We now look at a number of Rules for ‘getting new null sequences from old’.

Power Rule If {an} is a null sequence, where an� 0, for n¼ 1, 2, . . ., and

p> 0, then apn

� �

is a null sequence.

Thus, for example, the sequence�

1ffiffi

np�

is null – we simply apply the Power

Rule to the sequence 1n

� �

that we saw earlier to be null, using the positive

power p ¼ 12.

Remark

Notice that the Power Rule also holds without the requirement that an � 0, so

long as apn is defined – for example, if p ¼ 1

mwhere m is a positive integer.

Combination Rules If {an} and {bn} are null sequences, then the follow-

ing are also null sequences:

Sum Rule {anþ bn};

Multiple Rule {lan}, for any real number l;

Product Rule {anbn}.

Thus, for example, we may use known examples of null sequences to verify

that the following sequences are also null:

1nþ 1

n3

� �

– by applying the Sum Rule to the null sequences 1n

� �

and 1n3

� �

;47pn3

� �

– by applying the Multiple Rule to the null sequence 1n3

� �

with

l¼ 47p;

f 1n 2n�1ð Þg – by applying the Product Rule to the null sequences 1

n

� �

and

12n�1

� �

:

Problem 4 Use the above rules to show that the following sequences

are null:

(a) 1

2n�1ð Þ3n o

; (b) 6ffiffi

n5p þ 5

2n�1ð Þ7n o

; (c) f 1

3n4 2n�1ð Þ13g:

Our next rule, the Squeeze Rule, also enables us to ‘get new null sequences

from old’ – but in a slightly different way. To illustrate this rule, we look first at

the sequence diagrams of the two sequences�

1ffiffi

np�

and�

11þffiffi

np�

.

11 + n}}

1n }}

n

1

. . .

. . .

√

√

We shall prove these Ruleslater, in Sub-section 2.2.2.

Recall that apn ¼ anð Þp:

For 1ffiffi

np ¼ 1

n

� �12:

Note that the number l has tobe a fixed number that doesnot depend on n.

In your solution, you may useany of the sequences that youhave proved to be null so farin this sub-section.

46 2: Sequences

The points corresponding to the sequence�

11þffiffi

np�

are squeezed in between

the horizontal axis and the points corresponding to the null sequence�

1ffiffi

np�

,

since 11þffiffi

np < 1

ffiffi

np , for n¼ 1, 2, . . .. Hence, if from some point onwards all the

points corresponding to�

1ffiffi

np�

lie in a narrow strip in the sequence diagram of

half-width " about the axis, then (from the same point onwards) all the points

corresponding to�

11þffiffi

np�

will also lie in the same strip. So, since " may be any

positive number, it certainly looks from this sequence diagram argument that

the sequence�

11þffiffi

np�

must be a null sequence too.

Squeeze Rule If {bn} is a null sequence and

anj j � bn; for n ¼ 1; 2; . . .;

then {an} is a null sequence.

The trick in using the Squeeze Rule to prove that a given sequence {an} is

null is to think of a suitable sequence {bn} that dominates {an}and is itself null.

Thus, for example, since the sequence�

1ffiffi

np�

is null and 11þffiffi

np < 1

ffiffi

np , it follows

from the Squeeze Rule that�

11þffiffi

np�

is also a null sequence.

Proof of the Squeeze Rule We want to prove that {an} is null; that is: for

each positive number ", there is a number X such that

janj < "; for all n > X: (3)

We know that {bn} is null, so there is some number X such that

jbnj < "; for all n > X: (4)

We also know that janj � bn, for n¼ 1, 2, . . ., and hence it follows from (4) that

janj ð< jbnjÞ < "; for all n > X:

Thus inequality (3) holds, as required. &

Remark

In applying the Squeeze Rule, it is often useful to remember the following fact:

The behaviour of a finite number of terms does not matter: it is sufficient to

check that janj � bn holds eventually.

Example 3 Use the Squeeze Rule to prove that the sequence 12

� �n� �

is null.

Proof We want to prove that 12

� �n� �

is null.

What sequence can we find that dominates 12

� �n� �

? Well, we saw at the start

of this sub-section that 2n� n, for n¼ 1, 2, . . ., and it follows from this, by the

Reciprocal Rule for inequalities, that 12n � 1

n; for n ¼ 1; 2; . . .:

Thus the sequence 1n

� �

dominates the sequence 12

� �n� �

, and is itself null. It

follows from the Squeeze Rule that 12

� �n� �

is null. &

Problem 5 Use the inequality 2n� n2, for n� 4, and the Squeeze Rule

to prove that the sequence n 12

� �n� �

is null.

Example 4 Use the Squeeze Rule to prove that the sequence 10n

n!

� �

is null.

Solution We want to prove that 10n

n!

� �

is null.

With a bit of inspiration, we guess that the sequence 10n

n!

� �

is eventually

dominated by ln

� �

, for some constant l.

That is, {an} is dominated by{bn}.

And this is the X that we shalluse to verify (3).

That is, that {an} is eventuallydominated by {bn}.

Instances of these that wehave seen are

1ffiffiffi

np� �

;

1

2

� �n� �

;

n1

2

� �n� �

;

10n

n!

� �

;

n10

n!

� �

:

We suggest that you omitthese proofs on a first reading,and return to them when youare confident that youunderstand the basic ideas.

Recall that X need not be aninteger; any appropriate realnumber will serve.

Recall that anp¼ (an)p.

Writing out the expression 10n

n! in full, we see that

10n

n!¼ 10

1

� �

10

2

� �

. . .10

10

� �

� 10

11

� �

. . .10

n�1

� �

10

n

� �

< 3000� 10

n; for all n > 10;

¼ 30000

n:

In other words, 10n

n!

� �

is eventually dominated by ln

� �

, for l¼ 30000. This latter

sequence is null, by the Multiple Rule applied to the null sequence 1n

� �

:It follows, from the Squeeze Rule and the fact that 30000

n

� �

is a null sequence,

that the sequence 10n

n!

� �

is also null. &

Problem 6 Prove that the following sequences are null:

(a)�

1n2þn

�

; (b)�1ð Þnn!

n o

; (c) sin n2

n2þ2n

n o

:

We can now list a good number of generic types of null sequences, of which we

have seen specific instances already in this sub-section. We call these basic

null sequences, since we shall use them commonly together with the

Combination Rules and other rules for null sequences in order to prove that

particular sequences that we meet are themselves null.

Basic null sequences

(a) 1np

� �

; for p > 0;

(b) fcng; for jcj < 1;

(c) fnpcng; for p > 0 and jcj < 1;

(d) cn

n!

� �

; for any real c;

(e) np

n!

� �

; for p > 0:

2.2.2 Proofs

We now give a number of proofs which were omitted from the previous sub-

section so as not to slow down your gaining an understanding of the key ideas

there. First, recall the definition of a null sequence.

Definition A sequence {an} is null if


janj < "; for all n > X:

Proofs of the Power Rule and the Combination Rules

In the previous sub-section we proved the Squeeze Rule, but did not prove the

Power Rule or the Combination Rules. We now supply these proofs.

Power Rule If {an} is a null sequence, where an� 0, for n¼ 1, 2, . . ., and,

if p>0, then apn

� �

is a null sequence.

48 2: Sequences

Proof We want to prove that apn

� �

is null; that is:


apn < "; for all n > X: (5)

We know that {an} is null, so there is some number X such that

an < "1p; for all n > X: (6)

Taking the pth power of both sides of (6), we obtain the desired result (5), with

the same value of X. &

Remark

Notice how we used the (positive) number "1p in place of " in (6) in order to

obtain " in the final result (5). In the proofs which follow, we again apply the

definition of null sequence, using positive numbers which depend in some way

on ", in order to obtain " in the inequality that we are aiming to prove.

Sum Rule If {an} and {bn} are null sequences, then {anþ bn} is a null

sequence.

Proof We want to prove that the sum {anþ bn} is null; that is:


an þ bnj j < "; for all n > X: (7)

We know that {an} and {bn} are null, so there are numbers X1 and X2 such that

anj j <1

2"; for all n > X1;

and

bnj j <1

2"; for all n > X2:

Hence, if X¼max {X1, X2}, then both the two previous inequalities hold; so

if we add them we obtain, by the Triangle Inequality, that

an þ bnj j � anj j þ bnj j

<1

2"þ 1

2" ¼ "; for all n > X:

Thus inequality (7) holds, with this value of X. &

Before going on, first a comment about the number 12

that appears several

times in the above proof. It is used twice in expressions 12" at the start of the

proof in order to end up with a final inequality that says that some expression is

‘<"’. While this means that we end up with an inequality that shows at once

that the desired result holds, in fact it is not strictly necessary to end up with

precisely ‘<"’, as the following result shows:

Lemma The ‘Ke Lemma’ Let {an} be a sequence. Suppose that, for


anj j < K"; for all n > X;

where K is a positive real number that does not depend on " or X. Then {an}

is a null sequence.

For janpj ¼ an

p, since an� 0.

Here we use the number "1p in

place of " in the definition ofnull sequence.

We use 12" here rather than ",

in order to end up with thesymbol " on its owneventually in the desiredinequality (7).

You met the TriangleInequality in Sub-section 1.3.1.

In this case, the result that aparticular sequence is null.

Loosely speaking, we mayexpress this result as ‘K" isjust as good as "’ in thedefinition of null sequence.

For example, K might be 2 or p7

or 259, but it could not be 2nor X

259.

Proof We want to prove that the sequence {an} is null; that is:


anj j < "; for all n > X:

Now, whatever this positive number " may be, the number "K

is also a

positive number. It follows from the hypothesis stated in the Lemma that

therefore there is some number X such that

anj j < K � "

K

¼ "; for all n > X:

This is precisely the condition for {an} to be null. &

From time to time we shall use this Lemma in order to avoid arithmetic

complexity in our proofs.

Multiple Rule If {an} is a null sequence, then {lan} is a null sequence for

any real number l.

Proof We want to prove that the multiple {lan} is null; that is:


lanj j < "; for all n > X: (8)

If l¼ 0, this is obvious, and so we may assume that l 6¼ 0.

We know that {an} is null, so there is some number X such that

anj j <1

lj j "; for all n > X:

Multiplying both sides of this inequality by the positive number jlj, this gives

us that

lanj j < "; for all n > X:

Thus the desired result (8) holds. &

Product Rule If {an} and {bn} are null sequences, then {anbn} is a null

sequence.

Proof We want to prove that the product {anbn} is null; that is:


anbnj j < "; for all n > X: (9)

We know that {an} and {bn} are null, so there are numbers X1 and X2 such

that

anj j <ffiffiffi

"p; for all n > X1;

and

bnj j <ffiffiffi

"p; for all n > X2:

You may omit this proof on afirst reading.

We use 1lj j " here rather than "

in order to end up eventuallywith the symbol " on its ownin the desired inequality (8).

We useffiffiffi

"p

here rather than "in order to end up eventuallywith the symbol " on its ownin the desired inequality (9).

50 2: Sequences

Hence, if X¼max {X1, X2}, then both the two previous inequalities hold; so

if we multiply them we obtain that

anbnj j ¼ anj j � bnj j<

ffiffiffi

"p�

ffiffiffi

"p¼ "; for all n > X:

Thus inequality (9) holds with this value of X. &

Basic null sequences

At the end of the previous sub-section we gave a list of basic null sequences.

We end this sub-section by proving that these sequences are indeed null

sequences.

Basic null sequences The following sequences are null sequences:

(a) 1np

� �

; for p > 0;

(b) cnf g; for cj j < 1;

(c) npcnf g; for p > 0; cj j < 1;

(d) cn

n!

� �

; for any real c;

(e) np

n!

� �

; for p > 0.

Proof

(a) To prove that 1np

� �

is null, for p> 0, we simply apply the Power Rule to the

sequence 1n

� �

, which we know is null.

(b) To prove that {cn} is null, for jcj< 1, it is sufficient to consider only the

case 0� c< 1. If c¼ 0, the sequence is obviously null; so we may assume

that 0< c< 1.

With this assumption, we can write c in the form

c ¼ 1

1þ a; for some number a > 0:

Now, by the Binomial Theorem

1þ að Þn � 1þ na

� na; for n ¼ 1; 2; . . .;

and hence

cn ¼ 1

1þ að Þn

� 1

na; for n ¼ 1; 2; . . .:

Since 1n

� �

is null, we deduce that f 1nag is also null by the Multiple Rule;

hence {cn} is null, by the Squeeze Rule.

(c) To prove that {npcn} is null, for p > 0 and jcj < 1, we may again assume

that 0 < c < 1. Hence

c ¼ 1

1þ a; for some number a > 0:

For example

1

n10

� �

;

0:9ð Þnf g;

n3 0:9ð Þn� �

;

10n

n!

� �

;

n10

n!

� �

You met the BinomialTheorem in Sub-section 1.3.3.

Here we take l ¼ 1a

in theMultiple Rule.

First, we deal with the case p¼ 1. By the Binomial Theorem

1þ að Þn � 1þ naþ 1

2n n� 1ð Þa2

� 1

2n n� 1ð Þa2; for n ¼ 2; 3; . . .;

and hence

ncn ¼ n

1þ að Þn

� n12

n n� 1ð Þa2¼ 2

n� 1ð Þa2; for n ¼ 2; 3; . . .:

Since 1n�1

� �12

is null, we deduce that 2ðn�1Þa2

n o1

2is also null, by the

Multiple Rule. Hence {ncn} is null, by the Squeeze Rule. This proves

part (c) in the case p¼ 1.

To deduce that {npcn} is null for any p> 0 and 0< c< 1, we note that

npcn ¼ ndnð Þp; for n ¼ 1; 2; . . .;

where d ¼ c1p. Since 0< d< 1, we know that {ndn} is null, and so {npcn} is

null, by the Power Rule.

(d) To prove that cn

n!

� �

is null for any real c, we may assume that c> 0. If we

choose any integer m such that mþ 1> c, then we have, for n>mþ 1, that

cn

n!¼ c

1

� c

2

�

. . .c

m

�

� c

mþ 1

� �

. . .c

n� 1

� c

n

�

� c

1

� c

2

�

. . .c

m

�

� c

n

¼ K � c

n;

where K ¼ cm

m! is a constant.

Since 1n

� �

is null, we deduce that Kcn

� �

is also null, by the Multiple Rule; it

follows that cn

n!

� �

is also null, by the Squeeze Rule.

(e) To prove that np

n!

� �

is null, for p > 0, we write

np

n!¼ np

2n

� �

� 2n

n!

� �

; for n ¼ 1; 2; . . .:

Since np

2n

� �

and 2n

n!

� �

are both null sequences, by parts (c) and (d) respec-

tively, we deduce that np

n!

� �

is null, by the Product Rule. &

2.3 Convergent sequences

2.3.1 What is a convergent sequence?

In the previous section we looked at null sequences; that is, sequences which

converge to 0. We now turn our attention to sequences which converge to

limits other than 0.

Here we use the special caseof part (c) that we havealready proved,

For c is a constant and m issome constant, and hence cm

m! isalso some constant. Note thatwhat is varying is n, nothingelse.

52 2: Sequences

Problem 1 Consider the sequence an ¼ nþ1n; n ¼ 1; 2; . . ..

(a) Draw the sequence diagram for {an}, and describe (informally) how

this sequence behaves.

(b) What can you say (formally) about the behaviour of the sequence

bn ¼ an � 1; n ¼ 1; 2; . . .?

The terms of the sequence {an} in Problem 1 appear to get arbitrarily close to 1;

that is, the sequence {an} appears to converge to 1. If we subtract 1 from each

term an to form the sequence {bn}, then we obtain a null sequence. This

example suggests the following definition of a convergent sequence:

Definition The sequence {an} is convergent with limit ł, or converges to

the limit ł, if {an� ‘} is a null sequence. In this case, we say that {an}

converges to ł, and we write:

EITHER limn!1

an ¼ ‘,

OR an! ‘ as n!1.

The following are examples of convergent sequences:

every null sequence converges to 0;

every constant sequence {c} converges to c;

the sequencen

nþ1n

o

is convergent with limit 1.

Problem 2 For each of the following sequences {an}, draw its seque-

nce diagram and show that {an} converges to ‘ by considering an� ‘:(a) an ¼ n2�1

n2þ1; ‘ ¼ 1; (b) an ¼ n3þ �1ð Þn

2n3 ; ‘ ¼ 12:

The definition of convergence of a sequence is often given in the following

equivalent form:

Equivalent definition The sequence {an} is convergent with limit ł if:


an � ‘j j < "; for all n > X: (1)

Remarks

1. In terms of the sequence diagram for {an}, this definition states that, for

each positive number ", the terms an eventually lie inside the horizontal

strip from ‘� " up to ‘þ ".

� + ε{an}

|an – �| < ε

� – ε < an < � + ε

n

� – ε

X

�

These statements are read as:‘the limit of an, as n tends toinfinity, is ‘’ and ‘an tends to‘, as n tends to infinity’.Often, we omit ‘as n!1’.Do not let this use of thesymbol1 tempt you to thinkthat1 is a real number.

See Problem 1 above.

Note that here X need not bean integer; any appropriatereal number will serve.

2. Just as for null sequences, we can interpret ‘an! ‘ as n!1’ as an "�X

game in which player A chooses a positive number " and challenges player

B to find some number X such that (1) holds.

For example, consider the sequence {an} where an ¼ 8nþ62n

, n ¼ 1, 2, . . ..This sequence is convergent, with limit 4. For

an � 4j j ¼ 8nþ 6

2n� 4

¼ 6

2n¼ 3

n;

so that, if we make the choice X ¼ 3", it is certainly true that

an � 4j j < "; for all n > X:

In terms of the "�X game, this simply means that whatever choice

of " is made by player A, player B can always win by making the choice

X ¼ 3"!

B winsBA

I'll play ANY positive

ε

I'll play

X = 3ε

3. If a sequence is convergent, then it has a unique limit. We can see this by

using a sequence diagram. Suppose that the sequence {an} has two limits, ‘and m, where ‘ 6¼m. Then it is possible to choose a positive number " such

that horizontal strips from ‘� " up to ‘þ " and from m� " up to mþ " do

not overlap. For example, we can take " ¼ 13‘� mj j.

m + ε

� + ε

� – ε

� – m

n

m – ε

�

m

However, since ‘ and m are both limits of {an}, the terms an must eventually

lie inside both strips, and this is impossible.

A formal proof of this fact is given in the Corollary to Theorem 3, later in

this section.

4. If a given sequence converges to ‘, then this remains true if we add, delete

or alter a finite number of terms. This follows from the corresponding result

for null sequences.

5. Not all sequences are convergent. For example, the sequence {(�1)n} is not

convergent.

In this section we restrict our attention to sequences which do converge.

This is the case because, if

n >3

"½¼ X�;

then we have " > 3n

or 3n< ";

hence, for n>X, we have

an � 4j j ¼ 3

n<

3

X¼ ":

The vertical distance between‘ and m is j‘�mj, so that anychoice of " that is less thanhalf of this quantity will serve.

In other words ‘altering afinite number of terms doesnot matter’. See Sub-section 2.2.1.

We discuss non-convergentsequences in Section 2.4.

54 2: Sequences

2.3.2 Combination Rules for convergent sequences

So far you have tested the convergence of a given sequence {an} by calculating

an� ‘ and showing that {an� ‘} is null. This presupposes that you know in

advance the value of ‘. Usually, however, you are not given the value of ‘. You

are given only a sequence {an} and asked to decide whether or not it converges

and, if it does, to find its limit. Fortunately many sequences can be dealt with

by using the following Combination Rules, which extend the Combination

Rules for null sequences:

Theorem 1 Combination Rules

If limn!1

an ¼ ‘ and limn!1

bn ¼ m, then:

Sum Rule limn!1

an þ bnð Þ ¼ ‘þ m;

Multiple Rule limn!1

lanð Þ ¼ l‘; for any real number l;

Product Rule limn!1

anbnð Þ ¼ ‘m;

Quotient Rulelim

n!1

an

bn

� �

¼ ‘

m; provided that m 6¼ 0:

Remarks

1. In applications of the Quotient Rule, it may happen that some of the terms bn

take the value 0, in which case an

bnis not defined. However, we shall see (in

Lemma 1) that this can happen only for finitely many bn (because m 6¼ 0), and

so {bn} is eventually non-zero. Thus the statement of the rule does make sense.

2. The following rule is a special case of the Quotient Rule:

Corollary 1 If limn!1

an ¼ ‘ and ‘ 6¼ 0, then:

Reciprocal Rule limn!1

1

an

¼ 1

‘.

We shall prove the Combination Rules at the end of this sub-section, but first

we illustrate how to apply them.

Applying the Combination Rules

Example 1 Show that each of the following sequences {an} is convergent,

and find its limit:

(a) an ¼ 2nþ 1ð Þ nþ 2ð Þ3n2 þ 3n

; (b) an ¼ 2n2 þ 10n

n!þ 3n3 .

Solution Although the expressions for an are quotients, we cannot apply the

Quotient Rule immediately, because the sequences defined by the numerators

and denominators are not convergent. In each case, however, we can rearrange

the expressions for an and then apply the Combination Rules.

(a) In this case we divide both the numerator and denominator by n2 to give

an ¼2nþ 1ð Þ nþ 2ð Þ

3n2 þ 3n¼

2þ 1n

� �

1þ 2n

� �

3þ 3n

:

Sub-section 2.2.1.

A finite number of termsdo not matter.


Since 1n

� �

is a basic null sequence, we find by the Combination Rules that

limn!1

an ¼2þ 0ð Þ 1þ 0ð Þ

3þ 0¼ 2

3:

(b) This time we divide both the numerator and denominator by n! to give

an ¼2n2 þ 10n

n!þ 3n3¼

2n2

n! þ 10n

n!

1þ 3n3

n!

Since�

n2

n!

�

, 10n

n!

� �

and�

n3

n!

�

are all basic null sequences, we find by the

Combination Rules that

limn!1

an ¼0þ 0

1þ 0¼ 0: &

Remark

We simplified each of the above sequences by dividing both numerator and

denominator by the dominant term. In part (a), we divided by n2, which is the

highest power of n in the expression. In part (b), the choice of dominant term was

a little harder, but the choice of n! ensured that the resulting quotients in the

numerator and denominator were all typical terms of convergent sequences.

In choosing the dominant term the following ordering is often useful:

Domination HierarchyA factorial term n! dominates a power term cn for any c 2 R .

A power term cn dominates a term np for p> 0, jcj< 1.

The above examples illustrate the following general strategy:

Strategy To evaluate the limit of a complicated quotient:

1. Identify the dominant term, bearing in mind the basic null sequences.

2. Divide both the numerator and the denominator by the dominant term.

3. Apply the Combination Rules.

Problem 3 Show that each of the following sequences {an} is con-

vergent, and find its limit:

(a) an ¼ n3 þ 2n2 þ 32n3 þ 1

; (b) an ¼ n2 þ 2n

3n þ n3 ; (c) an ¼ n!þ �1ð Þn2n þ 3n! :

Proofs of the Combination Rules We prove the Sum Rule, the Multiple

Rule and the Product Rule by using the corresponding Combination Rules for

null sequences. Remember that

limn!1

an ¼ ‘

means that

an ! ‘f g is a null sequence:

See the list of basic nullsequences (introduced inSub-section 2.2.1) which isrepeated in the margin below.

Basic null sequences:1

np

� �

; for p> 0;

cnf g; for cj j < 1;

npcnf g; for p > 0; c > 1;

cn

n!

� �

; for c 2 R ;

np

n!

� �

; for p > 0:

Warning Notice that ‘3n!’means ‘3� (n!)’ and not‘(3n)!’.


56 2: Sequences

Sum Rule If limn!1


bn ¼ m, then

limn!1

an þ bnð Þ ¼ ‘þ m:

Proof By assumption, {an� ‘} and {bn�m} are null sequences; since

an þ bnð Þ � ‘þ mð Þ ¼ an � ‘ð Þ þ bn � mð Þwe deduce that {(anþ bn)� (‘þm)} is null, by the Sum Rule for null

sequences. &

Product Rule If limn!1


bn ¼ m, then

limn!1

anbnð Þ ¼ ‘m:

Proof The idea here is to express anbn� ‘m in terms of an� ‘ and bn�m

anbn � ‘m ¼ an � ‘ð Þ bn � mð Þ þ m an � ‘ð Þ þ ‘ bn � mð Þ:Since {an� ‘} and {bn�m} are null, we deduce that {anbn� ‘m} is null, by

the Combination Rules for null sequences. &

Remark

Note that the Multiple Rule is just a special case of the Product Rule in which

the sequence {bn} is a constant sequence.

To prove the Quotient Rule, we need to use the following lemma, which will

also be needed in the next sub-section:

Lemma 1 If limn!1

an ¼ ‘ and ‘> 0, then there is a number X such that

an >1

2‘, for all n > X:

Proof Since 12‘ > 0, the terms an must eventually lie within a distance 1

2‘ of

the limit ‘.

{an}

X n

|an – �| < 21�

21

23

� < an < �

�

23�

21�

In other words, there is a number X such that

an � ‘j j < 1

2‘; for all n > X:

Hence

� 1

2‘ < an � ‘ <

1

2‘; for all n > X;

and so the left-hand inequality gives1

2‘ < an; for all n > X;

as required. &

Here we are taking " ¼ 12‘ in

the definition of convergence.

Quotient Rule If limn!1


bn ¼ m, then

limn!1

an

bn

� �

¼ ‘

m, provided that m 6¼ 0:

Proof We assume that m> 0; the proof for the case m< 0 is similar. Once

again the idea is to write the required expression in terms of an� ‘ and bn�m

an

bn

� ‘

m¼ m an � ‘ð Þ � ‘ bn � mð Þ

bnm:

Now, however, there is a slight problem: {m(an� ‘)� ‘(bn�m)} is certainly a

null sequence, but the denominator is rather awkward. Some of the terms bn

may take the value 0, in which case the expression is undefined.

However, by Lemma 1, we know that that for some X we have

bn >1

2m; for all n > X:

Thus, for all n>X

an

bn

� ‘

m

¼ m an � ‘ð Þ � ‘ bn � mð Þj jbnm

� m an � ‘ð Þ � ‘ bn � mð Þj j12

m2

� mj j � an � ‘j j þ ‘j j � bn � mj j12

m2:

Since this last expression defines a null sequence, it follows by the Squeeze

Rule that�

an

bn� ‘

m

�

is null. &

2.3.3 Further properties of convergent sequences

There are several other theorems about convergent sequences which will be

needed in later chapters. The first is a general version of the Squeeze Rule.

Theorem 2 Squeeze Rule

If:

1. bn� an� cn, for n¼ 1, 2, . . .,

2. limn!1

bn ¼ limn!1

cn ¼ ‘,then lim

n!1an ¼ ‘:

Proof By the Combination Rules

limn!1

cn � bnð Þ ¼ ‘� ‘ ¼ 0 ;

so that cn � bnf g is a null sequence. Also, by condition 1

0 � an � bn � cn � bn; for n ¼ 1; 2; . . .;

and so an � bnf g is null, by the Squeeze Rule for null sequences.

In particular, this implies thatthe terms of {bn} areeventually positive.

Here we use Lemma 1.

Here we apply the TriangleInequality to the numerator.

You met the Squeeze Rulefor null sequences inSub-section 2.2.1.

That is, an � bnf g is squeezedby cn � bnf g.

58 2: Sequences

Now we write an in the form

an ¼ an � bnð Þ þ bn :

Hence by the Combination Rules

limn!1

an ¼ limn!1

an � bnð Þ þ limn!1

bn

¼ 0þ ‘ ¼ ‘: &

Remark

In applications of the Squeeze Rule, it is sufficient to check that condition 1

applies eventually. This is because the values of a finite number of terms do not

affect convergence.

The following example and problem illustrate the use of the Squeeze Rule and

the Binomial Theorem in the derivation of two important limits.

Example 2

(a) Prove that, if c> 0, then

1þ cð Þ1n� 1þ c

n; for n ¼ 1; 2; . . .:

(b) Deduce that

limn!1

a1n¼ 1; for any positive number a:

Solution

(a) Using the rules for inequalities, we obtain

1þ cð Þ1n� 1þ c

n, 1þ c � 1þ c

n

�n

; since c > 0 :

The right-hand inequality holds, because

1þ c

n

�n

� 1þ nc

n

�

¼ 1þ c ;

by the Binomial Theorem. It follows that the required inequality also holds.

(b) We consider the cases a> 1, a¼ 1, a< 1 separately.

If a> 1, then we can write a¼ 1þ c, where c> 0. Then, by part (a)

1 � a1n ¼ 1þ cð Þ

1n � 1þ c

n; for n ¼ 1; 2; . . .:

Since limn!1

1þ cn

� �

¼ 1, we deduce that

limn!1

a1n ¼ 1;

by the Squeeze Rule.

If a¼ 1, then a1n ¼ 1, for n¼ 1, 2, . . ., so

limn!1

a1n ¼ 1:

If 0< a< 1, then 1a> 1, so that lim

n!11a

� �1n¼ 1 by the first case in part (b).

Hence, by the Reciprocal Rule

limn!1

a1n ¼ 1

limn!1

1a

� �1n

¼ 1

1¼ 1: &

A finite number of terms donot matter.

We proved this inequality forthe case c ¼ 1 in Example 6(b)of Sub-section 1.3.3.

All the terms in the BinomialTheorem expansion ofð1þ c

nÞn are positive, so we

get a smaller number if weignore all the terms after thefirst two.

In this application of theSqueeze Rule, the ‘lower’sequence is {1}.

Problem 4

(a) Use the Binomial Theorem to prove that

n1n �1þ


2

n� 1

r

; for n ¼ 2; 3; . . .:

Hint: 1þ xð Þn� n n�1ð Þ2! x2; for n � 2; x � 0:

(b) Use the Squeeze Rule to deduce that

limn!1

n1n ¼ 1:

Next we show that taking limits preserves weak inequalities.

Theorem 3 Limit Inequality Rule

If limn!1


bn ¼ m, and also

an � bn; for n ¼ 1; 2; . . .;

then ‘�m.

Proof Suppose that an! ‘ and bn!m, where an� bn for n¼ 1, 2, . . ., but

that it is not true that ‘�m. Then ‘>m and so, by the Combination Rules

limn!1

an � bnð Þ ¼ ‘� m > 0:

Hence, by Lemma 1, there is an X such that

an � bn >1

2‘� mð Þ; for all n > X: (2)

However, we assumed that an� bn� 0, for n¼ 1, 2, . . ., so statement (2) is a

contradiction.

Hence it is true that ‘�m. &

Warning Taking limits does NOT preserve strict inequalities. That is, if

limn!1


bn ¼ m, and also an< bn, for n¼ 1, 2, . . ., then it follows

from Theorem 3 that ‘�m – but it does NOT necessarily follow that ‘<m.

For example, 12n< 1

n, for n¼ 1, 2, . . ., but lim

n!11

2n¼ lim

n!11n¼ 0.

We can now give the formal proof, promised earlier, that a convergent

sequence has only one limit.

Corollary 2 If limn!1


an ¼ m, then ‘¼m.

Proof Applying Theorem 3 with bn¼ an, we deduce that ‘�m and m� ‘.Hence ‘¼m. &

In Sub-section 2.2.1, we saw that a sequence {an} is null if and only if the

sequence jan jf g is null. Our next result is a partial generalisation of this fact.

Theorem 4 If limn!1

an ¼ ‘, then limn!1

anj j ¼ ‘j j.

This is a ‘proof bycontradiction’.

60 2: Sequences

Proof Using the ‘reverse form’ of the Triangle Inequality, we obtain

anj j � ‘j j

� an � ‘j j; for n ¼ 1; 2; . . .:

Since {an� ‘} is null, we deduce from the Squeeze Rule for null sequences that

anj j � ‘j jf g is null, as required. &

Remark

Theorem 4 is only a partial generalisation of the earlier result about null

sequences, because the converse statement does not hold. That is, if janj! j‘j,it does NOT necessarily follow that an! ‘. For example, consider the sequence

an¼ (�1)n, for n¼ 1, 2, . . .; in this case, janj! 1, but {an} does not even

converge.

2.4 Divergent sequences

2.4.1 What is a divergent sequence?

We have commented several times that not all sequences are convergent. We

now investigate the behaviour of sequences which do not converge.

Definition A sequence is divergent if it is not convergent.

Here are the sequence diagrams for �1ð Þnf g, {2n} and �1ð Þnnf g. Each of

these sequences is divergent but, as you can see, they behave differently.

It is rather tricky to prove, directly from the definition, that these sequences

are divergent.

The aim of this section is to obtain criteria for divergence, which avoid

having to argue directly from the definition. At the end of the section, we give

a strategy involving two criteria which deal with all types of divergence.

We obtain these criteria by establishing certain properties, which are neces-

sarily possessed by a convergent sequence; if a sequence does not have these

properties, then it must be divergent.

You met this in Sub-section1.3.1, expressed in the forma� bj j �

aj j � bj j

. Here weare writing an in place of a and‘ in place of b.

For example, to show that thesequence ð�1Þnf g isdivergent, we have to showthat ð�1Þnf g is notconvergent; that is, for everyreal number ‘, the sequenceð�1Þn � ‘f g is not null.


2.4.2 Bounded and unbounded sequences

One property possessed by a convergent sequence is that it must be bounded.

Definition A sequence {an} is bounded if there is a number K such that

anj j � K; for n ¼ 1; 2; . . .:

A sequence is unbounded if it is not bounded.

Thus a sequence {an} is bounded if all the terms an lie on the sequence diagram

in a horizontal strip from �K up to K, for some positive number K.

K

–K

{an}⏐an⏐ ≤ K

–K ≤ an ≤ Kn

For example, the sequence ð�1Þnf g is bounded, because

ð�1Þnj j � 1; for n ¼ 1; 2; . . .:

However the sequences {2n} and {n2} are unbounded, since, for each number K,

we can find terms of these sequences whose absolute values are greater than K.

Problem 1 Classify the following sequences as bounded or unbounded:

(a) 1þð�1Þnf g; (b) ð�1Þn nf g; (c) 2nþ1n

� �

; (d) 1� 1n

� �n� �

.

The sequence ð�1Þnf g shows that:

a bounded sequence is not necessarily convergent.

However we can prove that:

a convergent sequence is necessarily bounded.

Theorem 1 Boundedness Theorem

If {an} is convergent, then {an} is bounded.

Proof We know that an! ‘, for some real number ‘. Thus {an� ‘} is a null

sequence, and so there is a number X such that

an � ‘j j < 1; for all n > X:

For simplicity in the rest of the proof, we shall now assume that our initial

choice of X is a positive integer.

Nowanj j ¼ ðan � ‘Þ þ ‘j j� an � ‘j j þ ‘j j; by the Triangle Inquality:

It follows that

anj j � 1þ ‘j j; for all n > X:

This is the type of inequality needed to prove that {an} is bounded, but it does

not include the terms a1, a2, . . ., aX. To complete the proof, we let K be the

maximum of the numbers ja1j, ja2j, . . ., jaXj, 1þ j‘j.

Take "¼ 1 in the definition ofa null sequence.

K

X n

� + 1

� – 1

–K

�

That is, K¼max{ja1j,ja2j, . . ., jaXj, 1þ j‘j}.

62 2: Sequences

It follows that

anj j � K; for n ¼ 1; 2; . . .;

as required. &

From Theorem 1, we obtain the following test for the divergence of a

sequence:

Corollary 1 If {an} is unbounded, then {an} is divergent.

For example, the sequences {2n} and �1ð Þnnf g are both unbounded, so they

are both divergent.

Problem 2 Classify the following sequences as convergent or diver-

gent, and as bounded or unbounded:

(a)ffiffiffi

npf g; (b) n2 þ n

n2 þ 1

n o

; (c) �1ð Þnn2� �

; (d) n �1ð Þn� �

:

2.4.3 Sequences which tend to infinity

Although the sequences {2n} and �1ð Þnnf g are both unbounded (and hence

divergent), there is a marked difference in their behaviour. The terms of both

sequences become arbitrarily large, but those of the sequence {2n} become

arbitrarily large and positive. To make this precise, we must explain what we

mean by ‘arbitrarily large and positive’.

Definition The sequence {an} tends to infinity if:

for each positive number K, there is a number X such that

an > K; for all n > X:

In this case, we write

an !1 as n!1 :

Remarks

1. In terms of the sequence diagram for {an}, this definition states that,

for each positive number K, the terms an eventually lie above the line at

height K.

K

X n

{an}

2. If a sequence tends to infinity, then it is unbounded – and hence divergent,

by Corollary 1.

Often, we omit ‘as n!1’.


3. If a sequence tends to infinity, then this remains true if we add, delete or

alter a finite number of terms.

4. In the definition we can replace the phrase ‘for each positive number K’ by

‘for each number K’; for, if the inequality ‘an>K, for all n>X’ holds for each

number K, then in particular it certainly holds for each positive number K.

There is a version of the Reciprocal Rule for sequences which tend to

infinity. This enables us to use our knowledge of null sequences to identify

sequences which tend to infinity.

Theorem 2 Reciprocal Rule

(a) If the sequence {an} satisfies both of the following conditions:

1. {an} is eventually positive,

2. f 1ang is a null sequence,

then an!1.

(b) If an!1, then 1an! 0.

Proof of part (a) To prove that an!1, we have to show that:


an > K; for all n > X: (1)

Since {an} is eventually positive, we can choose a number X1 such that

an > 0; for all n > X1:

Since f 1ang is null, we can choose a number X2 such that

1

an

<1

K; for all n > X2:

Now, let X¼max {X1X2}; then

0 <1

an

<1

K; for all n > X:

This statement is equivalent to the statement (1), so an!1, as required. &

Example 1 Use the Reciprocal Rule to prove that the following sequences

tend to infinity:

(a) n3

2

n o

; (b) n!þ 10nf g; (c) n!� 10nf g

Solution

(a) Each term of the sequence�

n3

2

�

is positive and 1n3=2¼ 2

n3. Now, 1n3

� �

is a

basic null sequence and so 2n3

� �

is null, by the Multiple Rule. Hence�

n3

2

�

tends to infinity, by the Reciprocal Rule.

(b) Each term of the sequence n!þ 10nf g is positive and

limn!1

1

n!þ 10n¼ lim

n!1

1n!

1þ 10n

n!

¼ 0

1þ 0¼ 0;

by the Combination Rules.

Hence n!þ 10nf g tends to infinity, by the Reciprocal Rule.

A finite number of terms donot matter.

We prove only part (a): theproof of part (b) is similar.

Here we are taking " ¼ 1K

in thedefinition of a null sequence.

We make this choice of X sothat BOTH of the precedinginequalities hold for all n>X.

Notice that, in parts (b) and(c), n! is the dominant term.

Alternatively, since1

n!þ 10n � 1n!, the sequence

�

1n!þ 10n

�

is null, by the

Squeeze Rule for nullsequences.

64 2: Sequences

(c) First note that

n!� 10n ¼ n! 1� 10n

n!

� �

; for n ¼ 1; 2; . . .:

Since 10n

n!

� �

is a basic null sequence, we know that 10n

n! is eventually less

than 1, and so n!� 10n is eventually positive. Also

limn!1

1

n!� 10n¼ lim

n!1

1n!

1� 10n

n!

¼ 0

1� 0¼ 0;


Hence {n!� 10n} tends to infinity, by the Reciprocal Rule. &

There are also versions of the Combination Rules and Squeeze Rule for

sequences which tend to infinity. We state these without proof.


If {an} tends to infinity and {bn} tends to infinity, then:

Sum Rule {anþ bn} tends to infinity;

Multiple Rule {lan} tends to infinity, for l> 0;

Product Rule {anbn} tends to infinity.


If {bn} tends to infinity, and

an � bn; for n ¼ 1; 2; . . .;

then {an} tends to infinity.

Problem 3 For each of the following sequences {an}, prove that an!1:

(a) 2n

n

� �

; (b) 2n � n100� �

; (c) 2n

nþ 5n100

� �

; (d) 2n þ n2

n10 þ n

n o

.

We can also define {an} tends to minus infinity.

Definition The sequence {an} tends to minus infinity if

�an !1 as n!1:In this case, we write

an ! �1 as n!1:

For example, the sequences {�n2} and 10n � n!f g both tend to minus infinity,

because {n2} and n!� 10nf g tend to infinity. Sequences which tend to minus

infinity are unbounded, and hence divergent. However, the sequence �1ð Þnnf gshows that an unbounded sequence need not tend to infinity or to minus infinity.

2.4.4 Subsequences

We now give two useful criteria for establishing that a sequence diverges; both

of them involve the idea of a subsequence. For example, consider the bounded

divergent sequence �1ð Þnf g. This sequence splits naturally into two:

This follows by taking "¼ 1in the definition of a nullsequence.

This follows from the fact thatsequences which tend toinfinity are unbounded.


the even terms a2, a4, . . ., a2k, . . ., each of which equals 1;

the odd terms a1, a3, . . ., a2k�1, . . ., each of which equals �1.

Both of these are sequences in their own right, and we call them the evensubsequence {a2k} and the odd subsequence {a2k� 1}.

In general, given a sequence {an} we may consider many different subse-

quences, such as:

{a3k}, comprising the terms a3, a6, a9, . . .;

{a4kþ1}, comprising the terms a5, a9, a13, . . .;

{a2k!}, comprising the terms a2, a4, a12, . . ..

Definition The sequence ankf g is a subsequence of the sequence {an} if

{nk} is a strictly increasing sequence of positive integers; that is, if

n1 < n2 < n3 < . . .:

For example, the subsequence {a5kþ2} corresponds to the sequence of

positive integers

nk ¼ 5k þ 2; k ¼ 1; 2; . . .:

The first term of {a5kþ 2} is a7, the second is a12, the third is a17, and so on.

Notice that any strictly increasing sequence {nk} of positive integers must

tend to infinity, since it can be proved by Mathematical Induction that

nk � k; for k ¼ 1; 2; . . .:

Problem 4

(a) Let an¼ n2, for n¼ 1, 2, . . .. Write down the first five terms of each

of the subsequences ankf g, where:

(i) nk¼ 2k; (ii) nk¼ 4k� 1; (iii) nk¼ k2.

(b) Write down the first three terms of the odd and even subsequences of

the following sequence: an ¼ n �1ð Þn ; n ¼ 1; 2; . . .:

Our next theorem shows that certain properties of sequences are inherited by

their subsequences.

Theorem 5 Inheritance Property of Subsequences

For any subsequence ankf g of {an}:

(a) if an! ‘ as n!1, then ank! ‘ as k!1;

(b) if an!1 as n!1, then ank!1 as k!1.

Proof of part (a) We want to show that for each positive number ", there is a

number K such that

ank� ‘j j < "; for all k > K: (2)

However, since {an� ‘} is null, we know that there is a number X such that

an � ‘j j < "; for all n > X:

{(–1)n}

2 3 4 5 6 7 n1

1

–1 {a2k – 1}

{a2k}

The sequence ankf g is the

sequence

an1; an2

; an3; . . .:

Note that the sequence {an} isa subsequence of itself.

A result similar to part (b)holds for sequences wherean!�1.

You may omit this proof at afirst reading.We prove onlypart (a): the proof of part (b) issimilar.

66 2: Sequences

For simplicity in the rest of the proof, we shall now assume that our initial

choice of X is a positive integer.

So, if we take K so large that

nK � X ;then

nk > nK � X; for all k > K;

and soank� ‘j j < "; for all k > K;

which proves (2). &

The following criteria for establishing that a sequence is divergent are

immediate consequences of Theorem 5, part (a):

Corollary 2

1. First Subsequence Rule The sequence {an} is divergent if it has two

convergent subsequences with different limits.

2. Second Subsequence Rule The sequence {an} is divergent if it has a

subsequence which tends to infinity or a subsequence which tends to

minus infinity.

We can now formulate the strategy promised at the beginning of this section.

Strategy To prove that a sequence {an} is divergent:

EITHER

1. show that {an} has two convergent subsequences with different limits

OR

2. show that {an} has a subsequence which tends to infinity or a subse-

quence which tends to minus infinity.

For example, the sequence �1ð Þnf g has two convergent subsequences

which have different limits: namely, the even subsequence with limit 1 and

the odd subsequence with limit �1. So the sequence �1ð Þnf g is divergent, by

the First Subsequence Rule.

On the other hand, the sequence n �1ð Þn� �

has a subsequence (the even

subsequence) which tends to infinity. So n �1ð Þn� �

is divergent by the Second

Subsequence Rule.

Remark

In order to apply the above strategy successfully to prove that a sequence is

divergent, you need to be able to spot convergent subsequences with different

limits or subsequences which tend to infinity or to minus infinity. It is not

always easy to do this, and some experimentation may be required! If the

formula for an involves the expression (�1)n, it is a good idea to consider the

odd and even subsequences, although this may not always work. It may be

helpful to calculate the values of the first few terms in order to try to spot some

subsequences whose behaviour can be identified.

Problem 5 Use the above strategy to prove that each of the following

sequences {an} is divergent:

(a) �1ð Þnþ 1n

� �

; (b) 13

n� 13

n� ��

; (c) n sin 12

np� ��

.

We will occasionally makethis assumption if we want torefer to terms such as aX ratherthan mess about with nastyexpressions such as a[X],where [X] is the integral partof X.

Though we do not prove thefact, ANY divergent sequenceis of one (or both) of these twotypes.

The fact that this strategy willalways apply is simply areformulation of the resultmentioned in the abovemargin note.

In part (b) the square bracketsdenote ‘the integer part’function.

We end this section by giving a result about subsequences which will be

needed in later chapters.

Theorem 6 If the odd and even subsequences of {an} both tend to the

same limit ‘, then

limn!1

an ¼ ‘:

Proof We want to show that:


an � ‘j j < "; for all n > X: (3)

We know that there are integers K1 and K2 such that

a2k�1 � ‘j j < "; for all k > K1;

and

a2k � ‘j j < "; for all k > K2:

So we now let

X ¼ maxf2K1 � 1; 2K2g:With this choice of X, each n>X is either of the form 2k� 1, with k>K1, or of

the form 2k, with k>K2; it follows then that (3) holds with this value of X. &

2.5 The Monotone Convergence Theorem

2.5.1 Monotonic sequences

In Section 2.3, we gave various techniques for finding the limit of a convergent

sequence. As a result, you may be under the impression that, if we know that a

sequence converges, then there is some way of finding its limit explicitly.

However, it is sometimes possible to prove that a sequence is convergent, even

though we do not know its limit. For example, this situation occurs with a given

sequence {an}, which has the following two properties:

1. {an} is an increasing sequence;

2. {an} is bounded above; that is, there is a real number M such that

an � M; for n ¼ 1; 2; . . .:

Likewise, if {an} is a sequence which is decreasing and bounded below, then

{an} must be convergent.

We combine these two results into one statement.

Theorem 1 Monotone Convergence Theorem

If the sequence {an} is:

EITHER

1. increasing and 2. bounded above

OR

1. decreasing and 2. bounded below,

then {an} is convergent.

{a2k�1} and {a2k} are the oddand even subsequences of {an}.

68 2: Sequences

Proof of the first case Let {an} be a sequence that is increasing and bounded

above. Since {an} is bounded above, the set {an: n¼ 1, 2, . . .} has a least upper

bound, ‘ say; this is true by the Least Upper Bound Property of R . We now

prove that limn!1

an ¼ ‘.We want to show that:

for each positive number ", there is a number X such that:

an � ‘j j < "; for all n > X: (1)

We know that, if "> 0, then, since ‘ is the least upper bound of the set

{an: n¼ 1, 2, . . .}, there is an integer X such that

aX > ‘� ":Since {an} is increasing, an� aX, for n>X, and so

an > ‘� "; for all n > X:

Thus

an � ‘j j ¼ ‘� an < "; for all n > X;

which proves (1).

Hence {an} converges to ‘. &

The Monotone Convergence Theorem tells us that a sequence such as

1� 1n

� �

, which is increasing and bounded above (by 1, for example), must

be convergent. In this case, of course, we know already that 1� 1n

� �

is

convergent (with limit 1) without using the Monotone Convergence Theorem.

The Monotone Convergence Theorem is often used when we suspect that a

sequence is convergent, but cannot find the actual limit. It can also be used to

give precise definitions of numbers, such as p, about which we have only an

intuitive idea, and we do this later in this section.

For completeness, we point out that:

If {an} is increasing but is not bounded above, then an!1 as n!1.

Indeed, for any real number K, we can find an integer X such that

aX > K;

because {an} is not bounded above. Since {an} is increasing, an� aX for n>X,

and so

an > K; for all n > X:

Hence an!1 as n!1.

Similarly, we have the following result:

If {an} is decreasing but is not bounded below, then an!�1 as n!1.

We summarise these results about monotonic sequences in the following

useful theorem:

Theorem 2 Monotonic Sequence Theorem

If the sequence {an} is monotonic, then:

EITHER {an} is convergent

OR an!�1.

We prove only the increasingversion; the proof of thedecreasing version is similar.You met this Property inSub-section 1.4.3.

� + ε

� – ε

�

X n

This follows from thedefinition of least upperbound.

jan� ‘j ¼ ‘� an, becausean� ‘.

We will use the MonotoneConvergence Theorem inprecisely this way in Chapter 3.

Sub-section 2.5.4.

The expression ‘an!�1’ isshorthand for ‘either an!1or an!�1’.

Next we note a consequence of the Monotone Convergence Theorem that is

sometimes useful.

Corollary 1

(a) If a sequence {an} is increasing and bounded above, then it tends to its

least upper bound, sup {an: n ¼ 1, 2, . . .}.

(b) If a sequence {an} is decreasing and bounded below, then it tends to its

greatest lower bound, inf {an: n ¼ 1, 2, . . .}.

Problem 1 Prove part (b) of the Corollary.

We meet applications of these results in the rest of this section, to functions

defined recursively and to the study of the numbers e and p. But our first

application is to a result of considerable value in Analysis, Topology and other

parts of Mathematics.

The Bolzano–Weierstrass Theorem

We have seen that if a sequence is convergent, then certainly it must be

bounded. However if a sequence is bounded, it does not follow that it is

necessarily convergent. For example, the sequence {an}, where an ¼ 1n, is

bounded (by 1, since janj ¼ j 1n j � 1, 2, . . .) and is also convergent (to the

limit 0). Yet the sequence {an}, where an¼ (�1)n, is also bounded (by 1, since

anj j ¼

�1ð Þn

¼ 1, for n¼ 1, 2, . . .) but is divergent (since its odd and even

subsequences tend to different limits).

Our next result shows that, if a sequence is bounded, then, even though it

may diverge, it cannot behave ‘too badly’!

Theorem 3 Bolzano–Weierstrass Theorem

A bounded sequence must contain a convergent subsequence.

That is, if a sequence {an} is such that janj �M, for all n, then there exist

some number ‘ in [�M, M] and some subsequence ankf g such that ank

f gconverges to ‘ as k!1.

For example, the sequence {sin n} is bounded. With the tools at our disposal,

it is not at all clear what the behaviour of this sequence might be as n!1!

However, the Bolzano–Weierstrass Theorem asserts that there is at least one

number ‘ in [�1, 1] such that some subsequence {sin nk} converges to ‘. This is

itself a surprising result! In fact, however, using quite sophisticated mathe-

matics we can prove that every number ‘ in [�1, 1] has this property!

The proof of the Bolzano–Weierstrass Theorem that we give is of interest in

its own right; it depends on the notion of repeated bisection, which is a

standard technique in many areas of Analysis.

Proof We shall assume that the bounded sequence {an} has the property that

janj � M, for all n. Then all the terms an lie in the closed interval [�M, M],

which we will denote as [A1, B1].

Next, denote by p the midpoint of the interval [A1, B1]. Then at least one of

the two intervals [A1, p] and [p, B1] must contain infinitely many terms in the

The proof of part (a) iscontained in the proof ofTheorem 1 above.

This proof is similar to theearlier discussion.

We shall use it in Section 4.2.

This was the BoundednessTheorem – Theorem 1,Sub-section 2.4.2.

For jsin nj � 1.

Sadly, this is beyond the rangeof this book.

A1 ¼ �M and B1 ¼ M.

p ¼ 12

A1 þ B1ð Þ.

70 2: Sequences

sequence {an} – for otherwise the whole sequence would only contain finitely

many terms. If only one of the two intervals has this property, denote that

interval by the notation [A2, B2]; if both intervals have this property, choose the

left one (that is, [A1, p]) to be [A2, B2].

In either case, we obtain:

1. A2, B2½ � A1, B1½ �;2. B2 � A2 ¼ 1

2B1 � A1ð Þ;

3. [A1, B1] and [A2, B2] both contain infinitely many terms in the sequence {an}.

We now repeat this process indefinitely often, bisecting [A2, B2] to obtain

[A3, B3], and so on. This gives a sequence of closed intervals {[An, Bn]} for

which:

1. Anþ1;Bnþ1½ � An;Bn½ �, for each n2N;

2. Bn � An ¼ 12

� �n�1B1 � A1ð Þ, for each n2N;

3. each interval [An, Bn] contains infinitely many terms in the sequence {an}.

Property 1 implies that the sequence {An} is increasing and bounded above by

B1 ¼ M. Hence, by the Monotone Convergence Theorem, {An} is convergent;

denote by A its limit. By the Limit Inequality Rule, we must have that A � M.

Similarly, Property 1 implies that the sequence {Bn} is decreasing and

bounded below by A1¼�M. Hence, by the Monotone Convergence

Theorem, {Bn} is convergent; denote by B its limit. By the Limit Inequality

Rule, we must have that B� �M.

We may then deduce, by letting n!1 in Property 2 and using the

Combination Rules for sequences, that

B� A ¼ limn!1

Bn � limn!1

An ¼ limn!1

Bn � Anð Þ

¼ limn!1

1

2

� �n�1

B1 � A1ð Þ

¼ B1 � A1ð Þ limn!1

1

2

� �n�1

¼ 0:

In other words, the sequences {An} and {Bn} both converge to a common limit.

Denote this limit by ‘. Since ‘¼A¼B, we must have �M� ‘�M.

We find a suitable subsequence of {an} as follows. Choose any an1that lies in

[A1, B1]; this is possible since [A1, B1] contains infinitely terms in {an}. Next,

choose a term an2in the sequence, with n2> n1, such that an2

2 A2;B2½ �; this is

possible since [A2, B2] contains infinitely terms in {an}, by Property 3. And so on.

In this way, we construct a subsequence ankf g of {an}, with nkþ 1> nk, for

each k. Since Ak � ank� Bk, it follows from the Squeeze Rule for sequences

(that is, by letting k!1) that ankf g converges to ‘, as required. &

2.5.2 Sequences defined by recursion formulas

As we have seen, often sequences are defined by formulas; that is, for a given n,

we substitute that value of n into a formula and obtain the term an in the sequence

{an} Another way of specifying a sequence is to define its terms ‘inductively’, or

‘recursively’; here we specify the first term (or several terms) in the sequence,

and then have a formula that enables us to calculate all successive terms.

If both intervals containinfinitely many terms in thesequence, it does not matterwhich choice we take; but wespecify the left interval justfor definiteness.

Theorem 3, Sub-section 2.3.3.

That is, ‘2 [�M, M].

Since nkþ 1> nk, we musthave nk� k, for each k.

For Ak!A¼ ‘ andBk!B¼ ‘.


For example, the following sequences are defined recursively:

{an}, where a1¼ 2 and anþ1 ¼ 14

a2n þ 3

� �

, for n� 1;

{an}, where a1¼ 2, a2¼ 4 and an ¼ 12

an�1 þ an�2ð Þ, for n� 3.

Do these sequences converge? If they do converge, what are their limits? If we

choose a different value for a1 (or for a1 and a2, in the second sequence), do

their behaviours change? We can use the Monotone Convergence Theorem to

answer many such questions.

First, consider the sequence with recursion formula

anþ1 ¼1

4a2

n þ 3� �

; for n � 1: (2)

If indeed {an} is convergent, what value could its limit take? If we let ‘ denote

limn!1

an and let n!1 in equation (2), we obtain ‘ ¼ 14‘2 þ 3ð Þ. This can be

rearranged as 4‘¼ ‘2þ 3, and so as ‘2� 4‘þ 3¼ 0. Hence (‘� 1)(‘� 3)¼ 0,

so that ‘ = 1 or ‘ = 3.

This suggests that we should look at the differences anþ1� 1 and anþ1� 3.

Using equation (2), we deduce that

anþ1 � 1 ¼ 1

4a2

n � 1� �

¼ 1

4an þ 1ð Þ an � 1ð Þ; for n � 1; (3)

anþ1 � 3 ¼ 1

4a2

n � 9� �

¼ 1

4an þ 3ð Þ an � 3ð Þ; for n � 1: (4)

Next, it is useful to examine whether the sequence is monotonic by looking

at the difference anþ1� an

anþ1 � an ¼1

4a2

n � 4an þ 3� �

¼ 1

4an � 1ð Þ an � 3ð Þ: (5)

So the sign of anþ1� an depends on where an lies on the number-line in relation

to the numbers 1 and 3.

For example, it follows from (5) that

if 1 < an < 3, for some n � 1, then anþ1 � an < 0,

so that anþ1 < an:(6)

If an¼ 1, for some number n� 1, it follows from equation (3) that anþ1¼ 1.

Consequently, if our initial term is a1¼ 1, then it follows that an¼ 1, for all

n� 1; in other words, the sequence is simply the constant sequence {1}.

Similarly, if an¼ 3, for some number n� 1, it follows from equation (3) that

anþ1¼ 3. Consequently, if our initial term is a1¼ 3, then it follows that an¼ 3,

for all n� 1; in other words, the sequence is simply the constant sequence {3}.

Next, suppose that 1< an< 3, for some number n� 1. It follows from

equation (3) that anþ1� 1 is positive, so that 1< anþ1. On the other hand, it

follows from equation (4) that anþ1� 3 is negative, so that anþ1< 3.

Thus, if 1< an< 3, for some number n� 1, then 1< anþ 1< 3.

We can then prove by induction that, if 1< a1< 3, then 1< an< 3, for all

n> 1. (We omit the details.)

Next, it follows, from this last fact and (6) above that, if 1< a1< 3, then the

sequence {an} is strictly decreasing.

Hence, if 1< a1< 3 the sequence {an} is strictly decreasing and is bounded

below. Hence, by the Monotone Convergence Theorem, {an} is convergent.

Of course we have not yetproved that {an} isconvergent. This has been a‘what if?’ discussion so far,but a useful one.

Because 1 and 3 are thepossible limits.

We shall use this fact below.

In particular this discussioncovers the case a1¼ 2 that wementioned at the start of thissub-section.

Now comes the coup degrace!

72 2: Sequences

Since {an} is decreasing, whatever its limit might be (and we know that the only

two possibilities for the limit are 1 and 3), its limit must be less than or equal to its

first term a1. It follows that the limit of the sequence must be 1.

Problem 2 Let the sequence {an} be defined by the recursion formula

anþ1 ¼ 14

a2n þ 3

� �

, for n� 1.

(a) Prove that, if a1> 3, then an!1 as n!1.

(b) In the case that 0� a1< 1, determine whether {an} is convergent

and, if so, to what limit.

(c) Describe the behaviour of the sequence {an} in the case that a1< 0.

2.5.3 The number e

We will define e to be the limit of the sequence 1þ 1n

� �n� �

. If we plot the first

few terms of this sequence on a sequence diagram, then it seems that the

sequence is increasing and converges to a limit, which is less than 3.

To show that 1þ 1n

� �n� �

is convergent, using the Monotone Convergence

Theorem, we prove that the sequence is increasing and is bounded above.

1. 1þ 1n

� �n� �

is increasing

By the Binomial Theorem

1þ 1

n

� �n

¼ 1þ n1

n

� �

þ n n� 1ð Þ2!

1

n

� �2

þ . . . þ 1

n

� �n

:

The general term in this expansion is

n n� 1ð Þ . . . n� k þ 1ð Þk!

1

n

� �k

¼ 1

k!1� 1

n

� �

1� 2

n

� �

. . . 1� k � 1

n

� �

: (7)

If k is fixed, then each of the factors

1� 1

n

� �

; 1� 2

n

� �

; . . . ; 1� k � 1

n

� �

is increasing, and so the general term (7) increases as n increases. Since this

is true for each fixed k, the sequence ð1þ 1nÞn

� �

is increasing.

2. 1þ 1n

� �n� �

is bounded above

Here, note that the general term (7) satisfies the inequality

1

k!1� 1

n

� �

1� 2

n

� �

. . . 1� k � 1

n

� �

� 1

k!;

We saw that 1 and 3 were theonly possible limits at the startof the discussion.

For example, to three decimalplaces the first two terms are 2and 2.25, the sixth term is2.521, and the hundredth termis 2.704.

since each of the brackets is at most 1. Hence

1þ 1

n

� �n

� 1þ 1þ 1

2!þ 1

3!þ . . . þ 1

n!

� 1þ 1þ 1

21þ 1

22þ . . . þ 1

2n�1;

since k!¼ k (k� 1) . . . 2.1� 2k�1.

Now

1þ 1

21þ 1

22þ þ 1

2n�1¼ 1�

1� 12

� �n

1� 12

¼ 2� 1� 1

2

� �n� �

¼ 2� 1

2n�1;

and so

1þ 1

n

� �n

� 3� 1

2n�1; for n ¼ 1; 2; . . .:

Thus the sequence 1þ 1n

� �n� �

is bounded above, by 3.

It follows, by the Monotone Convergence Theorem, that the sequence

1þ 1n

� �n� �

is convergent with limit at most 3. This allows us to make the

following definition:

Definition e ¼ limn!1

1þ 1n

� �n:

For larger and larger values of n, the terms 1þ 1n

� �ngive better and better

approximate values for e. Unfortunately, the sequence 1þ 1n

� �n� �

converges

to e rather slowly, and we need to take very large integers n to get a reasonable

approximation to e¼ 2.71828. . ..Now we would like in general to make the definition of ex as

ex ¼ limn!1

1þ xn

� �n, but first we need to know that this limit actually exists.

Problem 3 Let x> 0.

(a) Prove that ð1þ xnÞn

� �

is an increasing sequence, by adapting the

method above where x¼ 1.

(b) Verify that 1þ kn� ð1þ 1

nÞk, for k¼ 1, 2, . . .. Using this fact, prove

that ð1þ xnÞn

� �

is bounded above.

(c) Deduce that ð1þ xnÞn

� �

is convergent.

Problem 4 By considering the product of the first n terms of the

sequence ð1þ 1nÞn

� �

, prove that n! > nþ1e

� �n, for n¼ 1, 2, . . ..

We now examine the convergence of the sequence ð1þ xnÞn

� �

, for x< 0.

Recall that, when x< 0, then �x> 0; in particular, from the above discus-

sion, the sequence ð1� xnÞn

� �

converges. We shall use Bernoulli’s inequality

(1þ c)n� 1þ nc, for c��1.

In investigating the convergence of the sequence ð1þ xnÞn

� �

, we are

only interested in what happens when n is large. So, we need consider

only the situation when n>�x. Then n2> x2, so that 1n2 <

1x2; thus x2

n2 < 1,

or � x2

n2 > �1.

For the sum of the geometricprogression aþ arþ ar2þ . . .

þ arn�1 is a 1�rn

1�r, if r 6¼ 1;

here we have a¼ 1, r ¼ 12:

For example

1þ 11000

� �1000¼ 2:716 . . .:

Then e¼ e1

So limn!1

1þ xn

� �nexists

for x> 0. Note that the limit isat least 1, and so cannot bezero.

You met Bernoulli’sInequality in Sub-section 1.3.3.

Recall that � x> 0.

74 2: Sequences

It follows that we may substitute � x2

n2 for c in Bernoulli’s Inequality. This

gives that

1� x2

n2

� �n

� 1þ n � x2

n2

� �

¼ 1� x2

n; for n > �x:

It follows that

1 � 1� x2

n2

� �n

� 1� x2

n; for n > �x: (8)

Now the sequence 1� x2

n

n o

converges to the limit 1, by the Combination

Rules. Hence, by applying the Squeeze Rule to inequality (8), the sequence

1� x2

n2

�nn o

converges to the limit 1 as n!1But

1þ x

n

�n

¼1þ x

n

� �n1� x

n

� �n

1� xn

� �n

¼1� x2

n2

�n

1� xn

� �n : (9)

We have just seen that the numerator is convergent, and we saw above that the

denominator is convergent (to a non-zero limit) for�x> 0; it follows from (9),

by the Quotient Rule, that the sequence ð1þ xnÞn

� �

is convergent.

So whatever value we choose for the real number x, the sequence ð1þ xnÞn

� �

is convergent. This means that we can now make the following legitimate

definition:

Definition For any real value of x,

ex ¼ limn!1

1þ x

n

�n

:

We can deduce more than this from equation (9), if we rewrite it in the

convenient form

1þ x

n

�n

1� x

n

�n

¼ 1� x2

n2

� �n

:

Letting n!1, we deduce from this last equation that

exe�x ¼ 1:

We have shown that this holds for x< 0; however, by simply interchanging x

and �x, it is clear that this holds for x> 0 also.

Theorem 4 Inverse Property of ex

For any real value of x, ex e�x¼ 1

The next property of the exponential function that we need to verify is that

ex ey¼ exþ y for any real values of x and y. We shall prove this fact later.

Here we use the fact that f1ng is

a basic null sequence.

Recall that here we areconsidering the case x< 0.

Note that when x¼ 0, thesequence is simply theconstant sequence {1}.

The functionx 7! ex, for x 2 R , is calledthe exponential function.

Here we also use the fact that

limn!1

1� x2

n2

� �n

¼ 1;

that we proved above.

In Sub-section 3.4.3.

2.5.4 The number p

One of the oldest mathematical problems is to determine the area of a disc of

radius r, and the length of its perimeter. It is well known that these magnitudes

are pr2 and 2pr, respectively. But what exactly is p? Is there a real number pwhich makes these formulas correct? – and, if so, how is it formally defined?

We now give a precise definition of p as the area of a disc of radius 1, using a

method originally devised by Archimedes to calculate approximate values for

the area of this disc. The idea is to calculate the areas of regular polygons

inscribed in the disc. Archimedes found an easy way to calculate the areas of

such regular polygons with 6 sides, 12 sides, 24 sides and so on. The results of

parts (a) and (b) of the following problem help to give the first two of these areas.

Problem 5 Verify that the following triangles have the stated areas:

Hint: In part (d), use the half-angle formula tan# ¼ 2 tan 12#

1�tan2 12#:

Let sn denote the number of sides of the nth such inner polygon (so s1¼ 6,

s2¼ 12, s3¼ 24, and, in general, sn¼ 3� 2n) and let an denote the area of the

nth inner polygon. Then

an ¼1

2sn sin

2psn

� �

; for n ¼ 1; 2; . . .: (10)

Geometrically, it is obvious that each time we double the number of sides of

the inner polygon the area increases, and so

a1 < a2 < a3 < . . . < an < anþ1 < . . .:

Hence {an} is (strictly) increasing. But is {an} convergent?

Here ‘inscribed’ means thatall the vertices of the polygonlie on the circle, so that theinside of the polygon iscontained in the inside of thecircle.

Here p is used only as asymbol to represent angles; itsvalue is not required.

For example, to three decimalplaces

a1 ¼ 2:598,

a2 ¼ 3,

. . .

a6 ¼ 3:141:

76 2: Sequences

Notice that each of the polygons lies inside a square of side 2, which has

area 4. This means that

an � 4; for n ¼ 1; 2; . . .;

and so {an} is increasing and bounded above (by 4). Hence, by the Monotone

Convergence Theorem, {an} is convergent.

Our intuitive idea of the area of the disc suggests that its area is greater than

each of the areas an, but ‘only just’! Put another way, the area of the disc should

be the limit of the increasing sequence {an}. This leads us to make the

following definition:

Definition p ¼ limn!1

an:

We shall explain in a moment how to calculate the terms an without

assuming a value for p.

First, however, we describe how to estimate the area of the disc using outer

polygons. Once again we start with a regular hexagon and repeatedly double

the number of sides. The results of part (c) and (d) of the last problem help to

give the first two such areas.

As before, let sn¼ 3� 2n, for n¼ 1, 2, . . ., and let bn denote the area of the

nth outer polygon. This nth outer polygon consists of sn isosceles triangles,

each of height 1 and base 2 tanðpsnÞ. Thus

bn ¼ sn tanpsn

� �

; for n ¼ 1; 2; . . .: (11)

Geometrically, it is obvious that each time we double the number of sides of

the outer polygon the area decreases, and so

b1 > b2 > b3 > > bn > bnþ1 > :

So the sequence {bn} is (strictly) decreasing and bounded below (by 0,

for example). Thus, by the Monotone Convergence Theorem, {bn} is also

convergent.

Intuitively, we expect that {bn} has the same limit as {an}, which we have

defined to be p. But how can we prove this?

This will enable us to squeezep between the area of theinner polygons and the area ofthe outer polygons.

For example, to three decimalplaces

b1 ¼ 3:464;

b2 ¼ 3:215;

. . .

b6 ¼ 3:142:


The terms an and bn can be calculated by using the following equations,

which are known jointly as the Euclidean algorithm

anþ1 ¼ffiffiffiffiffiffiffiffiffi

anbn

p

; for n ¼ 1; 2; . . .; (12)

and

bnþ1 ¼2anþ1bn

anþ1þ bn

; for n ¼ 1; 2; . . .: (13)

Starting with a1 ¼ 32

ffiffiffi

3p¼ 2:598 . . . and b1 ¼ 2

ffiffiffi

3p¼ 3:464 . . ., we use

these equations iteratively to calculate first a2 ¼ffiffiffiffiffiffiffiffiffi

a1b1

p, then b2 ¼ 2a2b1

a2þb1, and

so on. Here are (approximations to) the first few values of each sequence

obtained in this way:

sn 6 12 24 48 96 192

an 2.598 3 3.106 3.133 3.139 3.141

bn 3.464 3.215 3.160 3.146 3.143 3.142

It does appear that both sequences do converge to a common limit, namely p.

Remark

We have defined p using the areas of approximating polygons. An alternative

approach uses the perimeters of these polygons.

2.5.5 Proofs

We now prove several of the results referred to in the discussion of p in the

previous sub-section.

First, we prove equations (12) and (13)

anþ1 ¼ffiffiffiffiffiffiffiffiffi

anbn

p

; for n ¼ 1; 2; . . .; (12)

and

bnþ1 ¼2anþ1bn

anþ1 þ bn

; for n ¼ 1; 2; . . .: (13)

Using the half-angle formulas for sin # and cos #, it is easy to check that

sin1

2# ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

2sin# tan

1

2#

r

and

tan1

2# ¼ sin# tan#

sin#þ tan#:

Hence, since snþ1¼ 2sn

anþ1 ¼1

2snþ1 sin

2psnþ1

� �

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

2sn sin

2psn

� �

� sn tanpsn

� �

s

¼ffiffiffiffiffiffiffiffiffi

anbn

p

:

Equations (12) and (13)can be verified fromequations (10) and (11); wegive the details in Sub-section 2.5.5 below.

We prove this in Sub-section 2.5.5 below.

You may omit this sub-section at a first reading.

These comprise the Euclideanalgorithm.

We omit the details in bothcases.

This proves equation (12).

78 2: Sequences

Similarly

bnþ1 ¼ snþ1 tanp

snþ1

� �

¼2sn sin p

sn

�

� sn tan psn

�

sn sin psn

�

þ sn tan psn

�

¼ 2anþ1bn

anþ1 þ bn

:

Finally we prove that limn!1

bn ¼ limn!1

an.

To do this, we rearrange equation (12) as follows

bn ¼a2

nþ1

an

:

Then, by the Combination Rules, we can deduce that

limn!1

bn ¼limn!1

anþ1

�2

limn!1

an

¼ p2

p¼ p:

Then, since we defined p to be precisely limn!1

an, it follows that limn!1

bn ¼ limn!1

an,

as required.

Definition Let an denote the area of the regular polygon with 3� 2n sides

inscribed in a disc of radius 1, and bn the area of the regular polygon with

3� 2n sides circumscribing the disc. Then

p is the common value of the limits limn!1

an and limn!1

bn.

Remark

It can be shown that every two applications of equations (12) and (13) give an

extra decimal place in the decimal expansion for p. This decimal expansion is

now known to many millions of decimal places, using sequences which

converge to p much more rapidly than {an} or {bn}. Since p is irrational,

there is no possibility of the decimal expansion of p recurring.

2.6 Exercises

Section 2.1

1. Calculate the first five terms of each of the following sequences, and draw a

sequence diagram in each case:

(a) {n2� 4nþ 4}; (b)�1ð Þnþ1

n!

n o

; (c) sin 14

np� ��

.

This proves equation (13).

We shall return in Section 8.5to the question ofapproximating p, by the use ofpower series.

2.6 Exercises 79

2. Determine which of the following sequences are monotonic:

(a)�

nþ 1nþ 2

�

; (b)�1ð Þn

n

n o

; (c)�

21n

�

.

3. Prove that the following sequences are eventually monotonic:

(a) 5n

n!

� �

; (b) n þ 8n

� �

.

Section 2.2

1. For each of the following sequences {an} and numbers ", find a number X

such that janj<", for all n>X:

(a) an ¼ �1ð Þnn5 , "¼ 0.001; (b) an ¼ 1

2nþ 1ð Þ2, "¼ 0.002.

2. Use the definition of null sequence to prove that the two sequences in the

previous exercise are null.

3. Prove that the following sequences are not null:

(a)ffiffiffi

npf g; (b) 1 þ �1ð Þn

n

n o

.

4. Assuming that 1n

� �

is null, deduce that the following sequences are null.

State which rules you use.

(a) 2ffiffi

np þ 3

n7

n o

; (b) sin nn2 þ 1

n o

; (c)tan�1 2nð Þ

nþ 1þ sin n

n o

.

5. Prove thatffiffiffiffiffiffiffiffiffiffiffi

nþ 1p

�ffiffiffi

np� �

is null.

Hint: Use the fact that a� b ¼ a2 � b2

aþ b, for aþ b 6¼ 0.

6. Use the list of basic null sequences to prove that the following sequences are

null. State which rules you use.

(a) 34n þ 2n

3n

� �

; (b)�

6n10

n!

�

; (c)�

n1010n

n!

�

.

7. Prove that, if the sequence {an} of positive numbers is null, thenffiffiffiffiffi

anp� �

is

null.

Section 2.3

1. Show that the following sequences converge to 1, by calculating an� 1 in

each case:

(a)�

n� 1nþ 3

�

; (b)�

n2

n2þnþ1

�

.

2. Use the Combination Rules to find the limits of the following sequences:

(a) n3

2n3 þ 3nþ 4n

n o

; (b)�

n10 � 2n

2n þ n100

�

; (c)�

5n!þ 5n

n100 þ n!

�

.

3. (a) Prove that, if the sequence {an} of positive numbers is convergent with

limit ‘, where ‘> 0, thenffiffiffiffiffi

anp� �

is convergent with limitffiffi

‘p� �

.

You met this list in Sub-section 2.2.1.

80 2: Sequences

Hint: Use the fact that a� b ¼ a2 � b2

aþ b, for aþ b 6¼ 0.

(b) Prove that, if the sequence {ak} is convergent with limit ‘, ‘ 6¼ 0, and a1n

k

is defined, where n2N , then a1n

k

n o

is convergent with limit ‘1n.

Hint: Use the identity aq� bq¼ (a� b)(aq�1þ aq�2bþ aq�3b2þ þbq�1) with suitable choices for a, b and q.

Section 2.4

1. Classify the following sequences as convergent or divergent, and as

bounded or unbounded:

(a)�

n14

�

; (b)ð�1Þn100n

n!

n o

.

2. Use the Reciprocal Rule to prove that the following sequences tend to

infinity:

(a) n!n3

� �

; (b) n2 þ 2n� �

; (c) n2 � 2n� �

; (d) n!� n3 � 3n� �

.

3. Use the Subsequence Rules to prove that the following sequences are

divergent:

(a) �1ð Þn2nf g; (b)�1ð Þnn2

2n2 þ 1

n o

.

4. Each of the following general statements concerning sequences {an} and {bn}

is true or false. For each statement that is true, prove it. For each statement that

is false, write down examples of sequences to illustrate your assertion:

(a) If {an} and {bn} are divergent, then {anþ bn} is divergent.

(b) If {an} and {bn} are divergent, then {anbn} is divergent.

(c) If {an} is convergent and {bn} is divergent, then {anbn} is divergent.

(d) If a2n

� �

is convergent, then {an} is convergent.

(e) If a2n

� �

is convergent, and an> 0, then {an} is convergent.

(f) If {bn} is convergent, where bn¼ anþ1an, then {an} is convergent.

(g) If an! 0 as n!1, then anþ1þanþ2þþa2nffiffi

np

n o

is convergent.

5. Prove that every sequence has a monotonic subsequence.

Section 2.5

1. Prove that, if {an} is increasing and has a subsequence ankf g which is

convergent, then {an} is convergent.

2. Let anf g1n¼1 be a sequence for which anþ1 ¼ 3an þ 1anþ3

, for n� 1.

(a) Prove that, if a1>�1, the sequence {an} converges.

Hint: consider the cases �1< a1< 1, a1¼ 1 and a1> 1 separately.

(b) If a1��1, does the sequence converge or diverge?

The Power Rule in Sub-section 2.2.1 and thesubsequent remark there showthat this result holds in thecase that ‘¼ 0.

2.6 Exercises 81

3. Let anf g1n¼1 be a sequence for which anþ1 ¼ 12

an þ ‘2

an

�

, for n� 1, where

a1> 0 and ‘> 0.

(a) Prove that {an} is decreasing for n� 2, and hence that an! ‘ as n!1.

(b) With ‘2¼ 2 and a1¼ 1.5, calculate the terms a2, a3, a4 and a5.

4. Obtain formulas for the perimeters of the inner and outer polygons intro-

duced in Sub-section 2.5.4. Prove that these sequences tend to 2p as n!1.

5. Bernoulli’s Inequality states that (1þ x)n� 1þ nx, for x��1, n¼ 1, 2, . . ..

(a) Apply Bernoulli’s Inequality, with x ¼ �1n2 , to give an alternative proof

that 1þ 1n

� �n� �

is increasing.

(b) Apply Bernoulli’s Inequality, with x ¼ 1n2�1

, to prove that 1þ 1n

� �nþ1n o

is decreasing.

(c) Deduce from parts (a) and (b) that

1þ 1

n

� �n

� e � 1þ 1

n

� �nþ1

; for n ¼ 1; 2; . . .:

6. Let an; bn½ �f g1n¼1 be a sequence of closed intervals with the property that

anþ1; bnþ1½ � � an; bn½ �, for n¼ 1, 2, . . ..

(a) Prove that there exists some point c that lies in all the intervals [an, bn],

for n¼ 1, 2, . . ..

(b) Prove that, if bn� an! 0 as n!1, then there exists only one such

point c.

You met this in Sub-section 1.3.3.

This result is sometimescalled The Nested IntervalsTheorem.

82 2: Sequences

3 Series

The Ancient Greek philosopher Zeno of Elea proposed a number of paradoxes

of the infinite, which are said to have had a profound influence on Greek

mathematics. For example, Zeno claimed that it is impossible for an object to

travel a given distance, since it must first travel half the distance, then half

of the remaining distance, then half of what remains, and so on. There must

always remain some distance left to travel, and so the journey cannot be

completed.

This paradox relies partly on the intuitive feeling that it is impossible to add

up an infinite number of positive quantities and obtain a finite answer.

However, the following illustration of the paradox suggests that such an

infinite sum is plausible.

0 1

12

14

18

78

34

12

116

1516

The distance from 0 to 1 can be split up into the infinite sequence of

distances 12

, 14

, 18, . . ., and so it seems reasonable to write

1

2þ 1

4þ 1

8þ 1

16þ � � � ¼ 1:

This chapter is devoted to the study of such expressions, which are called

infinite series.

The following example shows that infinite series need to be treated with

care. Suppose that it is possible to add up 2, 4, 8, . . ., and that the answer is s

2þ 4þ 8þ 16þ � � � ¼ s:

If we multiply through by 12, we obtain

1þ 2þ 4þ 8þ � � � ¼ 1

2s;

which we can rewrite in the form

1þ s ¼ 1

2s:

It follows from this equation that s¼�2, which is obviously non-sensical.

We can avoid reaching such absurd conclusions by performing arithmetical

operations only with convergent infinite series.

In Section 3.1 we define the concept of convergent infinite series in terms

of convergent sequences, and give some examples. We also describe various

properties which are common to all convergent series.

Because the left-hand side ofthe previous equation can beexpressed as

1þ ð2þ 4þ 8þ � � �Þ¼ 1þ s:

83

Section 3.2 is devoted to series with non-negative terms, and we give several

tests for the convergence of such series.

In Section 3.3 we deal with the much harder problem of determining

whether a series is convergent when it contains both positive and negative

terms. We also look at what can happen if we rearrange the terms of a

convergent series – does the new series still converge? This section is quite

long, so you might wish to tackle it in several sessions.

In Section 3.4 we explain how ex can be represented as an infinite series of

powers of x, and we use this representation to prove that the number e is

irrational, and that exey ¼ exþy.

3.1 Introducing series

3.1.1 What is a convergent series?

We begin by defining precisely what is meant by the statement that

1

2þ 1

4þ 1

8þ 1

16þ � � � ¼ 1:

Let sn be the sum of the first n terms on the left-hand side. Then

s1 ¼1

2;

s2 ¼1

2þ 1

4¼ 3

4;

s3 ¼1

2þ 1

4þ 1

8¼ 7

8; . . .; and so on:

In general, using the formula for the sum of a geometric progression, with

a ¼ r ¼ 12, we obtain

sn ¼1

2þ 1

4þ 1

8þ � � � þ 1

2n¼ 1

2þ 1

22þ 1

23þ � � � þ 1

2n

¼ 1

2�

1� 12

� �n

1� 12

¼ 1� 1

2

� �n

:

The sequence 12

� �n� �

is a null sequence, and so

limn!1

sn ¼ limn!1

1� 1

2

� �n� �

¼ 1� limn!1

1

2

� �n

¼ 1:

It is the precise mathematical statement that sn! 1 as n!1which justifies

our earlier (intuitive) statement that

1

2þ 1

4þ 1

8þ 1

16þ � � � ¼ 1:

We use this approach to define a convergent infinite series.

You first met ex in Section 2.5.

The sum of a geometricprogression aþ arþar 2 þ � � � þ arn�1 is a 1�rn

1�r,

if r 6¼ 1.

Here we use the CombinationRules for sequences.

84 3: Series

Definition Given a sequence {an} of real numbers, the expression

a1 þ a2 þ a3 þ � � �is called an infinite series, or simply a series. The nth partial sum of this

series is

sn ¼ a1 þ a2 þ a3 þ � � � þ an:

The behaviour of the infinite series a1þ a2þ a3þ � � � is determined by the

behaviour of {sn}, its sequence of partial sums.

Definition The series a1þ a2þ a3þ � � � is convergent with sum s if the

sequence {sn} of partial sums converges to s. In this case, we say that the

series converges to s, and we write

a1 þ a2 þ a3 þ � � � ¼ s:

The series diverges if the sequence {sn} diverges.

Thus we prove results about series by applying results for sequences proved

in Chapter 2 to the sequence of partial sums {sn}.

Example 1 For each of the following infinite series, calculate its nth partial

sum, and determine whether the series is convergent or divergent:

(a) 1þ 1þ 1þ � � �; (b) 13þ 1

32 þ 133 þ � � �; (c) 2þ 4þ 8þ � � �.

Solution

(a) In this case

sn ¼ 1þ 1þ � � � þ 1 ¼ n:

The sequence {sn} tends to infinity, and so this series is divergent.

(b) Using the formula for summing a finite geometric progression, with

a ¼ r ¼ 13, we obtain

sn ¼1

3þ 1

32þ 1

33þ � � � þ 1

3n

¼ 1

3�

1� 13

� �n

1� 13

¼ 1

21� 1

3

� �n� �

:

Since 13

� �n� �

is a basic null sequence

limn!1

sn¼1

2;

and so this series is convergent, with sum 12.

(c) In this case

sn ¼ 2þ 4þ 8þ � � � þ 2n

¼ 22n � 1

2� 1

� �

¼ 2nþ1 � 2:

The sequence {2nþ1� 2} tends to infinity, and so this series is divergent.&


Sigma notation

Next, we explain how to use the sigma notation to represent infinite series.

Recall that a finite sum such as

1

2þ 1

4þ 1

8þ � � � þ 1

1024¼ 1

21þ 1

22þ 1

23þ � � � þ 1

210

can be represented using sigma notation as

X

10

n¼1

1

2n:

This notation can be adapted to represent infinite series, as follows

X

1

n¼1

an¼ a1 þ a2 þ a3 þ � � �:

Convention When using the sigma notation to represent the nth partial

sum sn of a series, we write

sn ¼ a1 þ a2 þ a3 þ � � � þ an ¼X

n

k¼1

ak:

We have written k as the ‘dummy variable’ here, to avoid using n for two

different purposes in the same expression. We could have used any letter (other

than n) for the dummy variable; for example, the expressionsP

n

i¼1

ai andP

n

r¼1

ar

also stand for a1þ a2þ a3þ � � � þ an.

If we need to begin a series with a term other than a1, then we write, for example

X

1

n¼0

an¼ a0 þ a1 þ a2 þ � � � orX

1

n¼3

an¼ a3 þ a4 þ a5 þ � � �:

For any series, the nth partial sum sn is obtained by adding the first n terms. For

example, the nth partial sums of the above two series are

sn ¼ a0 þ a1 þ a2 þ � � � þ an�1 ¼X

n�1

k¼0

ak and

sn ¼ a3 þ a4 þ a5 þ � � � þ anþ2 ¼X

nþ2

k¼3

ak:

Problem 1 For each of the following infinite series, calculate the nth

partial sum sn, and determine whether the series is convergent or divergent:

(a)P

1

n¼1

�13

� �n; (b)

P

1

n¼0

�1ð Þn; (c)P

1

n¼�1

12

� �n:

Remark

Notice that inserting into a series, or omitting or altering, a finite number of

terms does not affect the convergence of the series, but may affect the sum. For

example, since the series 12þ 1

4þ 1

8þ � � � converges with limit 1, it follows that

� is the Greek capital letter‘sigma’.

For exampleX

1

n¼1

1

2n¼ 1

2þ 1

4þ 1

8þ � � � ;

X

1

n¼1

1

n¼ 1þ 1

2þ 1

3þ 1

4þ � � � :

For the above examples

sn¼1

2þ1

4þ1

8þ�� þ 1

2n

¼X

n

k¼1

1

2k;

sn¼ 1þ1

2þ1

3þ1

4þ�� þ1

n

¼X

n

k¼1

1

k:

For example, we writeX

1

n¼3

1

n!� n

for the series

1

3!� 3þ 1

4!� 4þ � � �:

Recall that the convergence ordivergence of sequences wasnot affected by omitting oraltering a finite number ofterms in the sequence.

86 3: Series

the series 8þ 4þ 2þ 1þ 12þ 1

4þ 1

8þ � � � ¼ 8þ 4þ 2þ 1ð Þ þ 1

2þ 1

4þ 1

8þ � � �

converges with limit (8þ 4þ 2þ 1)þ 1¼ 16.

3.1.2 Geometric series

All the series considered so far are examples of geometric series. The standard

geometric series with first term a and common ratio r isX

1

n¼0

arn¼ aþ ar þ ar2 þ � � �:

The following theorem enables us to decide whether any given geometric

series is convergent or divergent.

Theorem 1 Geometric Series

(a) If jrj< 1, thenP

1

n¼0

arn is convergent, with sum a1�r

.

(b) If jrj � 1 and a 6¼ 0, thenP

1

n¼0

arn is divergent.

Proof

(a) If r 6¼ 1, then the nth partial sum sn is given by

sn ¼ aþ ar þ ar2 þ � � � þ arn�1 ¼ a1� rn

1� r:

Now, if rj j < 1, then rnf g is a basic null sequence, and so

limn!1

sn ¼ limn!1

a1� rn

1� r¼ a

1� rlim

n!11� rnð Þ

¼ a

1� r;

by the Combination Rules for sequences. Thus, if jrj< 1, thenP

1

n¼0

arn is

convergent, with sum a1�r

.

(b) Part (b) can be deduced immediately from the Non-null Test, which

appears later in this section. We defer the proof until then. &

Decimal representation of rational numbers

Geometric series provide another interpretation of the decimal representation

of rational numbers. Recall that rational numbers have terminating, or recur-

ring, decimal representations. For example

3

10¼ 0:3 and

1

3¼ 0:333 . . . ¼ 0:3:

Another way to interpret the symbol 0.333 . . . is as an infinite series

3

101þ 3

102þ 3

103þ � � �:

This is a geometric series, with first term a ¼ 310

and common ratio r ¼ 110

.

Since 0 < 110< 1, this series is convergent, with sum

a

1� r¼

310

1� 110

¼ 3

9¼ 1

3:

For 12þ 1

4þ 1

8þ � � � ¼ 1:

This value for sn is easilyverified by MathematicalInduction.

Sub-section 3.1.5, Example 3.

Sub-section 1.1.3.

Thus we have another method of finding the fraction which is equal to a given

recurring decimal, by calculating the sum of the corresponding geometric series.

Problem 2 Interpret the following decimals as infinite series, and

hence represent them as fractions:

(a) 0.111 . . .; (b) 0.86363 . . .; (c) 0.999 . . ..

3.1.3 Telescoping series

Geometric series are easy to deal with because it is possible to find a formula

for the nth partial sum sn. The next problem deals with another series for which

we can calculate sn explicitly.

Problem 3 Calculate the first four partial sums of the following series,

giving your answers as fractions

X

1

n¼1

1

n nþ 1ð Þ ¼1

1� 2þ 1

2� 3þ 1

3� 4þ � � �:

The results obtained in Problem 3 suggest the general formula

sn ¼1

1� 2þ 1

2� 3þ 1

3� 4þ � � � þ 1

n nþ 1ð Þ ¼n

nþ 1:

This formula can be proved by Mathematical Induction, or we can use the identity

1

n nþ 1ð Þ ¼1

n� 1

nþ 1; for n ¼ 1; 2; . . .;

which implies that

sn ¼1

1� 2þ 1

2� 3þ 1

3� 4þ � � � þ 1

n� 1ð Þnþ1

n nþ 1ð Þ

¼ 1

1� 1

2

� �

þ 1

2� 1

3

� �

þ 1

3� 1

4

� �

þ � � � þ 1

n� 1� 1

n

� �

þ 1

n� 1

nþ 1

� �

¼ 1� 1

nþ 1

¼ n

nþ 1:

Since

limn!1

sn ¼ limn!1

n

nþ 1¼ lim

n!1

1

1þ 1n

¼ 1;

we deduce that the given series is convergent, with sum

X

1

n¼1

1

n nþ 1ð Þ ¼ 1:

Problem 4 Find the nth partial sum of the seriesP

1

n¼1

1n nþ2ð Þ, using the

fact that

2

n nþ 2ð Þ ¼1

n� 1

nþ 2; for n ¼ 1; 2; . . .:

Deduce that this series is convergent, and find its sum.

You should at least read thesolution to part (c), even ifyou do not tackle it first.

This cancellation of adjacentterms explains why this seriesis said to be telescoping.

88 3: Series

3.1.4 Combination Rules for convergent series

At the start of this chapter we saw that performing arithmetical operations on

the divergent series 2þ 4þ 8þ � � � can lead to absurd conclusions. However,

the following result shows that there are Combination Rules for convergent

series, which follow directly from the Combination Rules for convergent

sequences:


IfP

1

n¼1

an ¼ s andP

1

n¼1

bn ¼ t, then:

Sum RuleP

1

n¼1

an þ bnð Þ ¼ sþ t;

Multiple RuleP

1

n¼1

lan ¼ ls, for l2R .

Proof Consider the sequences of partial sums {sn} and {tn}, where

sn ¼X

n

k¼1

ak and tn ¼X

n

k¼1

bk:

By assumption, sn! s and tn! t as n!1.

Sum Rule

The nth partial sum of the seriesP

1

n¼1

an þ bnð Þ is

X

n

k¼1

akþbkð Þ¼ a1þb1ð Þþ a2þb2ð Þþ �� þ anþbnð Þ

¼ a1þa2þ�� þanð Þþ b1þb2þ�� þbnð Þ¼ snþ tn:

By the Sum Rule for sequences

limn!1

sn þ tnð Þ ¼ limn!1

sn þ limn!1

tn ¼ sþ t;

and so the sequence {snþ tn} of partial sums ofP

1

n¼1

an þ bnð Þ has limit sþ t.

Hence this series is convergent and

X

1

n¼1

an þ bnð Þ ¼ sþ t:

Multiple Rule

The nth partial sum of the seriesP

1

n¼1

lan is

X

n

k¼1

lak ¼ la1 þ la2 þ � � � þ lan

¼ l a1 þ a2 þ � � � þ anð Þ ¼ lsn:

By the Multiple Rule for sequences

limn!1

lsnð Þ ¼ l limn!1

sn¼ ls;

We may rearrange the termshere, because this is the sumof a finite number of terms.


and so the sequence {lsn} of partial sums ofP

1

n¼1

lan has limit ls. Hence this

series is convergent and

X

1

n¼1

lan ¼ ls: &

Example 2 Prove that the following series is convergent and calculate its sum

X

1

n¼1

1

2nþ 3

n nþ 1ð Þ

� �

:

Solution We know thatP

1

n¼1

12n is convergent, with sum 1, and that

P

1

n¼1

1n nþ1ð Þ is

convergent, with sum 1.

Hence, by the Sum Rule and the Multiple Rule

X

1

n¼1

1

2nþ 3

n nþ 1ð Þ

� �

is convergent; with sum 1þð3� 1Þ ¼ 4: &

Problem 5 Prove that the following series is convergent and calculate

its sum

X

1

n¼1

3

4

� �n

� 2

n nþ 1ð Þ

� �

:

3.1.5 The Non-null Test

For all the infinite series we have so far considered, it is possible to derive a

simple formula for the nth partial sum. For most series, however, this is very

difficult or even impossible.

Nevertheless, it may still be possible to decide whether such series are

convergent or divergent by applying various tests. The first test that we give

arises from the following result:

Theorem 3 IfP

1

n¼1

an is a convergent series, then {an} is a null sequence.

Proof Let sn ¼P

n

k¼1

ak denote the nth partial sum ofP

1

n¼1

an. BecauseP

1

n¼1

an is

convergent, we know that {sn} is convergent. Suppose that limn!1

sn ¼ s.

We want to deduce that {an} is null. To do this, we write

sn ¼ sn�1 þ an;

and so

an ¼ sn � sn�1:

Thus, by the Combination Rules for sequences

limn!1

an ¼ limn!1

sn � sn�1ð Þ ¼ limn!1

sn � limn!1

sn�1

¼ s� s ¼ 0;

and so {an} is a null sequence, as required. &

You met the second series inSub-section 3.1.3.

For example, consider theseries

X

1

n¼1

1

n¼ 1þ 1

2þ 1

3þ � � � ;

and

X

1

n¼1

n2

2n2 þ 1¼ 1

3þ 4

9þ 9

19þ � � � :

90 3: Series

The following useful test for divergence is an immediate corollary of

Theorem 3:

Corollary Non-null Test If {an} is not a null sequence, thenP

1

n¼1

an is

divergent.

The virtue of the Non-null Test is its ease of application. For example, it

enables us to decide immediately thatX

1

n¼1

1;X

1

n¼1

�1ð Þn andX

1

n¼1

n are divergent;

because the corresponding sequences {1}, {(�1n)} and {n} are not null.

There are various ways to show that a given sequence {an} does not tend to

zero. For example, we can show that

an ! ‘ as n!1;where ‘ 6¼ 0, or that {an} tends to infinity or to minus infinity. More generally,

we can use the result about subsequences which states that

if an ! 0 as n!1; then ank! 0 as k!1;

for any subsequence ankf g of {an}. This leads to the following strategy:

Strategy To show thatP

1

n¼1

an is divergent, using the Non-null Test:

EITHER

1. show that {an} has a convergent subsequence with non-zero limit;

OR

2. show that {an} has a subsequence which tends to infinity, or a subse-

quence which tends to minus infinity.

This strategy can be used to prove part (b) of Theorem 1, as follows:

Example 3 Prove that, if jrj � 1 and a 6¼ 0, thenP

1

n¼0

arn is divergent.

Solution We want to show that, if jrj � 1 and a 6¼ 0, then {arn} is not a null

sequence. Now, the even subsequence

ar2k� �

¼ a; ar2; ar4; ar6; . . .� �

converges to a 6¼ 0 if jrj ¼ 1, and tends to infinity or to minus infinity if jrj � 1.

Thus, {ar2k} is not null, and so {arn} is not null.

Hence, by the Non-null Test,P

1

n¼0

arn is divergent if jrj> 1 and a 6¼ 0. &

Warning The converse of the Non-null Test is FALSE. In other words:

If {an} is a null sequence, it is not necessarily true thatP

1

n¼1

an is convergent.

For example, the sequence f1ng is a null sequence, but

P

1

n¼1

1n¼1þ 1

2þ 1

3þ � � �

is divergent.

You met this InheritanceProperty of subsequences inSub-section 2.4.4, Theorem 5,for any limit ‘.

It tends to infinity if a> 0 andto minus infinity if a< 0.

We shall prove this rathersurprising fact in Sub-section 3.2.1.

The important thing to remember is that you can never use the Non-null Test

to prove that a series is convergent, although you may be able to use it to prove

that a series is divergent.

Problem 6 Prove that the seriesP

1

n¼1

n2

2n2 þ 1is divergent.

3.2 Series with non-negative terms

In this section we consider only seriesP

1

n¼1

an with non-negative terms. In other

words we assume that an� 0, for n¼ 1, 2, . . ..

It follows that the partial sums ofP

1

n¼1

an, which are given by

s1 ¼ a1; s2 ¼ a1 þ a2; s3 ¼ a1 þ a2 þ a3; . . .;

sn ¼ a1 þ a2 þ a3 þ � � � þ an; . . .;

form an increasing sequence {sn}.

The fact that {sn} is increasing makes it easier to deal with series having

non-negative terms. If we can prove that the sequence {sn} of partial sums is

bounded above, then {sn} is convergent, by the Monotone Convergence

Theorem, and soP

1

n¼1

an is convergent. On the other hand, if we can prove

that the sequence {sn} of partial sums is not bounded above, then {sn} cannot

be convergent, and so we must have that sn!1 as n!1. We can rephrase

these facts in the following convenient way:

Boundedness Theorem for series A seriesP

1

n¼1

an of non-negative terms is

convergent if and only if its sequence {sn} of partial sums is bounded above.

We shall use this, for example, to prove the Cauchy Condensation Test in

Sub-section 3.2.1.

3.2.1 Tests for convergence

We now explore several tests for the convergence of series with non-negative

terms.

Problem 1 Use your calculator to find the first eight partial sums of

each of the following series

X

1

n¼1

1

n2¼ 1þ 1

22þ 1

32þ � � � and

X

1

n¼1

1

n¼ 1þ 1

2þ 1

3þ � � �;

giving your answers to 2 decimal places. Plot your answers on a

sequence diagram.

Example 1 Prove that the seriesP

1

n¼1

1n2 ¼ 1þ 1

22 þ 132 þ � � � is convergent.

Sub-section 2.5.1, Theorem 1.

This is essentially the sameresult as that of theBoundedness Theorem forsequences, Sub-section 2.4.2,Theorem 1.

92 3: Series

Solution All the terms of the series are positive, so we shall use the

Monotone Convergence Theorem.

Let sn be the nth partial sum of the series. Then

sn ¼ 1þ 1

22þ 1

32þ 1

42þ � � � þ 1

n2

< 1þ 1

1� 2þ 1

2� 3þ 1

3� 4þ � � � þ 1

n� 1ð Þ � n

¼ 1þ 1

1� 1

2

� �

þ 1

2� 1

3

� �

þ 1

3� 1

4

� �

þ � � � þ 1

n� 1� 1

n

� �

¼ 2� 1

n< 2:

It follows that {sn} is both increasing and bounded above (by 2), so that by

the Monotone Convergence Theorem {sn} is convergent. Hence the series

itself is also convergent. &

Remark

In fact the sum of the series isP

1

n¼1

1n2 ¼ p2

6, surprisingly. However a proof of this

goes beyond the scope of this book.

Not all series are convergent, though!


1

n¼1

1n¼ 1þ 1

2þ 1

3þ � � � is divergent.

Solution All the terms of the series are positive.

Let sn be the nth partial sum of the series. Then

sn ¼ 1þ 1

2þ 1

3þ 1

4þ 1

5þ 1

6þ 1

7þ 1

8þ 1

9þ 1

10þ � � �

¼ 1þ 1

2þ 1

3þ 1

4

� �

þ 1

5þ 1

6þ 1

7þ 1

8

� �

þ � � �

� � � þ 1

2k þ 1þ 1

2k þ 2þ � � � þ 1

2k þ 2k

� �

þ � � �:

It follows that a subsequence of partial sums is

s1 ¼ 1;

s2 ¼ 1þ 1

2;

s4 ¼ 1þ 1

2þ 1

3þ 1

4

� �

> 1þ 1

2þ 1

2;

s8 ¼ 1þ 1

2þ 1

3þ 1

4

� �

þ 1

5þ 1

6þ 1

7þ 1

8

� �

> 1þ 1

2þ 1

2þ 1

2;

..

.

s2k > 1þ 1

2þ 1

2þ � � � þ 1

2|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}

¼ 1þ 1

2k:

k terms

For any k> 1, 1k2< 1

ðk�1Þk.

Here we have telescopingcancellation.

The proof depends on use oftrigonometric series or ofComplex Analysis.

This series is often called theharmonic series, since itsterms are proportional to thelengths of strings that produceharmonic tones in music.

We think of n as being fairlylarge, in order to see thepattern.

Here each bracket adds up tomore than 1

2.

Hence the sequence {sn} is increasing and not bounded above. Therefore it

cannot be convergent, so that by the Monotone Convergence Theorem the

sequence {sn} is divergent. Hence the series itself is also divergent. &

We can extend this type of domination approach to test further series for

convergence, in the following way:

Theorem 1 Comparison Test

(a) If 0 � an � bn, for n ¼ 1, 2, . . ., andP

1

n¼1

bn is convergent, thenP

1

n¼1

an is

convergent.

(b) If 0 � bn � an, for n¼ 1, 2, . . ., andP

1

n¼1

bn is divergent, thenP

1

n¼1

an is

divergent.

Remark

As with the Squeeze Rule for non-negative null sequences, it is sufficient to

prove that the necessary inequalities in parts (a) and (b) hold eventually.

In applications, we use this result in the following way:

Strategy To test a seriesP

1

n¼1

an of non-negative terms for convergence

using the Comparison Test:

EITHER find a convergent seriesP

1

n¼1

bn of non-negative terms where the bn

dominate the an,

OR find a divergent seriesP

1

n¼1

bn of non-negative terms where the an

dominate the bn.


1

n¼1

1n3 is convergent.

Solution We shall use the Comparison Test.

Let bn ¼ 1n2. Then 0 � 1

n3 � bn, for n¼ 1, 2, . . ., and we know that the seriesP

1

n¼1

1n2 is convergent (this was Example 1 above). It follows from part (a) of the

Comparison Test that the seriesP

1

n¼1

1n3 is convergent. &


1

n¼1

1ffiffi

np is divergent.

Solution We shall use the Comparison Test.

Let bn ¼ 1n. Then 0 � bn � 1

ffiffi

np , for n¼ 1, 2, . . ., and we know that the series

P

1

n¼1

1n

is divergent (this was Example 2 above). It follows from part (b) of the

Comparison Test that the seriesP

1

n¼1

1ffiffi

np is divergent. &

Next, consider the seriesP

1

n¼1

1ffiffi

npþ3

ffiffi

n4pþ1

. Its terms are somewhat similar to

those of the seriesP

1

n¼1

1ffiffi

np , since we can express 1

ffiffi

npþ3

ffiffi

n4pþ1

in the form

In the proof of part (a), in Sub-section 3.2.2, we shall in factsee that

X

1

n¼1

an �X

1

n¼1

bn:

That is, it is sufficient that0� an� bn or 0� bn� an forall n>X, for some number X.

That is, where an� bn.

That is, where bn� an.

For 1n3 � 1

n2 :

For 1n� 1

ffiffi

np :

In other words, the termffiffiffi

np

dominates the term 3ffiffiffi

n4pþ 1.

94 3: Series

1ffiffi

np � 1

1þ3n�1

4þn�1

2

, where the second fraction tends to 1 as n!1. So, since the

seriesP

1

n¼1

1ffiffi

np is divergent, it seems likely that the series

P

1

n¼1

1ffiffi

npþ3ffiffi

n4pþ1

is also

divergent.

We can pin down the underlying idea here in the following useful result:

Theorem 2 Limit Comparison Test

Suppose thatP

1

n¼1

an andP

1

n¼1

bn have positive terms, and that

an

bn

! L as n!1; where L 6¼ 0:

(a) IfP

1

n¼1


1

n¼1

an is convergent.

(b) IfP

1

n¼1


1

n¼1

an is divergent.

So, take an ¼ 1ffiffi

npþ3ffiffi

n4pþ1

and bn ¼ 1ffiffi

np , for n¼ 1, 2, . . .. Then both an and bn

are positive, and

an

bn

¼ 1=ffiffiffi

npþ 3

ffiffiffi

n4pþ 1ð Þð Þ

1=ffiffiffi

np

ð Þ

¼ffiffiffi

np

ffiffiffi

npþ 3

ffiffiffi

n4pþ 1

¼ 1

1þ 3n�14 þ n�

12

! 1 as n!1:

It then follows from part (b) of the Limit Comparison Test that, since the

seriesP

1

n¼1

1ffiffi

np is divergent, the series

P

1

n¼1

1ffiffi

npþ3ffiffi

n4pþ1

is also divergent.

Example 5 Use the Limit Comparison Test to prove that the seriesP

1

n¼1

n2�3nþ45n4�n

is convergent.

Solution Set an ¼ n2�3nþ45n4�n

and bn ¼ 1n2, for n¼ 1, 2, . . .. Then both an and bn

are positive, and

an

bn

¼ n2 � 3nþ 4

5n4 � n� n2

1

¼ n4 � 3n3 þ 4n2

5n4 � n

¼ 1� 3n�1 þ 4n�2

5� n�3! 1

5as n!1:

:

Since 156¼ 0 and the series

P

1

n¼1

1n2 is known to be convergent, then the series

P

1

n¼1

n2�3nþ45n4�n

is also convergent. &

By Example 4.

In other words, bn behaves‘rather like an’ for large n.Note that it is IMPORTANT thatL is non-zero.

Since 1 6¼ 0.

By Example 4.

We make this choice for bn

since, for large n, the

expression n2�3nþ45n4�n

is ‘more or

less the same as’ n2

5n4 ¼ 15n2 –

where the factor ‘5’ will notaffect our argument.

Dividing numerator anddenominator by the dominantterm n4.

By Example 1.


Problem 2 Use the Comparison Test or the Limit Comparison Test to

determine the convergence of the following series:

(a)P

1

n¼1

1n3þn

; (b)P

1

n¼1

1nþffiffi

np ; (c)

P

1

n¼1

nþ42n3�nþ1

; (d)P

1

n¼1

cos2 2nð Þn3 :

Our next test is motivated in part by the geometric series – recall that the

geometric series aþ ar þ ar2 þ � � � ¼P

1

n¼0

arn, where a 6¼ 0, is convergent if

jrj< 1 but divergent if jrj �1. The reason that this converges, if jrj< 1, is that

the terms are then forced to tend to 0 so quickly that the partial sums sn increase

so slowly as n increases that they remain bounded above. On the other hand, if

jrj � 1, the terms do not tend to 0, so that the series must diverge.

Theorem 3 Ratio Test

Suppose thatP

1

n¼1

an has positive terms, and that anþ1

bn! ‘ as n!1.

(a) If 0� ‘< 1, thenP

1

n¼1

an is convergent.

(b) If ‘> 1, thenP

1

n¼1

an is divergent.

Remark

If ‘¼ 1, then the test gives us no information on whether the series converges.

For example, if an ¼ 1n, then anþ1

an¼ n

nþ1¼ 1

1þ1n

! 1 as n!1; we have seen

that the seriesP

1

n¼1

an ¼P

1

n¼1

1n

diverges. On the other hand, if an ¼ 1n2 then

anþ1

an¼ n2

nþ1ð Þ2 ¼1

1þ2nþ 1

n2

! 1 as n!1; but the seriesP

1

n¼1

an ¼P

1

n¼1

1n2 converges.

Example 6 Use the Ratio Test to determine the convergence of the following

series:

(a)P

1

n¼1

n2n; (b)

P

1

n¼1

10n

n! .

Solution

(a) Let an ¼ n2n, n¼ 1, 2, . . .. Then an is positive, and

anþ1

an

¼ nþ 1

2nþ1� 2n

n

¼1þ 1

n

2! 1

2as n!1:

Since 0 < 12< 1, it follows from the Ratio Test that the series

P

1

n¼1

an ¼P

1

n¼1

n2n is convergent.

(b) Let an ¼ 10n

n! , n¼ 1, 2, . . .. Then an is positive, and

anþ1

an

¼ 10nþ1

nþ 1ð Þ!�n!

10n

¼ 10

nþ 1! 0 as n!1:

By the Non-null Test.

Note that with the Ratio Test,we concentrate on the seriesP

1

n¼1

an itself and do not need to

‘think of’ some other seriesP

1

n¼1

bn.

Part (b) includes the caseanþ1

an!1.

That is, the Ratio Test isinconclusive if ‘¼ 1.

By Example 2.

By Example 1.

96 3: Series

It follows from the Ratio Test that the seriesP

1

n¼1

an ¼P

1

n¼1

10n

n! is

convergent. &

Problem 3 Use the Ratio Test to determine whether the following

series are convergent

(a)P

1

n¼1

n3

n!; (b)P

1

n¼1

n22n

n! ; (c)P

1

n¼1

2nð Þ!nn .

We have now seen examples of many different convergent and divergent

series, including examples of various generic types. We call these basic series,

since we shall use them commonly together with various other tests in order to

prove that particular series are themselves convergent or divergent.

Basic series The following series are convergent:

(a)P

1

n¼1

1np ; for p � 2;

(b)P

1

n¼1

cn; for 0 � c < 1;

(c)P

1

n¼1

npcn; for p > 0; 0 � c < 1;

(d)P

1

n¼1

cn

n!; for c � 0.

The following series is divergent:

(e)P

1

n¼1

1np ; for p � 1.

Another test for convergence

When we examined the convergence or divergence of the seriesP

1

n¼1

1n, we found

it convenient to group the terms into blocks that contained in turn 1, 2, 4, 8, . . .successive individual terms of the series. In fact, a similar technique applies to

a wide range of series, and the following result is thus often invaluable in order

to save repeating that same type of argument:

Theorem 4 Cauchy Condensation Test

Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2na2n , for

n¼ 1, 2, . . .

X

1

n¼1

an is convergent if and only ifX

1

n¼1

bn is convergent:

For example, consider the seriesP

1

n¼1

1n. Here, let an ¼ 1

n, so that

bn ¼ 2na2n ¼ 2n � 1

2n¼ 1:

Then the Condensation Test tells us that the seriesP

1

n¼1

1n

is convergent if and

only if the seriesP

1

n¼1

bn ¼P

1

n¼1

1 is convergent. This is divergent, by the Non-

null Test, so that the original seriesP

1

n¼1

1n

must itself have been divergent.

Instances of these that wehave seen areP

1

n¼1

1n2;

Geometric Series,P

1

n¼1

12

� �n;

P

1

n¼1

n2n ¼

P

1

n¼1

n 12

� �n;

P

1

n¼1

10n

n!;

P

1

n¼1

1ffiffi

np ¼

P

1

n¼1

1n1=2

.

You will see this techniqueused in the proof ofTheorem 4, in Sub-section 3.2.2.

For, if an ¼ 1n, then a2n ¼ 1

2n.

This use of the Condensation Test is much simpler than setting out to

perform a grouping and estimation exercise on such occasions! The

Condensation Test is also often useful when the individual terms of a series

include expressions such as loge n.

Example 7 Use the Condensation Test to determine the convergence of the

seriesP

1

n¼2

1

nffiffiffiffiffiffiffiffi

loge np .

Solution Let an ¼ 1


loge np , n¼ 2, . . .. Then an is positive; and, since

nffiffiffiffiffiffiffiffiffiffiffiffi

loge np� �

is an increasing sequence, anf g ¼n

1


loge np

o

is a decreasing

sequence.

Next, let bn ¼ 2na2n ; thus

bn ¼ 2n � 1

2nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

logeð2nÞp

¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n loge 2p ¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffi

loge 2p � 1

n12

:

SinceP

1

n¼2

1

n12

is a basic divergent series, it follows by the Multiple Rule that

P

1

n¼2

bn must be divergent. Hence by the Condensation Test, the original series

P

1

n¼2

1


loge np must also be divergent. &

Problem 4 Use the Condensation Test to determine the convergence

of the following series:

(a)P

1

n¼2

1n loge n

; (b)P

1

n¼2

1

n loge nð Þ2.

3.2.2 Proofs

In the previous sub-section we gave a number of tests for the convergence of

series with non-negative terms. We now supply the proofs of these tests.

Theorem 1 Comparison Test

(a) If 0� an� bn, for n¼ 1, 2, . . ., andP

1

n¼1


1

n¼1

an is

convergent.

(b) If 0� bn� an, for n¼ 1, 2, . . ., andP

1

n¼1


1

n¼1

an is

divergent.

Proof

(a) Consider the nth partial sums

sn ¼ a1 þ a2 þ � � � þ an; for n ¼ 1; 2; . . .;

Note that here we only sumfrom n¼ 2, since it makes no

sense to talk about 1

1ffiffiffiffiffiffiffiffi

loge 1p , as

loge 1 ¼ 0. Again we areusing the fact that theconvergence of a series is notaffected if we add or alter afinite of terms.


98 3: Series

and

tn ¼ b 1 þ b2 þ � � � þ bn ; for n ¼ 1; 2; . . .:

We know that

a1 � b1 ; a2 � b2 ; . . .; an � bn ;

and so

sn � tn ; for n ¼ 1; 2; . . .:

We also know thatP

1

n ¼ 1

bn is convergent , and so the incr easing seque nce

{tn} is conver gent, with limi t t, say. Hence

sn � tn � t ; for n ¼ 1; 2; . . .;

and so the increasing sequence {sn} i s b ou nd e d a b ov e b y t. By t he M ono tone

C o nv er g e nc e T h eo r e m , { sn} i s t he refore also c onv erg ent, and soP

1

n ¼ 1

an is a

con verg ent se ri es.

Note : Moreover , by the Limit Inequ ality Rule

X

1

n¼ 1

an �X

1

n ¼ 1

bn :

(b) We can deduce part (b) from part (a), as follows. Suppose thatP

1

n ¼ 1

an is

conver gent. Th en, by part (a),P

1

n ¼ 1

bn mus t also be conver gent. Howeve r,P

1

n ¼ 1

bn is assumed to be dive rgent, and soP

1

n¼ 1

an mus t also be dive rgent. &

The orem 2 Limit Compari son Test

Sup pose thatP

1

n ¼ 1

an andP

1

n¼ 1

bn have posi tive terms , and that

an

bn

! L as n !1; where L 6¼ 0:

(a) IfP

1

n¼ 1

bn is convergent , thenP

1

n ¼ 1

an is conver gent.

(b) IfP

1

n ¼ 1

bn is divergen t, thenP

1

n¼ 1

an is divergen t.

Proof

(a) Becaus e�

an

bn

�

is convergent , it mus t be bounded . Thus there is a const ant K

such that

an

bn

� K ; for n ¼ 1 ; 2 ; . . .;

and so

an � Kbn ; for n ¼ 1 ; 2; . . .:

By the Multiple Rule,P

1

n¼1

Kbn is convergent. Hence, by the Comparison

Test,P

1

n¼1

an is also convergent.

By adding up the previousinequalities.

You met this Rule inSub-section 2.3.3, Theorem 3.This additional inequality willsometimes be useful.

Such ‘proofs bycontradiction’ can sometimessave a great deal of detailedarguments.

See Sub-section 2.4.2,Theorem 1 (the BoundednessTheorem).


(b) We can deduce part (b) from part (a), as follows.

Suppose thatP

1

n¼1

an is convergent. Then, by part (a),P

1

n¼1

bn must also be

convergent, because

bn

an

! 1

Las n!1;

by the Quotient Rule for sequences.

However,P

1

n¼1

bn is assumed to be divergent, and soP

1

n¼1

an must also be

divergent. &

Theorem 3 Ratio Test

Suppose thatP

1

n¼1

an has positive terms, and that anþ1

an! ‘ as n!1.

(a) If 0� ‘< 1, thenP

1

n¼1

an is convergent.

(b) If ‘> 1, thenP

1

n¼1

an is divergent.

Proof

(a) Because ‘< 1, we can choose "> 0 so that

r ¼ ‘þ " < 1:

Hence there is a positive number X, which we may assume to be a positive

integer, such that

anþ1

an

� r; for all n � X:

Thus, for n�X, we have

an

aX

¼ an

an�1

� �

an�1

an�2

� �

. . .aXþ1

aX

� �

� rn�X;

since each of the n�X brackets is less than or equal to r. Hence

an � aXrn�X; for all n � X: (1)

Now

X

1

n¼X

aXrn�X ¼ aX þ aXr þ aXr2 þ � � �;

which is a geometric series with first term aX and common ratio r. Since

0< r< 1, this series is convergent, and so, by inequality (1) and the

Comparison Test,P

1

n¼X

an is also convergent. HenceP

1

n¼1

an is convergent.

(b) Since

anþ1

an

! ‘ as n!1;

and ‘> 1, there is a positive number X, which we may assume to be a

positive integer, such that

anþ1

an

� 1; for all n � X:

We are following a ‘proof bycontradiction’ approach here.

Remember that L 6¼ 0.

Remember that part (b)includes the case anþ1

an!1.

For example, take" ¼ 1

21� ‘ð Þ.

Recall that we occasionallymake the convenientassumption that X is an integerto avoid notationalcomplications.

1

X n

� + ε

� – ε�

�1

X n

100 3: Series

(This also holds if anþ1

an!1 as n!1.)

Thus, for n�X, we have

an

aX

¼ an

an�1

� �

an�1

an�2

� �

. . .aXþ1

aX

� �

� 1;

since each of the brackets is greater than or equal to 1. Hence

an � aX > 0; for all n � X;

and so {an} cannot be a null sequence. Thus, by the Non-null Test,P

1

n¼X

an is

divergent. HenceP

1

n¼1

an is also divergent. &

Theorem 4 Cauchy Condensation Test

Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2na2n for

n¼ 1, 2, . . .

X

1

n¼1

an is convergent if and only ifX

1

n¼1

bn is convergent:

Proof Denote by sn and tn the nth partial sums of the seriesP

1

n¼1

an andP

1

n¼1

bn,

respectively

sn ¼ a1 þ a2 þ � � � þ an and tn ¼ 21a2 þ 22a4 þ � � � þ 2na2n :

Since the terms an are decreasing and a1� 0, it follows that

s2n ¼ a1 þ a2ð Þ þ a3 þ a4ð Þ þ � � � þ a2n�1þ1 þ � � � þ a2nð Þ� 0þ a2ð Þ þ a4 þ a4ð Þ þ � � � þ a2n þ � � � þ a2nð Þ¼ 0þ a2ð Þ þ 2a4ð Þ þ � � � þ 2n�1a2n

� �

¼ 1

2tn: ð2Þ

We may also write s2n in another way

s2n ¼ a1 þ a2 þ a3ð Þ þ a4 þ a5 þ a6 þ a7ð Þ þ � � �þ a2n�1 þ � � � þ a2n�1ð Þ þ a2n

� a1 þ a2 þ a2ð Þþ a4 þ a4 þ a4 þ a4ð Þ þ � � � þ a2n�1 þ � � � þ a2n�1ð Þ þ a2n

¼ a1 þ 2a2ð Þ þ 4a4ð Þ þ � � � þ 2n�1a2n�1

� �

þ a2n

� a1 þ tn: ð3Þ

Now suppose thatP

1

n¼1

an is convergent. Then, by the Boundedness Theorem for

series, its sequence of partial sums {sn} must be bounded above and so the sequence

s2nf g must also be bounded above – since it is a subsequence of {sn}. It therefore

follows from inequality (2) that the sequence 12

tn

� �

must be bounded above, so the

sequence {tn} must also be bounded above. Another application of the Boundedness

Theorem shows that therefore the seriesP

1

n¼1

bn must be convergent.

The kth bracket contains 2k�1

terms.

The kth bracket contains 2k�1

occurrences of a2k.

The kth bracket contains 2k

terms.

The kth bracket contains 2k

occurrences of a2k.

In obtaining (3), we havereplaced the term a2n in the pre-vious line by a larger term 2na2n.

The Boundedness Theoremfor series appeared inSection 3.2, above.


Next, suppose thatP

1

n¼1

an is divergent. Then, by the Boundedness Theorem

for series, its sequence of partial sums {sn} must be unbounded above, and so

the sequence s2nf gmust also be unbounded above – since it is a subsequence of

{sn}. It therefore follows from inequality (3) that the sequence {a1þ tn} must

be unbounded above, so the sequence {tn} must also be unbounded above.

Another application of the Boundedness Theorem shows that therefore the

seriesP

1

n¼1

bn must be divergent. &

We end this section by proving the convergence/divergence of the basic

series given earlier.

Basic series The following series are convergent:

(a)P

1

n¼1

1np; for p � 2;

(b)P

1

n¼1

cn; for 0 � c < 1;

(c)P

1

n¼1

npcn; for p > 0; 0 � c < 1;

(d)P

1

n¼1

cn

n!; for c � 0.

The following series is divergent:

(e)P

1

n¼1

1np; for p � 1.

Proof

(a) This series is convergent, by the Comparison Test, since, if p� 2, then

1

np� 1

n2; for n ¼ 1; 2; . . .;

and the seriesP

1

n¼1

1n2 is convergent.

(b) The seriesP

1

n¼1

cn is a geometric series with common ratio c, and so it

converges if 0� c< 1.

(c) Let an¼ npcn, for n¼ 1, 2, . . .. Then, for c 6¼ 0

anþ1

an

¼ nþ 1ð Þpcnþ1

npcn¼ 1þ 1

n

� �p

c:

Thus, if k is any integer greater than or equal to p, then

c � anþ1

an

� 1þ 1

n

� �k

c:

Now, by the Product Rule for sequences

1þ 1

n

� �k

! 1 as n!1;


In Chapter 7 we shall prove

thatP

1

n¼1

1np is convergent, for

all p> 1.

Sub-section 3.2.1, Example 1.

Sub-section 3.1.2.

If c¼ 0, the series is clearlyconvergent.

We introduce the integer khere, so that we can use theProduct Rule for sequences.

102 3: Series

and so, by the Squeeze Rule for sequencesanþ1

an

! c as n!1:

Since 0� c< 1, we deduce, from the Ratio Test, thatP

1

n¼1

npcn is

convergent.

(d) Let an ¼ cn

n!, for n¼ 1, 2, . . .. Then, for c 6¼ 0,

anþ1

an

¼ cnþ1

nþ 1ð Þ!

cn

n!¼ c

nþ 1:

Thus anþ1

an! 0 as n!1, and we deduce from the Ratio Test that

P

1

n¼1

cn

n! is

convergent.

(e) We saw earlier that the seriesP

1

n¼1

1n

is divergent. It follows that the seriesP

1

n¼1

1np is also divergent, by the Comparison Test, since if p� 1 then 1

np � 1n,

for n¼ 1, 2, . . .. &

3.3 Series with positive and negative terms

The study of seriesP

1

n¼1

an, with an� 0 for all values of n, is relatively straight-

forward, because the sequence of partial sums {sn} is increasing. Similarly, if

an� 0 for all values of n, then {sn} is decreasing.

However, it is harder to determine the behaviour of a series with both

positive and negative terms, because {sn} is neither increasing nor decreasing.

However, if the sequence {an} contains only finitely many negative terms, then

the sequence {sn} is eventually increasing, and we can apply the methods of

Section 3.2. Similarly, if {an} contains only finitely many positive terms, then

the sequence {sn} is eventually decreasing, and we can again apply the

methods of Section 3.2, after making a sign change.

In this section we look at series such as

1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � and 1þ 1

22� 1

32þ 1

42þ 1

52� 1

62þ � � �;

which contain infinitely many terms of either sign. We give several methods

which can often be used to prove that such series are convergent.

3.3.1 Absolute convergence

Suppose that we want to determine the behaviour of the infinite series

X

1

n¼1

�1ð Þnþ1

n2¼ 1� 1

22þ 1

32� 1

42þ 1

52� 1

62þ � � �: (1)

We know that the series

X

1

n¼1

1

n2¼ 1þ 1

22þ 1

32þ 1

42þ 1

52þ 1

62þ � � � (2)

is convergent. Does this mean that the series (1) is also convergent? In fact it

does, as we now prove.

If c¼ 0, the series is clearlyconvergent.

In Sub-section 3.2.1,Example 2.

For example, the convergenceof

1þ 2þ 3� 1

42� 1

52� 1

62� � � �

follows from that ofP

1

n¼1

1n2 :

The series (2) is a basicconvergent series.

Consider the two related series

1þ 0þ 1

32þ 0þ 1

52þ 0þ � � �

and

0þ 1

22þ 0þ 1

42þ 0þ 1

62þ � � � :

Each of these series is dominated by the series (2), and so they are both

convergent, by the Comparison Test. Applying the Multiple Rule with l¼�1,

and the Sum Rule, we deduce that the series

1� 1

22þ 1

32� 1

42þ 1

52� 1

62þ � � � is convergent:

The argument just given is the basis for a concept called absolute conver-

gence, which we now define.

Definitions

A seriesP

1

n¼1

an is absolutely convergent ifP

1

n¼1

anj j is convergent.

A seriesP

1

n¼1

an that is convergent but not absolutely convergent is condi-

tionally convergent.

For example, the series (1) is absolutely convergent, because the seriesP

1

n¼1

1n2

is convergent.

However, the series

X

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � �

is not absolutely convergent, because the seriesP

1

n¼1

1n

is divergent; the series is,

in fact, conditionally convergent.

Theorem 1 Absolute Convergence Test

IfP

1

n¼1

an is absolutely convergent, thenP

1

n¼1

an is convergent.

It follows that the series (1) is convergent (as we have already seen), and also

that the series

1þ 1

22� 1

32þ 1

42þ 1

52� 1

62þ � � �

is convergent, because the seriesP

1

n¼1

1n2 is convergent. Indeed, no matter how

we distribute the plus and minus signs among the terms of 1n2

� �

, the resulting

series is convergent.

However, the Absolute Convergence Test tells us nothing about the

behaviour of the series

1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � (3)

and

1þ 1

2� 1

3þ 1

4þ 1

5� 1

6þ 1

7þ 1

8� 1

9þ � � � : (4)

If the terms an are all non-negative, then absoluteconvergence and convergencehave the same meaning.

We shall examine thebehaviour of conditionallyconvergent series later, inSub-sections 3.3.2 and 3.3.3.

You saw this in Example 2,Sub-section 3.2.1.

The proofs of the results inthis sub-section appear inSub-section 3.3.6.

In fact, series (3) is convergent(by the Alternating Test,which we introduce in Sub-section 3.3.2), and series (4) isdivergent (see Exercise 2(a)for Section 3.3).

104 3: Series

The seriesP

1

n¼1

1n

is divergent, and so the two series (3) and (4) are not

absolutely convergent.

Example 1 Prove that the following series are convergent:

(a)P

1

n¼1

�1ð Þnþ1

n3 ; (b) 1þ 12� 1

4þ 1

8þ 1

16� 1

32� � �.

Solution

(a) Let an¼ �1ð Þnþ1

n3 , for n¼ 1, 2, . . .; then anj j ¼ 1n3. We know that

P

1

n¼1

1n3 is

convergent, so it follows thatP

1

n¼1

�1ð Þnþ1

n3 is absolutely convergent. Hence,

by the Absolute Convergence Test,P

1

n¼1

�1ð Þnþ1

n3 is convergent.

(b) The seriesP

1

n¼0

12n is a convergent geometric series, so that the series

1þ 1

2� 1

4þ 1

8þ 1

16� 1

32� � �

is absolutely convergent. Then, by the Absolute Convergence Test, this

series is also convergent. &

Problem 1 Prove that the following series are convergent:

(a)P

1

n¼1

ð�1Þnþ1n

n3þ1; (b)

P

1

n¼1

cos n2n .

The Absolute Convergence Test states that, ifP

anj j is convergent, thenP

an is also convergent, but it does not indicate any explicit connection

between the sums of these two series. Clearly,P

an is less thanP

anj j if any

of the terms an are negative.

For example,P

1

n¼1

12n ¼ 1

2þ 1

4þ 1

8þ � � � ¼ 1, whereas

P

1

n¼1

�1ð Þnþ1

2n ¼12� 1

4þ 1

8� � � � ¼ 1

3:

The following result relates the values ofP

an andP

anj j:

Triangle Inequality (infinite form)

IfP

1

n¼1


1

n¼1

an

�

�

�

�

�

�

�

�

�P

1

n¼1

anj j.

Problem 2 Show that the series 12þ 1

4� 1

8þ 1

16þ 1

32� 1

64þ � � � is con-

vergent, and that its sum lies in [�1,1]. (You do NOT need to find the sum

of the series.)

3.3.2 The Alternating Test

Suppose that we want to determine the behaviour of the following infinite

series, in which the terms have alternating signsX

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � �:

ForP

1

n¼1

1n3 is an example of a

basic convergent series.

Recall that you met theTriangle Inequality fora1; a2; . . .; an

X

n

k¼1

ak

�

�

�

�

�

�

�

�

�

�

�X

n

k¼1

akj j;

in Sub-section 1.3.1.


The Absolute Convergence Test does not help us with this series, becauseP

1

n¼1

1n

is divergent.

In fact, the seriesP

1

n¼1

�1ð Þnþ1

nis convergent. To see why, we first calculate

some of its partial sums and plot them on a sequence diagram

s1 ¼ 1;

s2 ¼ 1� 1

2¼ 0:5;

s3 ¼ 1� 1

2þ 1

3¼ 0:83;

s4 ¼ 1� 1

2þ 1

3� 1

4¼ 0:583;

s5 ¼ 1� 1

2þ 1

3� 1

4þ 1

5¼ 0:783;

s6 ¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6¼ 0:616:

The sequence diagram for {sn} suggests that

s1 � s3 � s5 � . . . � s2k�1 � . . .

and

s2 � s4 � s6 � . . . � s2k � . . . ;

for all k. In other words:

the odd subsequence {s2k�1} is decreasing

and:

the even subsequence {s2k} is increasing.

Also, the terms of {s2k�1} all exceed the terms of {s2k}, and both subsequences

appear to converge to a common limit s, which lies between the odd and even

partial sums.

To prove this, we write the even partial sums {s2k} as follows

s2k ¼ 1� 1

2

� �

þ 1

3� 1

4

� �

þ � � � þ 1

2k � 1� 1

2k

� �

:

All the brackets are positive, and so the subsequence {s2k} is increasing.

We can also write

s2k ¼ 1� 1

2� 1

3

� �

� 1

4� 1

5

� �

� � � � � 1

2k � 2� 1

2k � 1

� �

� 1

2k:

Again, all the brackets are positive, and so {s2k} is bounded above, by 1.

Hence {s2k} is convergent, by the Monotone Convergence Theorem.

Let

limk!1

s2k ¼ s:

Since

s2k ¼ s2k�1 �1

2k

and 12k

� �

is null, we have

You met subsequencesearlier, in Sub-section 2.4.4.

106 3: Series

limk!1

s2k�1 ¼ limk!1

s2k þ1

2k

� �

¼ limk!1

s2k þ limk!1

1

2k

� �

¼ sþ 0 ¼ s;

by the Combination Rules for sequences. Thus the odd and even subsequences

of {sn} both tend to the same limit s, and so {sn} itself tends to s. Hence

X

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � is convergent, with sum s:

The same method can be used to prove the following general result:

Theorem 2 Alternating Test

If

an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;

where {bn} is a decreasing null sequence with positive terms, then

X

1

n¼1

an ¼ b1 � b2 þ b3 � b4 þ � � � is convergent:

When we apply the Alternating Test, there are a number of conditions which

must be checked. We now describe these in the form of a strategy.

Strategy To prove thatP

1

n¼1

an is convergent, using the Alternating Test,

check that

an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;

where:

1. bn� 0, for n¼ 1, 2, . . .;

2. {bn} is a null sequence;

3. {bn} is decreasing.

Remark

It is often easiest to check that {bn} is decreasing by verifying that�

1bn

�

is

increasing.

Here are some examples.

Example 2 Determine which of the following series are convergent:

(a)P

1

n¼1

�1ð Þnþ1

ffiffi

np ; (b)

P

1

n¼1

�1ð Þnþ1

n4 ; (c)P

1

n¼1

�1ð Þnþ1.

Solution

(a) The sequence�1ð Þnþ1

ffiffi

np

n o

has terms of the form an ¼ �1ð Þnþ1bn, where

bn ¼1ffiffiffi

np ; n ¼ 1; 2; . . .:

By Theorem 6 ofSub-section 2.4.4.

In fact, s¼ loge 2 ’ 0.69.

This test is sometimes calledthe Leibniz Test.

To show that {bn} is null, usethe techniques introduced inSub-section 2.2.1.


Now:

1. bn ¼ 1ffiffi

np � 0, for n¼ 1, 2, . . .;

2. bnf g ¼�

1ffiffi

np�

is a basic null sequence;

3. bnf g ¼�

1ffiffi

np�

is decreasing, because�

1bn

�

¼ffiffiffi

npf g is increasing.

Hence, by the Alternating Test,P

1

n¼1

�1ð Þnþ1

ffiffi

np is convergent.

(b) The sequence�1ð Þnþ1

n4

n o

has terms of the form an ¼ �1ð Þnþ1bn, where

bn ¼1

n4; n ¼ 1; 2; . . .:

Now:

1. bn ¼ 1n4 � 0, for n¼ 1, 2, . . .;

2. bnf g ¼ 1n4

� �


3. bnf g ¼ 1n4

� �

is decreasing, because�

1bn

�

¼�

n4�

is increasing.


1

n¼1

�1ð Þnþ1

n4 is convergent.

(c) The sequence�

�1ð Þnþ1�

is not a null sequence. Hence, by the Non-null

TestX

1

n¼1

�1ð Þnþ1is divergent. &

Problem 3 Determine which of the following series are convergent:

(a)P

1

n¼1

�1ð Þnþ1

n3 ; (b)P

1

n¼1

�1ð Þnþ1 nnþ2

; (c)P

1

n¼1

�1ð Þnþ1

n1=3 þ n1=2.

3.3.3 Rearrangement of a series

In the last sub-section we saw that the series 1� 12þ 1

3� 1

4þ 1

5� 1

6þ � � � is

convergent. Let us denote by ‘ the sum of this series.

If we temporarily ignore the need for careful mathematical proof and simply

‘move terms about’, look what we get

‘ ¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � (5)

¼ 1� 1

2� 1

4þ 1

3� 1

6� 1

8þ 1

5� 1

10� 1

12þ 1

7� 1

14� 1

16þ � � � ð6Þ

¼ 1� 1

2

� �

� 1

4þ 1

3� 1

6

� �

� 1

8þ 1

5� 1

10

� �

� 1

12

þ 1

7� 1

14

� �

� 1

16þ � � �

Alternatively, we can showthat this series is convergentby the Absolute ConvergenceTest.

Notice that the fact that thenth term includes a factor(�1)nþ1 does not imply at allthat a series is convergent!

In fact this series isconditionally convergent,since the harmonic series1þ 1

2þ 1

3þ 1

4þ 1

5þ 1

6þ � � � is

divergent.

This is the definition of ‘.

The pattern in (6) is to followthe next available positiveterm by the two next availablenegative terms.

Here we insert some brackets.

108 3: Series

¼ 1

2� 1

4þ 1

6� 1

8þ 1

10� 1

12þ 1

14� 1

16þ � � �

¼ 1

21� 1

2þ 1

3� 1

4þ 1

5� 1

6þ 1

8� � � �

� �

¼ 1

2‘:

It follows from the fact that ‘ ¼ 12‘ that ‘ ¼ 0. However this is impossible,

since the partial sum s2 of the original series is 12, and the even partial sums s2k

are increasing.

What has gone wrong? We assumed that operations that are valid for sums of

a finite number of terms also hold for sums of an infinite number of terms – we

rearranged (infinitely many of) the terms in the series (5) to obtain the series

(6) – without any justification. This rearrangement operation is not valid in

general!

We now give a precise definition of what we mean by a rearrangement.

Loosely speaking, the seriesP

1

n¼1

bn ¼ b1 þ b2 þ � � � is a rearrangement of the

seriesP

1

n¼1

an ¼ a1 þ a2 þ � � � if precisely the same terms appear in the

sequences {bn}¼ {b1, b2, . . .} and {an}¼ {a1, a2, . . .}, though they may occur

in a different order.

Definition A seriesP

1

n¼1

bn ¼ b1 þ b2 þ � � � is a rearrangement of the seriesP

1

n¼1

an ¼ a1 þ a2 þ � � � if there is a bijection ƒ such that

f : N ! N

bn 7! af ðnÞ:

Example 2 Assuming that the sum of the seriesP

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3� 1

4þ

15� 1

6þ � � � is loge 2, prove that the series

P

1

n¼1

an¼ 1þ 13� 1

2þ 1

5þ 1

7� 1

4þ 1

9þ

111� 1

6þ � � � converges to 3

2loge 2.

Solution Let sn and tn denote the nth partial sums of the series

1þ 13� 1

2þ 1

5þ 1

7� 1

4þ 1

9þ 1

11� 1

6þ � � � and 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � �,

respectively.

We now introduce an extra piece of notation, Hn; we denote by Hn the nth

partial sumP

n

k¼1

1k¼ 1þ 1

2þ � � � þ 1

nof the harmonic series.

Now, the terms of the seriesP

1

n¼1

an come ‘in a natural way’ in threes, so it

seems sensible to look at the partial sums s3n

s3n ¼ 1þ 1

3� 1

2þ 1

5þ 1

7� 1

4þ 1

9þ 1

11� 1

6þ � � �

þ 1

4n� 3þ 1

4n� 1� 1

2n

Here we insert the value ofeach bracket.

Here we pull out a commonfactor of 1

2.

Recall that, since the seriesconverges to ‘, then s2k

converges to ‘ also.

In fact the series (6) doesconverge; we ask you tosupply a careful proof of thisin Exercise 2 on Section 3.3.

Recall that a bijection is aone–one onto mapping.

This will enable us to tacklethings much more easily!

From the definition of s3n.


¼ 1þ 1

3� 1

2

� �

þ 1

5þ 1

7� 1

4

� �

þ 1

9þ 1

11� 1

6

� �

þ � � �

þ 1

4n� 3þ 1

4n� 1� 1

2n

� �

¼ 1þ 1

3þ 1

5þ � � � þ 1

4n� 1

� �

� 1

2þ 1

4þ 1

6þ � � � þ 1

2n

� �

¼ 1þ 1

2þ 1

3þ � � � þ 1

4n

� �

� 1

2þ 1

4þ 1

6þ � � � þ 1

4n

� ��

� 1

2Hn

¼ 1þ 1

2þ 1

3þ � � � þ 1

4n

� �

� 1

21þ 1

2þ 1

3þ � � � þ 1

2n

� ��

� 1

2Hn

¼ H4n �1

2H2n �

1

2Hn: (7)

Furthermore

t2n ¼ 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � þ 1

2n� 1� 1

2n

¼ 1þ 1

2þ 1

3þ � � � þ 1

2n

� �

� 21

2þ 1

4þ 1

6þ � � � þ 1

2n

� �

¼ H2n � 1þ 1

2þ 1

3þ � � � þ 1

n

� �

¼ H2n � Hn: (8)

We now eliminate the H’s from equations (7) and (8) as follows

s3n¼ H4n �1

2H2n �

1

2Hn

¼ H4n� H2nð Þ þ 1

2H2n� Hnð Þ

¼ t4n þ1

2t2n: (9)

But it follows from the hypotheses of the example that tn! loge 2, and so

t2n! loge 2 and t4n! loge 2, as n!1. Hence, letting n!1 in equation (9),

we see that

s3n! loge 2þ 1

2loge 2

¼ 3

2loge 2:

Next

s3n�1¼ s3n þ1

2n

! 3

2loge 2

and

s3n�2¼ s3n þ1

2n� 1

4n� 1

! 3

2loge 2 as n!1:

It follows from the above three results that sn! 32

loge 2 as n!1: &

We insert some brackets in afinite sum, for convenience.

We separate out the positiveand negative terms.

We add some terms to thecontents of the first bracket,then subtract them again; thelast bracket is simply 1

2Hn.

By the definition of t2n.

We insert some positive terms,then remove them again.

By the definition of H2n.

By the definition of Hn.

110 3: Series

Problem 4 Prove that the seriesP

1

n¼1

an ¼ 1� 12� 1

4þ 1

3� 1

6� 1

8þ 1

5�

110� 1

12þ 1

7� 1

14� 1

16þ � � � converges to 1

2loge 2. [You may assume that

the sum of the seriesP

1

n¼1

�1ð Þnþ1

n¼1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � is loge 2.]

In order to better understand the behaviour of series with positive and

negative terms, we introduce some notation. For a conditionally convergent

series

X

1

n¼1

an ¼ a1 þ a2 þ � � �;

we define the quantities aþn and a�n as follows

aþn ¼an; if an � 0,

0; if an < 0,

�

and a�n ¼0; if an � 0,

�an; if an < 0.

�

In other words, the sequence aþn� �

picks out the non-negative terms and the

sequence a�n� �

picks out the non-positive terms (and discards their sign).

In particular, an¼ aþn � a�n , and aþn � 0 and a�n � 0.

Next, since the seriesP

1

n¼1

an is convergent, an! 0 as n!1 . It follows that

aþn ¼1

2anj j þ anð Þ ! 0 as n!1

and

a�n ¼1

2anj j � anð Þ ! 0 as n!1:

Now, sinceP

1

n¼1

an is conditionally convergent, it must contain infinitely

many negative terms, since otherwiseP

1

n¼1

an andP

1

n¼1

anj j would be the same

apart from a finite number of terms. Since altering a finite number of terms

does not affect the convergence of a series, it would then follow that the seriesP

1

n¼1

anj j would be convergent – which we know is not the case.

Similarly, there must be infinitely many positive terms inP

1

n¼1

an.

Finally, we show that, if the seriesP

1

n¼1

an is conditionally convergent, then

the corresponding seriesP

1

n¼1

aþn (the ‘positive part’ ofP

1

n¼1

an) andP

1

n¼1

a�n (the

‘negative part’ ofP

1

n¼1

an) are both divergent.

For, suppose thatP

1

n¼1

aþn were convergent. Then, since an¼ aþn � a�n , we

have a�n ¼ aþn � an; it follows that the seriesP

1

n¼1

a�n must also be convergent.

And then, since anj j ¼ aþn þ a�n , it would also follow thatP

1

n¼1

anj j would be

convergent – which we know is not the case.

A similar argument shows thatP

1

n¼1

a�n cannot be convergent.

In particular, it follows from the divergence of the seriesP

1

n¼1

aþn andP

1

n¼1

a�n ,

whose terms are non-negative, that their partial sums must tend to 1 as

n!1.

For example, for the series

X

1

n¼1

an¼1� 1

2þ 1

3� 1

4

þ 1

5� 1

6þ � � �;

we have

X

1

n¼1

aþn ¼ 1� 0þ 1

3� 0

þ 1

5� 0þ � � �

and

X

1

n¼1

a�n ¼ 0þ 1

2þ 0þ 1

4

þ 0þ 1

6� � �:

For the seriesP

1

n¼1

an is

assumed to be onlyconditionally convergent.

For the seriesP

1

n¼1

an is

assumed to be onlyconditionally convergent.

By the Boundedness Theoremfor series.

In general it is not true that any rearrangement of a conditionally convergent

series converges. The situation is much more interesting!

Theorem 3 Riemann’s Rearrangement Theorem

Let the seriesP

1

n¼1

an ¼ a1 þ a2 þ � � � be conditionally convergent, with nth

partial sum sn. Then there are rearrangements of the series that have the

following properties:

1. For any given number s, the rearranged series converges to s;

2. For any given numbers x and y, one subsequence of the sn converges to x

and another subsequence of the sn converges to y;

3. The sequence sn converges to1.

4. One subsequence of the sn converges to1 and another subsequence of

the sn converges to �1.

We will not prove Theorem 3. Instead, we illustrate case (1) using the facts that

we have already discovered about conditionally convergent series. Using

similar ideas, we can prove the various parts of the theorem.

Example 3 Find a rearrangement of the seriesP

1

n¼1

�1ð Þnþ1

n¼1� 1

2þ 1

3� 1

4þ

15� 1

6þ � � � that converges to the sum 3.

Solution Since the ‘positive’ part of the seriesP

1

n¼1

�1ð Þnþ1

nhas partial sums that

tend to1, we start to construct the desired rearranged series as follows. Take

enough of the ‘positive’ terms 1, 13

, 15

, . . ., 12N1�1

so that the sum

1þ 13þ 1

5þ � � � þ 1

2N1�1is greater than 3, choosing N1 so that it is the first

integer such that this sum is greater than 3. Then these N1 terms will form

the first N1 terms in our desired rearranged series.

Next, take enough of the ‘negative’ terms 12

, 14

, 16

, . . ., 12N2

so that the

expression

1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1

� �

� 1

2þ 1

4þ 1

6þ � � � þ 1

2N2

� �

is less than 3, choosing N2 so that it is the first integer such that this sum is less

than 3. These N1þN2 terms form the first N1þN2 terms in our desired rear-

ranged series.

Now add in some more ‘positive’ terms 12N1þ1

, 12N1þ3

, . . ., 12N3�1

so that

the sum

1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1

� �

� 1

2þ 1

4þ 1

6þ � � � þ 1

2N2

� �

þ 1

2N1 þ 1þ 1

2N1 þ 3þ � � � þ 1

2N3 � 1

� �

is greater than 3, choosing N3 so that it is the first available integer such that

this sum is greater than 3. Then these N2þN3 terms will form the first N2þN3

terms in our desired rearranged series.

Some rearrangementsconverge, some diverge.

This is not a complete list ofall the possibilities that canoccur!

Recall that the sum of thisseries is loge 2 orapproximately 0.69.

This is possible since

1þ 1

3þ 1

5þ � � � ! 1;

as n!1.

112 3: Series

We then add in just enough ‘negative’ terms to make the next sum of

N3þN4 terms less than 3, and so on indefinitely.

In each set of two steps in this process we must use at least one of the

‘positive’ terms and one of the ‘negative’ terms, so that eventually all the

‘positive’ terms and all the ‘negative’ terms of the original series will be taken

exactly once in the new series, which we denote byP

1

n¼1

bn. So certainly the

seriesP

1

n¼1

bn is a rearrangement of the original seriesP

1

n¼1

�1ð Þnþ1

n.

But how do we know that the sum of the rearranged series is 3? As we keep

making the partial sums ofP

1

n¼1

bn swing back and forth from one side of 3 to the

other, we need the ‘radius’ of the swings to tend to 0. We achieve this by

always changing the sign of the terms fromP

1

n¼1

�1ð Þnþ1

nthat we choose next, once

the partial sum ofP

1

n¼1

bn has crossed beyond the value 3. For this ensures that

(from 12N2

onwards) the difference between a partial sum and 3 will always be

less than the absolute value of the last term 12Nk�1

or 12Nk

used, and we know that

this last term tends to 0 as we go further out along the original series.

It follows that the rearranged seriesP

1

n¼1

bn converges to 3, as required. &

Problem 5 Find a rearrangement of the seriesP

1

n¼1

�1ð Þnþ1

n¼1� 1

2þ 1

3�

14þ 1

5� 1

6þ � � � whose partial sums tend to1.

The situation is very much simpler, though, in the case of absolutely con-

vergent series.

Theorem 4 Let the seriesP

1

n¼1

an ¼ a1 þ a2 þ � � � be absolutely convergent.

Then any rearrangementP

1

n¼1

bn of the series also converges absolutely, and

P

1

n¼1

bn ¼P

1

n¼1

an.

For example, the series

1� 1

22þ 1

32� 1

42þ 1

52� 1

62þ 1

72� 1

82þ 1

92� � � �

is absolutely convergent. It follows from Theorem 4 that the following series is

also absolutely convergent – and to the same sum

1|{z}

� 1

22� 1

42|fflfflfflfflfflffl{zfflfflfflfflfflffl}

þ 1

32þ 1

52þ 1

72|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

� 1

62� 1

82� 1

102� 1

122|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

þ � � �

1 term 2 terms 3 terms 4 terms

:

3.3.4 Multiplication of series

We now look at the question of how we might multiply together two infinite

seriesP

1

n¼0

an andP

1

n¼0

bn to obtain a single seriesP

1

n¼0

cn with the property that

At each step there are always‘positive’ terms or ‘negative’terms left to choose, sincethere are infinitely many ofeach.

For clearly 1Nk� 1

k, since we

use at least 1 term in theoriginal series at each step ofthe rearrangement process.

For the seriesP

1

n¼1

�1ð Þnþ1

n2

�

�

�

�

�

�¼P

1

n¼1

1n2 , which is

a basic convergent series.

In this sub-section we sum ourseries from n¼ 0 rather thanfrom n¼ 1, simply because inmany applications ofTheorem 5 (below) this is theversion that we shall require.


X

1

n¼0

cn ¼X

1

n¼0

an

!

�X

1

n¼0

bn

!

:

Now, a first thought might be that cn¼ anbn, so thatP

1

n¼0

anbn ¼P

1

n¼0

an

� �

�P

1

n¼0

bn

� �

: However this is not the case!

For example, let an ¼ 12

� �nand bn ¼ 1

3

� �n. Then

X

1

n¼0

an ¼X

1

n¼0

1

2

� �n

¼ 2 andX

1

n¼0

bn ¼X

1

n¼0

1

3

� �n

¼ 3

2;

butX

1

n¼0

anbn ¼X

1

n¼0

1

6

� �n

¼ 6

5:

We can get a clue to what the correct formula for cn might be by, tempora-

rily, abandoning our rigorous approach and simply ‘pushing symbols about’!

If we multiply out both brackets and collect terms in a convenient way, we get

X

1

n¼0

an

!

�X

1

n¼0

bn

!

¼ a0þa1þa2þa3þ�� ð Þ� b0þb1þb2þb3þ��ð Þ

¼ a0 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ a1 b0 þ b1 þ b2 þ b3 þ � � �ð Þþ a2 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ a3 b0 þ b1 þ b2 þ b3 þ � � �ð Þ þ � � �

¼ a0b0 þ a0b1 þ a0b2 þ a0b3 þ � � � þ a1b0 þ a1b1 þ a1b2 þ a1b3 þ � � �þ a2b0 þ a2b1 þ a2b2 þ a2b3 þ � � � þ a3b0 þ a3b1 þ a3b2 þ a3b3 þ � � �

¼ a0b0 þ a0b1 þ a1b0ð Þ þ a0b2 þ a1b1 þ a2b0ð Þþ a0b3 þ a1b2 þ a2b1 þ a3b0ð Þ þ � � �:

Notice that every term that we would expect to be in such a product appears

exactly once in this final expression. We have grouped the terms in the final

expression in a particularly convenient way, so that we can see a pattern

emerging from this non-rigorous argument.

We can formalise this discussion as follows:

Theorem 5 Product Rule

Let the seriesP

1

n¼0

an andP

1

n¼0

bn be absolutely convergent, and let

cn ¼ a0bn þ a1bn�1 þ a2bn�2 þ � � � þ an b0 ¼X

n

k¼0

akbn�k:

Then the seriesP

1

n¼0

cn is absolutely convergent, and

X

1

n¼0

cn ¼X

1

n¼0

an

!

�X

1

n¼0

bn

!

:

Let us now return to the two seriesP

1

n¼0

an ¼P

1

n¼0

12

� �nand

P

1

n¼0

bn ¼P

1

n¼0

13

� �nthat

we considered earlier, whose sums were 2 and 32, respectively. With the for-

mula in Theorem 5 for cn, we have

These are both geometricseries.

Note that 656¼ 2� 3

2!

The pattern here is that in thekth bracket, the sum of thesubscripts of each term isprecisely k.

114 3: Series

cn ¼X

n

k¼0

akbn�k ¼X

n

k¼0

1

2k� 1

3n�k

¼ 1

3n

X

n

k¼0

3

2

� �k

¼ 1

3n�

1� 32

� �nþ1

1� 32

¼ 1

3n� �2ð Þ � 1� 3

2

� �nþ1 !

¼ 3

2n� 2

3n:

It follows that

X

1

n¼0

cn ¼ 3�X

1

n¼0

1

2n� 2�

X

1

n¼0

1

3n

¼ 3� 2� 2� 3

2

� �

¼ 6� 3 ¼ 3:

Since 2� 32¼ 3; we see that the sum of the series

P

1

n¼0

cn is what Theorem 5

predicts it should be.

Remark

Notice that the theorem requires that the seriesP

1

n¼0

an andP

1

n¼0

bn are absolutely

convergent. Without this assumption, the result may be false.

For example, let an ¼ bn ¼ �1ð Þnþ1

ffiffiffiffiffiffi

nþ1p . Then both

P

1

n¼0

an andP

1

n¼0

bn are conver-

gent, by the Alternating Test – but they are not absolutely convergent. Now, by

the formula for cn, we have

cn ¼X

n

k¼0

akbn�k ¼X

n

k¼0

�1ð Þkþ1


k þ 1p � �1ð Þn�kþ1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n� k þ 1p

¼ �1ð ÞnX

n

k¼0

1ffiffiffiffiffiffiffiffiffiffiffi

k þ 1p

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n� k þ 1p ;

so that

cnj j ¼X

n

k¼0


k þ 1p

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n� k þ 1p

�X

n

k¼0


nþ 1p

�ffiffiffiffiffiffiffiffiffiffiffi

nþ 1p ¼ 1:

Since we therefore do not have cn! 0 as n!1, it follows that the seriesP

1

n¼0

cn

is divergent.

We shall use Theorem 5 in Section 3.4 to give a definition of the exponential

function x 7! ex in terms of series and to examine its properties, and in

Section 8.4 on power series.

Yet again, this demonstratesthat the hypotheses of ourtheorems really do matter!


3.3.5 Overall strategy for testing for convergence

We now give an overall strategy for testing a seriesP

an for convergence, as a

flow chart.

We suggest that, given a series which you wish to test for convergence or

absolute convergence, you use the tests in the order indicated in the chart.

Naturally, you may be able to short-circuit this strategy in certain cases.

Is {an} null?

NO

NO

YES

YES

YES Does Alternating Test

apply?

Try usingFirst

Principles

NO or

CAN’T DECIDE

Try using:Basic seriesRatio TestLimit Comparison TestComparison TestCondensation Test

Is Σ⏐an⏐ convergent?

Σ an is (absolutely) convergent

Σ an is convergent

(Alternating Test)

Σan is divergent

(Non-null Test)

Remark

We have labelled the final box ‘First Principles’, to indicate that, if the various

tests do not produce a result, then it may be possible to work directly with the

sequence of partial sums {sn}.

Problem 6 Test the following series for convergence and absolute

convergence:

(a)P

1

n¼1

12

n; (b)P

1

n¼1

5nþ2n

3n ; (c)P

1

n¼1

32n3�1

;

(d)P

1

n¼1

�1ð Þnþ1

n13

; (e)P

1

n¼1

�1ð Þnþ1n2

n2þ1; (f)

P

1

n¼1

�1ð Þnþ1n

n3þ5;

(g)P

1

n¼1

2n

n6; (h)P

1

n¼1

�1ð Þnþ1n

n2þ2; (i)

P

1

n¼2

1

n loge nð Þ34

:

There is a variety of further tests for convergence or divergence which can

be applied to series with non-negative terms. In this book we give only one

further test, called the Integral Test. In particular it will enable us to prove that

X

1

n¼1

1

npis convergent; for all p > 1:

For example, if you wish toexamine a series whose termsalternate in sign, it is best togo straight to the AlternatingTest.

You will meet the IntegralTest in Section 7.4.

116 3: Series

3.3.6 Proofs

We now supply the proofs omitted earlier in this section.

Theorem 1 Absolute Convergence Test

IfP

1

n¼1


1

n¼1

an is convergent.

Proof We know thatP

1

n¼1

anj j is convergent, and we want to prove thatP

1

n¼1

an

is convergent.

To do this, we define two new sequences

aþn ¼an; if an � 0,

0; if an < 0,

�

and a�n ¼0; if an � 0,

�an; if an < 0.

�

Both the sequences aþn� �

and a�n� �

are non-negative, and

an ¼ aþn � a�n ; for n ¼ 1; 2; . . .:

Also

aþn � anj j; for n ¼ 1; 2; . . .; (10)

and

a�n � anj j; for n ¼ 1; 2; . . .: (11)

SinceP

1

n¼1

anj j is convergent, we deduce from (10) and (11) thatP

1

n¼1

aþn and

P

1

n¼1

a�n are convergent, by the Comparison Test. Thus

X

1

n¼1

an ¼X

1

n¼1

aþn � a�n� �

¼X

1

n¼1

aþn �X

1

n¼1

a�n (12)

is convergent, by the Combination Rules for series. &

Triangle Inequality (infinite form)

IfP

1

n¼1


1

n¼1

an

�

�

�

�

�

�

�

�

�P

1

n¼1

anj j.

Proof Let

sn ¼ a1 þ a2 þ � � � þ an; n ¼ 1; 2; . . .;

and

tn ¼ a1j j þ a2j j þ � � � þ anj j; n ¼ 1; 2; . . .:

Then, by the Absolute Convergence Test

limn!1

sn ¼X

1

n¼1

an exists;

also

limn!1

tn ¼X

1

n¼1

anj j:

You may omit these proofs ona first reading.

For example, ifX

1

n¼1

an ¼ 1� 1

22þ 1

32

� 1

42þ � � �;

then

X

1

n¼1

aþn ¼ 1þ 0þ 1

32þ 0þ ��

and

X

1

n¼1

a�n ¼ 0þ 1

22þ 0

þ 1

42þ�� :

Sub-section 3.1.4.

Alternatively, the infiniteform of the TriangleInequality can be deducedfrom statements (10), (11) and(12) in the proof of theAbsolute Convergence Test.

Now, by the Triangle Inequality

snj j ¼ a1 þ a2 þ � � � þ anj j� a1j j þ a2j j þ � � � þ anj j ¼ tn;

and so

�tn � sn � tn:

Thus, by the Limit Inequality Rule for sequences

� limn!1

tn � limn!1

sn � limn!1

tn;

that is

�X

1

n¼1

anj j �X

1

n¼1

an �X

1

n¼1

anj j;

and so

X

1

n¼1

an

�

�

�

�

�

�

�

�

�

�

�X

1

n¼1

anj j: &

Theorem 2 Alternating Test

If

an ¼ �1ð Þnþ1bn; n ¼ 1; 2; . . .;

where {bn} is a decreasing null sequence with positive terms, then

X

1

n¼1

an ¼ b1 � b2 þ b3 � b4 þ � � � is convergent:

Proof We can write the even partial sums s2k ofP

1

n¼1

an as follows

s2k ¼ b1 � b2ð Þ þ b3 � b4ð Þ þ � � � þ b2k�1 � b2kð Þ:

Since {bn} is decreasing, all the brackets are non-negative, and so the even

subsequence of partial sums, {s2k}, is increasing.

We can also write the even partial sums s2k as

s2k ¼ b1 � b2 � b3ð Þ � b4 � b5ð Þ � � � � � b2k�2 � b2k�1ð Þ � b2k:

Again, all the brackets are non-negative, and so s2k is bounded above, by b1.

Hence {s2k} is convergent, by the Monotone Convergence Theorem.

Now let

limk!1

s2k ¼ s:

Since s2k ¼ s2k�1 � b2k so that s2k�1 ¼ s2k þ b2k, and {bn} is null, we have

limk!1

s2k�1 ¼ limk!1

s2k þ b2kð Þ

¼ limk!1

s2k þ limk!1

b2k ¼ s;

In Remark 3 following theproof of the TriangleInequality in Sub-section 1.3.1,we commented that theTriangle Inequality for twonumbers, a and b, can beextended in the obvious way toany finite sum of numbers.

You met this in Sub-section 2.3.3, Theorem 3.

118 3: Series

by the Sum Rule for sequences. Thus the odd and even subsequences of {sn}

both tend to the same limit s, and so {sn} tends to s. Hence

X

1

n¼1

an ¼ b1� b2þ b3� b4þ � � � is convergent; with sum s: &

Theorem 4 Let the seriesP

1

n¼1

an ¼ a1 þ a2 þ � � � be absolutely conver-

gent. Then any rearrangementP

1

n¼1

bn of the series also converges absolutely,

andP

1

n¼1

bn ¼P

1

n¼1

an:

Proof First of all, we prove the result in the special case that ak� 0 for

all k.

Choose any positive integer n. Then choose a positive integer N such that

b1, b2, . . ., bn all occur among the terms a1, a2, . . ., aN of the original series.

It follows that

X

n

k¼1

bk �X

N

k¼1

ak

�X

1

k¼1

ak: (13)

Hence the partial sumsP

n

k¼1

bk of the rearranged seriesP

1

k¼1

bk are bounded

above, so that the rearranged series must be convergent. It also follows from

(13) thatP

1

k¼1

bk �P

1

k¼1

ak.

By reversing the roles of the terms a1, a2, . . . and b1, b2, . . . in the above

argument, the same argument shows thatP

1

k¼1

ak �P

1

k¼1

bk. It thus follows that

the two seriesP

1

k¼1

ak andP

1

k¼1

bk must in fact have the same sum.

To complete our proof, we now drop the condition that ak� 0 for all k.

We then define the quantities aþk , a�k , bþk and b�k as follows

aþk ¼ak; if ak � 0,

0; if ak < 0,

�

a�k ¼0; if ak � 0,

�ak; if ak < 0,

�

bþk ¼bk; if bk � 0,

0; if bk < 0,

�

b�k ¼0; if bk � 0,

�ak; if bk < 0.

�

Now,P

1

k¼1

aþk is convergent, since aþk ¼ 12

akj j þ akð Þ and bothP

1

k¼1

ak

�

�

�

� andP

1

k¼1

ak

converge.P

1

k¼1

bþk is a rearrangement ofP

1

k¼1

aþk , and both series have non-

negative terms. It follows from the first part of the proof that both series

converge, and have the same sum.

Similarly,P

1

k¼1

a�k andP

1

k¼1

b�k both converge, and have the same sum.

By Theorem 6 , Sub-section 2.4.4.

In particular, N� n.

For the right-hand side issimply the left-hand side withpossibly some more non-negative terms inserted.

For the partial sums of a seriesof non-negative terms arealways less than or equal thesum of the whole series.

By the Combination Rules forseries.

But bk

�

�

�

� ¼ bþk þ b�k , and so the seriesP

1

k¼1

bk

�

�

�

� converges, by the Sum Rule

for series. In other words, the rearranged seriesP

1

k¼1

bk is absolutely convergent.

Finally

X

1

k¼1

bk ¼X

1

k¼1

bþk �X

1

k¼1

b�k

¼X

1

k¼1

aþk �X

1

k¼1

a�k

¼X

1

k¼1

ak: &


Let the seriesP

1

n¼0

an andP

1

n¼0

bn be absolutely convergent, and let

cn ¼ a0bn þ a1bn�1 þ a2bn�2 þ � � � þ anb0 ¼X

n

k¼0

akbn�k:

Then the seriesP

1

n¼0

cn is absolutely convergent, and

X

1

n¼0

cn ¼X

1

n¼0

an

!

�X

1

n¼0

bn

!

:

Proof First, we introduce some notation, for n� 0

sn ¼P

n

k¼0

ak; tn ¼P

n

k¼0

bk; un ¼P

n

k¼0

ck;

s0n ¼P

n

k¼0

akj j; t0n ¼P

n

k¼0

bkj j; u0n ¼P

n

k¼0

ckj j;

s ¼P

1

k¼0

ak; t ¼P

1

k¼0

bk;

s0 ¼P

1

k¼0

akj j; t0 ¼P

1

k¼0

bkj j:

In particular, we know that sn! s, tn! t, sn0 ! s0 and tn

0 ! t0 as n!1.

Also, by the Product Rule for sequences, sn tn! st and sn0 tn0 ! s0t0, as n!1.

Hence, from the definition of limit, it follows that, for any positive number ",there is some integer N for which

sntn � stj j < 1

3" and s0nt0n � s0t0

�

�

�

� <1

3"; for all n � N:

It follows that, for n�N

s0nt0n � s0Nt0N�

�

�

� ¼ s0nt0n � s0t0� �

� s0Nt0N � s0t0� �

�

�

�

�

� s0nt0n � s0t0�

�

�

�þ s0Nt0N � s0t0�

�

�

�

<1

3"þ 1

3"

¼ 2

3":

And, in factP

1

k¼1

bk

�

�

�

� ¼P

1

k¼1

bþk þP

1

k¼1

b�k .

Here we use an integer Nrather than a general numberX and a weak inequality n�Nrather than a strict inequality;this simplifies the notation inthe rest of the argument alittle.

120 3: Series

Now let n be such that n� 2N, and consider the terms ai bj that occur in the

expression un� sNtN. Every such term has iþ j� n, since un consists of all such

terms; but none of these terms has both i�N and j�N.

Hence, for every term ai bj in un� sNtN, a corresponding term jaibjjwill occur in the expression sn

0 tn0 � sN

0 tN0 – for this last expression consists

of all terms jaibjj with i� n and j� n, but not with both i�N and j�N.

(Note that we made the requirement that n� 2N in order that every term in

sNtN appears in un.)

a0b0 a0b1 a0b2 ...

a1b0 a1b1 a1b2 ...

a2b0 a2b1 a2b2 ...

...

0

N

N 2N n

2N

n

sNtN

un – sNtN

In the above diagram, we set out the terms ai bj in rows and columns, where

the term ai bj occurs in the (iþ 1)th row and the ( jþ 1)th column. Then the

small square contains all the terms ai bj with 0� i�N and 0� j�N; these add

up to sNtN.

Hence, for all n� 2N, we have

un � stj j ¼ un � sNtNð Þ þ sNtN � stð Þj j� un � sNtNj j þ sNtN � stj j� s0nt0n � s0Nt0N�

�

�

�þ sNtN � stj j

<2

3"þ 1

3"

¼ ";it follows, from this inequality, that un! st as n!1.

Finally, we have

u0n � s0nt0n� s0t0:

For terms with both i�N andj�N are all in sNtN.

By the Triangle Inequality.

By the discussions aboveconcerning un� sNtN and

sn0 tn0 � sN

0 tN0.

For the product sn0 tn0 contains

more non-negative terms thandoes un

0 ; and the sequences{sn0} and {tn

0} are increasing.

Hence, the increasing sequence {un0} is bounded above, and so tends to a

limit as n!1. Thus the seriesP

1

n¼0

cn is absolutely convergent. &

3.4 The exponential function x j!ex

Earlier, we defined e¼ 2.71828 . . . to be the limit

e ¼ limn!1

1þ 1

n

� �n

;

and we defined ex to be the limit

ex ¼ limn!1

1þ x

n

� �n

; for any real x:

Here we show that the formula

ex ¼X

1

n¼0

xn

n!; for any real x;

is an equivalent definition of the quantity ex.

Remark

The seriesP

1

n¼0

xn

n! converges for all values of x. For

xnþ1

nþ 1ð Þ!

�

�

�

�

�

�

�

�

xn

n!

�

�

�

�

�

�

�

�

¼ xj jnþ 1

! 0 as n!1;

so that, by the Ratio Test, the series is absolutely convergent for all x, and so

is convergent for all x.

3.4.1 The definition of ex as a power series, for x> 0

If we plot the partial sum functions of the infinite series of powers of x

X

1

n¼0

xn

n!¼ 1þ xþ x2

2!þ x3

3!þ � � �; (1)

the resulting graph appears to be that of ex. (A series of multiples of increasing

powers of x is called a power series.) In particular, when x ¼ 1, the sum of

the series

X

1

n¼0

1

n!¼ 1þ 1þ 1

2!þ 1

3!þ � � �

is approximately 2.71828 . . ..


This series is a basic series oftype (d), in the case that x� 0.

122 3: Series

y

e

y = ex

y = 1 + x

y = 1

x

y = 1 + x + x2

2!

y = 1 + x + x2

2!x3

3!+

10–1

However the fact that the sequence of partial sums of the series (1) appears

to converge to ex does not constitute a proof of this fact. Our first aim is to

supply this proof in the particular case that x> 0.

Theorem 1 If x> 0, thenP

1

n¼0

xn

n! ¼ ex.

Proof We defined ex to be ex ¼ limn!1

1þ xn

� �n, and so we have to show that

X

1

n¼0

xn

n!¼ lim

n!11þ x

n

� �n

; for x > 0:

The (n + 1)th partial sum of the above series is

snþ1 ¼ 1þ xþ x2

2!þ x3

3!þ � � � þ xn

n!:

By the Binomial Theorem

1þ x

n

� �n

¼ 1þ nx

n

� �

þ n n� 1ð Þ2!

x

n

� �2

þ � � � þ x

n

� �n

:

A typical term in this expansion is

n n�1ð Þ . . . n� kþ1ð Þk!

x

n

� �k

¼ xk

k!1�1

n

� �

1�2

n

� �

� � � 1� k�1

n

� �

� xk

k!;

since each bracket is less than 1.

Thus

1þ x

n

� �n

� 1þ xþ x2

2!þ x3

3!þ � � � þ xn

n!

¼ snþ1;

and so, by the Limit Inequality Rule for sequences

limn!1

1þ x

n

� �n

� limn!1

snþ1;

We shall deal with the casex< 0 in Sub-section 3.4.3.

Hence, for x> 0, the seriesP

1

n¼0

xn

n! gives an equivalent

definition of ex.


3.4 The exponential function x j! ex 123

so that

ex �X

1

n¼0

xn

n!: (2)

On the other hand, for any integers m and n with m� n, we have

1þ x

n

� �n

� 1þ nx

n

� �

þ n n� 1ð Þ2!

x

n

� �2

þ� � � þ n n� 1ð Þ . . . n�mþ 1ð Þm!

x

n

� �m

¼ 1þ xþ x2

2!1� 1

n

� �

þ � � � þ xm

m!1� 1

n

� �

1� 2

n

� �

. . . 1�m� 1

n

� �

:

Now keep m fixed and let n!1. By the Limit Inequality Rule for

sequences, we obtain

limn!1

1þ x

n

� �n

� 1þ xþ x2

2!þ x3

3!þ � � � þ xm

m!;

and so

ex � smþ1:

Now let m!1; by the Limit Inequality Rule for sequences, we obtain

ex � limm!1

smþ1;

so that

ex �X

1

n¼0

xn

n!(3)

Combining inequalities (2) and (3), we obtain

ex ¼X

1

n¼0

xn

n!; for x > 0: &

Problem 1 Estimate e2 (to 3 decimal places) by calculating the seventh

partial sum of the series (1) when x = 2.

3.4.2 Calculating e

The representation of e by the infinite series

e ¼X

1

n¼0

1

n!¼ 1þ 1þ 1

2!þ 1

3!þ � � � (4)

provides a much more efficient way of calculating approximate values for e

than the equation e ¼ limn!1

1þ 1n

� �n. This is illustrated by the following table of

approximate values:

n 1 2 3 4 5

1þ 1n

� �n2 2.25 2.37 2.44 2.49

P

n

k¼0

1k!

2 2.50 2.67 2.71 2.717

This follows since {sn} and{snþ1} both tend to the samelimit, the sum of the infiniteseries.

The calculation of e via thelimit is a new calculation eachtime, whereas via the seriesinvolves only adding oneextra term to the previousapproximation.

124 3: Series

We can estimate how quickly the sequence of partial sums

sn ¼X

n�1

k¼0

1

k!¼ 1þ 1þ 1

2!þ 1

3!þ � � � þ 1

n� 1ð Þ! ; n ¼ 1; 2; . . .;

converges to e as follows. The difference between e and sn is given by

e� sn ¼X

1

k¼n

1

k!¼ 1

n!þ 1

nþ 1ð Þ!þ1

nþ 2ð Þ!þ � � �

¼ 1

n!1þ 1

nþ 1ð Þ þ1

nþ 1ð Þ nþ 2ð Þ þ � � ��

<1

n!1þ 1

nþ 1

n2þ � � �

�

:

The last expression in square brackets is a geometric series with first term

1 and common ratio 1n, and so its sum is 1

1�1n

¼ nn�1

. Hence

0 < e� sn <1

n!� n

n� 1

¼ 1

n� 1ð Þ!�1

n� 1; for n ¼ 1; 2; . . .: (5)

For example, this estimate shows that

0 < e� s6 <1

5!� 1

5¼ 1

600¼ 0:0016:

Since the partial sum

s6 ¼ 1þ 1þ 1

2!þ 1

3!þ 1

4!þ 1

5!

¼ 1þ 1þ 1

2þ 1

6þ 1

24þ 1

120¼ 163

60¼ 2:716;

we deduce very easily that

2:716 < e < 2:716þ 0:0016 ¼ 2:7183 :

Problem 2 Estimate how many terms in the series (4) are needed to

determine e to ten decimal places.

The inequality (5) can also be used to show that e is irrational.

Theorem 2 The number e is irrational.

Proof Suppose that e ¼ mn, where m and n are positive integers.

Now, by the above estimate (5)

0 < e� snþ1<1

n!� 1

n;

and so

0 < n! e� snþ1ð Þ< 1

n:

Here we replace various termsby larger terms (because theirdenominators are smaller); sothe new sum is greater.

In fact,

e ¼ 2:71828182845 . . .:


3.4 The exponential function x j! ex 125

Since e ¼ mn, we have

0 < n!m

n� 1þ 1þ 1

2!þ 1

3!þ � � � þ 1

n!

� ��

<1

n:

But n! mn� 1þ 1þ 1

2!þ 13!þ � � � þ 1

n!

� ��

is an integer, since n! times each

expression in the square bracket is itself an integer. So we have found an

integer which lies strictly between 0 and 1. This is clearly impossible, and so

e cannot be rational after all. &

3.4.3 The definition of ex as a power series, for all real x

Earlier we saw thatP

1

n¼0

xn

n!¼ ex, for all x> 0. We now show that this formula is

valid for all real values of x, and we will also verify the fundamental property

of the exponential function, namely that exey¼ exþy for all real values of x

and y. The Product Rule for series will be our crucial tool in this work.

First we check the following.

Lemma 1 For any x 2 R ,P

1

n¼0

xn

n! �P

1

n¼0

�xð Þnn! ¼ 1:

Proof By the Product Rule for series,P

1

n¼0

xn

n! �P

1

n¼0

�xð Þnn! ¼

P

1

n¼0

cn, where, for

n� 1, we have

cn ¼X

n

k¼0

xk

k!� �xð Þn�k

n� kð Þ!

¼X

n

k¼0

xk �xð Þn�k

k! n� kð Þ!

¼ 1

n!

X

n

k¼0

n!

k! n� kð Þ!xk �xð Þn�k

¼ 1

n!xþ �xð Þð Þn¼ 0:

Since c0¼ 1� 1¼ 1, the result then follows. &

We can then use the result in Lemma 1 to prove thatP

1

n¼0

xn

n!¼ ex, for x< 0 as

well as x> 0. Since, obviously,P

1

n¼0

xn

n! ¼ ex when x¼ 0, this completes the

proof thatP

1

n¼0

xn

n!¼ ex, for all x 2 R .

Theorem 3 If x< 0, thenP

1

n¼0

xn

n! ¼ ex.

Proof For x< 0, the following chain of equalities holds

X

1

n¼0

xn

n!¼ 1

,

X

1

n¼0

�xð Þn

n!ðby Lemma 1Þ

Here we have simplysubstituted m

nfor e in the

previous expression.

This was Theorem 1 inSub-section 3.4.1.

This was Theorem 5 inSub-section 3.3.4.

We may apply the Product

Rule since the seriesP

1

n¼0

xn

n! and

P

1

n¼0

�xð Þnn! are both absolutely

convergent.

Here we apply the BinomialTheorem to xþ �xð Þð Þn.

In other words, the definitionof ex as a power series isequivalent to its definitionas a limit.

Here the definition of ex onthe right is that

ex ¼ limn!1

1þ x

n

� �n

.

126 3: Series

¼ 1.

limn!1

1þ�x

n

� �n

ðby Theorem 1; since �x > 0Þ

¼ 1=e�x ðthis is the definition of e�xÞ¼ ex ðby the Inverse Property of exÞ:

This completes the proof. &

In order to be crystal clear where we have reached, we now combine the results

of Theorem 1 and Theorem 3 into the following:

Theorem 4 For all x 2 R ,P

1

n¼0

xn

n!¼ lim

n!11þ x

n

� �n¼ ex.

Finally, we verify the Fundamental Property of the exponential function,

which was left as unfinished business at the end of Sub-section 2.5.3.

Theorem 5 Fundamental Property of the exponential function

For all x, y2R , exey ¼ exþy.

Proof By the Product Rule for series, ex � ey ¼P

1

n¼0

xn

n! �P

1

n¼0

yn

n! ¼P

1

n¼0

cn,

where, for n � 1, we have

cn ¼X

n

k¼0

xk

k!� yn�k

n� kð Þ!

¼X

n

k¼0

xkyn�k

k! n� kð Þ!

¼ 1

n!

X

n

k¼0

n!

k! n� kð Þ! xkyn�k

¼ xþ yð Þn

n!:

Since c0¼ 1� 1¼ 1, the result then follows. &

3.5 Exercises

Section 3.1

1. Prove thatP

1

n¼1

�34

� �nis convergent, and find its sum.

2. Interpret 0:12 as an infinite series, and hence find the value of 0:12 as a

fraction.

The Inverse Property of ex

was Theorem 4 of Sub-section 2.5.3.

3.5 Exercises 127

3. Prove that

1

4n2 � 1¼ 1

2

1

2n� 1� 1

2nþ 1

� �

; for n ¼ 1; 2; . . .;

and deduce that

X

1

n¼1

1

4n2 � 1¼ 1

2:

4. Determine whether the following series converge:

(a)P

1

n¼1

45

� �nþ 4n nþ2ð Þ

� �

; (b)P

1

n¼1

1þ 12

� �n� �

; (c)P

1

n¼1

1ffiffi

np � 1

ffiffiffiffiffiffi

nþ1p

� �

.

Section 3.2


(a)P

1

n¼1

cos 1=nð Þ2n2þ3

; (b)P

1

n¼1

n2

2n3�n; (c)

P

1

n¼1

ffiffiffiffi

2np

4n3þnþ2;

(d)P

1

n¼1

nþ1ð Þ52n ; (e)

P

1

n¼1

n23n

n! ; (f)P

1

n¼1

n!ð Þ22nð Þ!.

2. (a) Use the Ratio Test to prove thatP

1

n¼1

2nn!nn converges, but that

P

1

n¼1

3nn!nn

diverges.

(b) For which positive values of c can you use the Ratio Test to prove

thatP

1

n¼1

cnn!nn is convergent?

3. Prove that

1ffiffiffi

np � 1


nþ 1p ¼ 1

ffiffiffi

np ffiffiffiffiffiffiffiffiffiffiffi

nþ 1p ffiffiffiffiffiffiffiffiffiffiffi

nþ 1p

þffiffiffi

np� � ; for n ¼ 1;2; . . .;

and use Exercise 4(c) on Section 3.1 to deduce thatP

1

n¼1

1

n32

is convergent.


(a)P

1

n¼3

1n logenð Þ loge logenð Þð Þ; (b)

P

1

n¼2

ffiffiffiffiffiffi

nþ1p

2n2�3nþ1ð Þ logenþ logenð Þ2ð Þ.

Section 3.3

1. Test the following series for convergence and absolute convergence:

(a)P

1

n¼1

�1ð Þnþ1

1þffiffi

np ; (b)

P

1

n¼1

sin nn2 ; (c)

P

1

n¼1

�1ð Þnþ1n!

n4þ3; (d)

P

1

n¼1

nþ2n

3nþ5.

2. (a) Prove that

1

3n� 2þ 1

3n� 1� 1

3n>

1

3n; for n ¼ 1; 2; . . .;

and deduce that

1þ 1

2� 1

3þ 1

4þ 1

5� 1

6þ � � � is divergent:

128 3: Series

(b) Prove that

1

n� 1

2n� 1

2nþ 2

�

�

�

�

�

�

�

�

<1

n2; for n ¼ 1; 2; . . .;

and deduce that

1� 1

2� 1

4þ 1

3� 1

6� 1

8þ 1

5� 1

10� 1

12þ 1

7� 1

14� 1

16þ � � �

is convergent:

(c) Prove that

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2n� 1p � 1

2n� 1

2n; for n ¼ 1; 2; . . .;

and deduce that

1� 1

2þ 1

ffiffiffi

3p � 1

4þ 1

ffiffiffi

5p � 1

6þ � � �

is divergent:

3. Prove that there exists a rearrangement of the seriesP

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3�

14þ 1

5� 1

6þ � � � for which one subsequence of its partial sums tends to 1 and

another subsequence of its partial sums tends to�1. Alternatively, prove that

no such rearrangement exists.

3.5 Exercises 129

4 Continuity

The graphs of many functions that we come across look to be smooth curves,

without any jumps.

1y = ex

– 12 π

–1

1 y = sin x

12 π

On the other hand, the graphs of some functions that arise naturally, or that

we construct for various purposes, do not appear to be smooth, but to contain

jumps.

y = [x]

1

–2 –1 1

–1

221– π

21 π

y = tan x

However there are some functions where the graphs appear to have an

unreasonable number of jumps!

How can we describe this complicated situation? The key underlying idea is

that of continuity. Loosely speaking, a function is said to be continuous if we

can draw its graph without taking our pencil off the page. This is the same as

the graph ‘having no jumps’. The concept is an important one in Analysis

since, in many situations, the crucial step in proving that a function has some

property is to prove that the function is continuous. This is the first of two

chapters that study continuous functions.

In Section 4.1, we define the phrase:

the function f is continuous at the point c,

and we give a number of rules which state, for example, that various combina-

tions and compositions of continuous functions are themselves continuous.

130

Using these rules, together with a list of basic continuous functions, we can

deduce that many functions are continuous at each point of their domains. For

example, the functions x 7! xþ 1x

and x 7! x sin 1x

are continuous at each point

of R � {0}, and the trigonometric and exponential functions are continuous

at each point of their domains.

Section 4.2 is devoted to the properties of continuous functions. The two

fundamental properties of continuous functions are the Intermediate Value

Theorem and the Extreme Value Theorem. We shall see a useful applica-

tion of the Intermediate Value Theorem to finding the zeros of various

functions.

In Section 4.3, we discuss the Inverse Function Rule; this rule gives condi-

tions under which a continuous function f has a continuous inverse function

f�1. We then use the Inverse Function Rule to obtain the inverses of standard

functions. Some of these inverse functions will be familiar to you already, but

the Inverse Function Rule enables us to establish their properties by a rigorous

argument.

Finally, in Section 4.4, we shall provide a rigorous definition of the expo-

nential function x 7! ax, where x2R and a> 0.

4.1 Continuous functions

4.1.1 What is continuity?

To accord with our intuitive idea of what we should mean by ‘the function f is

continuous at the point c’, we wish to define this concept in such a way that the

following two functions are continuous at the point c:

y1.

c x

f (c)

y2.

c x

f (c)

On the other hand, we wish to formulate our definition so that the following

two functions are not continuous at the point c:

y3.

c x

f (c)

y4.

f (c)

c x

In particular, to findingsolutions of polynomialequations.


So, our definition must say, in a precise way, that:

if x tends to c, then f (x) tends to f (c).

There are several ways of making this idea precise. In this chapter we adopt a

definition which involves the convergence of sequences, as this will enable us

to use the results about sequences that we met in Chapter 2.

Definitions A function f defined on an interval I that contains c as an

interior point is continuous at c if:

for each sequence {xn} in I such that xn! c, then f (xn)! f (c).

If f is not continuous at the point c in I, then it is discontinuous at c.

Thus, for example, in graph 3 above, if {xn} is a strictly decreasing

sequence that tends to c as n!1, then {f (xn)} is a strictly decreasing

sequence that tends to {f (c)}. On the other hand, if {xn} is a strictly

increasing sequence that tends to c as n!1, then {f (xn)} is a strictly

increasing sequence that tends to a limit as n!1 – but that limit is some

number less than f (c).

However, in graph 4, if {xn} is a strictly increasing sequence that tends to c

as n!1, then {f (xn)} is a strictly increasing sequence that tends to f (c); but, if

{xn} is a strictly decreasing sequence that tends to c as n!1, then the

sequence {f (xn)} may not tend to any limit as n!1.

On the other hand, in graphs 1 and 2, no matter how the sequence {xn} tends

to c, then for sure we do have that {f (xn)} tends to f (c).

For these examples, then, the definition agrees with what we believe the

essence of continuity should be.

Example 1 Prove that the function f (x)¼ x3, x2R , is continuous at the point 12.

Solution Let {xn} be any sequence in R that converges to 12; that is, xn ! 1

2.

Then, by the Combination Rules for sequences, it follows that

f xnð Þ ¼ x3n !

1

2

� �3

¼ 1

8as n!1;

while f 12

� �

¼ 18. In other words, {f(xn)} converges to f 1

2

� �

as n!1.

It follows that f is continuous at 12, as required. &

Example 2 Prove that the function f xð Þ ¼ 1; x < 0;2; x � 0;

�

is discontinuous at 0.

Solution Let {xn} be any sequence in R that converges to 0 from the right. By

looking at the graph y¼ f(x), it is clear that for such a sequence f(xn)! 2¼ f(0).

This will not help us to show that f is discontinuous at 0!

However, if we let {xn} be any non-constant sequence in R that converges

to 0 from the left, the situation will be very different. By looking at the

graph y¼ f(x), it is clear that for such a sequence f(xn)! 1 6¼ f(0). We now

make this precise.

In Chapter 5 we shall meet adifferent definition, and seethat the two definitions are infact equivalent.

18

12

y

y = x3

x

y

2

y = {1, x < 0,2, x ≥ 0.

x

1

To prove that a function isdiscontinuous, all we need todo is to find just one sequencefor which our definition ofcontinuity at the relevantpoint does not hold.

132 4: Continuity

Choose the sequence xnf g ¼ � 1n

� �

; n ¼ 1; 2; . . .. For this sequence

f xnð Þ ¼ 1! 1 6¼ f 0ð Þ as n!1:It follows that the function f cannot be continuous at 0. &

Problem 1

(a) Determine whether the function f xð Þ ¼ x3 � 2x2; x 2 R , is contin-

uous at 2.

(b) Determine whether the function f xð Þ ¼ x½ �; x 2 R , is continuous at 1.

Problem 2

(a) Prove that the function f xð Þ ¼ 1; x 2 R , is continuous on R .

(b) Prove that the function f xð Þ ¼ x; x 2 R , is continuous on R .

However, as we know, not every function is defined on the whole of R – but we

still want to be able to discuss the continuity of such functions. For example:

� f xð Þ ¼ffiffiffi

xp

, where the domain of f is the interval [0,1);

� f xð Þ ¼ x�12 þ ð1� xÞ

12, where the domain of f is the half-closed interval (0, 1];

� f xð Þ ¼ 1x, where the domain of f is R � {0}.

How can we deal with these various different situations?

The first two situations are dealt with by introducing the notion of one-sided

continuity.

Suppose that f is defined on some set S that contains an interval [c, cþ r) for

some r> 0, but where f is not necessarily defined on any interval (c� r, cþ r)

that contains c as an interior point. Then it seems reasonable to say that f is

continuous on one side of c. In particular, that f is continuous on the right at the

point c of S.

Similarly, if f is defined on some set S that contains an interval (c� r, c] for

some r> 0, but f is not necessarily defined on any interval (c� r, cþ r) that

contains c as an interior point, then it seems reasonable to say that f is

continuous on one side of c. In particular, that f is continuous on the left at

the point c of S.

The case of the function f xð Þ ¼ 1x, where the domain of f is R – {0}, is

different. The domain of f is the whole of R , with just one point omitted. That

is, it is the union of two open intervals (�1, 0) and (0,1) – to both of which

the definition of continuity applies in its originally stated form. So it makes

sense to simply drop the original restriction that the domain of f must be an

open interval in R .

These considerations lead us to making the following less restrictive defini-

tion of continuity than our original definition:

Definitions A function f defined on a set S in R that contains a point c is

continuous at c if:

for each sequence {xn} in S such that xn! c, then f(xn)! f(c).

If f is continuous on the whole of its domain, we often simply say that f is

continuous (without explicit mention of the points of continuity).

If f is not continuous at the point c in S, then it is discontinuous at c.

Here we make a specificchoice for the sequence {xn}to prove that f cannot becontinuous at 0.

Here [�] is the integer partfunction.

By ‘continuous on R’, wemean ‘continuous at eachpoint of R’.

y = x1

y

x

Here f is continuous on S.


In addition, we have the following related definitions:

Definitions A function f whose domain contains an interval [c, cþ r) for

some r> 0 is continuous on the right at c if:

for each sequence {xn} in [c, cþ r) such that xn! c, then f(xn)! f(c).

A function f whose domain contains an interval (c� r, c] for some r> 0 is

continuous on the left at c if:

for each sequence {xn} in (c� r, c] such that xn! c, then f(xn)! f(c).

A function f whose domain contains an interval I is continuous on I if it is

continuous at each interior point of I, continuous on the right at the left end-

point of I (if this belongs to I), and continuous on the left at the right end-

point of I (if this belongs to I).

The connection between the definitions of continuity and of one-sided

continuity is rather obvious.

Theorem 1 A function f whose domain contains an interval I that contains

c as an interior point is continuous at c if and only if f is both continuous on

the left at c and continuous on the right at c.

Example 3 Determine whether the function f given by f xð Þ ¼ffiffiffi

xp; x � 0, is

continuous on x : x 2 R ; x � 0f g.

Solution The domain of f is the interval I ¼ x : x � 0f g.Thus, we have to show that for each c in I:

for each sequence {xn} in I such that xn! c, thenffiffiffiffiffi

xnp !

ffiffiffi

cp

.

First, if c¼ 0, then we know already that, for any null sequence {xn} in I,ffiffiffiffiffi

xnp� �


Next, let c> 0. We have to prove that if xn � cf g is a null sequence, then so

isffiffiffiffiffi

xnp �

ffiffiffi

cp� �

. Now, from the identity

ffiffiffiffiffi

xn

p �ffiffiffi

cp¼ xn � c

ffiffiffiffiffi

xnp þ

ffiffiffi

cp ;

we see that, since c 6¼ 0

ffiffiffiffiffi

xn

p �ffiffiffi

cp! 0

2ffiffiffi

cp ¼ 0 as n!1:

In other words,ffiffiffiffiffi

xnp �

ffiffiffi

cp� �

is indeed a null sequence. This completes the

proof. &

Problem 3 Prove that the nth root function

f xð Þ ¼ x1n; where n 2 N and

x � 0; if n is even;x 2 R ; if n is odd;

�

is continuous.

We omit a proof of thisstraight-forward result.

In fact, f is continuous on theright at 0, and continuous atall other points of its domain.

y

x0

y = √x

Recall that c 6¼ 0, so that 1ffiffi

cp is

indeed defined.

You may omit this Problem ifyou are short of time.

We defined a1n in

Sub-section 1.5.3.

134 4: Continuity

Hints: Use the result of Exercise 3(b) on Section 2.3, in Section 2.6, and

the identity aq � bq ¼ a� bð Þ aq�1 þ aq�2bþ aq�3b2 þ � � � þ bq�1ð Þwith suitable choices for a, b and q. In your solution, be careful not to

use the letter n for two different purposes.

Example 4 Determine whether the function f xð Þ ¼ 1x; x 2 R � 0f g, is

continuous.

Solution The domain of f is the set R � {0}, the union of the two open

intervals (�1, 0) and (0,1).

Let c be any point of R � {0}, and let {xn} be any sequence in R � {0} that

converges to c. Then, by the Quotient Rule for sequences, it follows that

f xnð Þf g ¼�

1xn

�

! 1c

as n!1; in other words, that f xnð Þ ! f cð Þ ¼ 1c

as

n!1. So f is continuous at c.

Since c is an arbitrary point of R � {0}, it follows that f is continuous

on R � {0}. &

Our work so far on continuity illustrates the following general strategy:

Strategy for continuity

� To prove that a function f : S!R is continuous at a point c of S, prove

that:

for each sequence {xn} in S such that xn! c, then f(xn)! f(c).

� To prove that a function f : S!R is discontinuous at a point c of S:

find one sequence {xn} in S such that xn! c but f(xn) 6! f(c).

Problem 4 Prove that the following functions are continuous on R :

(a) f(x)¼ c, x2R ;

(b) f(x)¼ xn, x2R , n2N;

(c) f(x)¼ jxj, x2R .

Problem 5 Determine the points of continuity and discontinuity of the

signum function

f xð Þ ¼�1; x < 0,

0; x ¼ 0,

1; x > 0.

8

<

:

Remarks

The definition of function that we are using involves a mapping that we call

f (say) from a set in R , the domain A (say), to another set in R , the codomain

B (say).

1. Let f be continuous at a point c in A, and assume that, for some set A0 � A,

c2A0. Suppose that another function g has domain A0 on which g(x)¼ f(x).

Technically g is a different function from f, for sure. However if f is

continuous at c, then it is a simple matter of some definition checking to

Example 3 above was thespecial case of Problem 3,where n¼ 2.

Sub-section 2.3.2.

Recall that just one suchsequence suffices.

y

x

y = f(x)

–1

1

You can think of the mappingf as being a formula x 7! y thatmaps a point x of A onto to apoint y of B, and we writey¼ f (x) to indicate this.

verify that g too is continuous at c. So, restriction of a function to a smaller

domain does not affect its continuity at a point.

2. Let f and g be functions defined on sets in R that contain an open interval I,

and c2 I. Then, if f(x)¼ g(x) for all x2 I, f is continuous at c if g is

continuous at c, and f is discontinuous at c if g is discontinuous at c.

Again, we may simply ignore any difference between the domains of the

functions when we are studying continuity at a point.

3. The underlying point here is that continuity at a point is a local property. It

is only the behaviour of the function near that point that determines whether

it is continuous at the point.

4.1.2 Rules for continuous functions

We have seen how to recognise whether a given function is continuous at a

point. However, it would be tedious to have to go back to first principles to

determine on each occasion whether a complicated function is continuous.

As usual, we avoid such problems by having a set of rules that enable us to

construct continuous functions.

Combination Rules for continuous functions

If f and g are functions that are continuous at a point c, then so are:

Sum Rule fþ g;

Multiple Rule lf, for l2R ;

Product Rule fg;

Quotient Rule fg, provided that f (c) 6¼ 0.

For example, any polynomial p(x)¼ a0þ a1xþ � � � þ anxn, x2R , is contin-

uous at all points of R since we can build up the expression for p by successive

applications of the Combination Rules for continuous functions.

Similarly, any rational function rðxÞ ¼ pðxÞqðxÞ, where p and q are polynomials

and the domain of r is R minus the points where q vanishes, is continuous at all

points of its domain, since we can build up the expression for r by successive

applications of the Combination Rules for continuous functions.

The Combination Rules above are all natural analogues of the corresponding

results for sequences. However, we can combine functions in more ways than

we can combine sequences – for example, we can compose functions f and g to

obtain the function g � f . This approach too will often enable us to obtain new

continuous functions.

Composition Rule Let f be continuous on a set S1 that contains a point c,

and g be continuous on a set S2 that contains the point f(c). Then g � f is

continuous at c.

For example, we know that the function f(x)¼ x2þ 1, x2R , is continuous on R

and that the function g xð Þ ¼ffiffiffi

xp; x � 0, is continuous on I¼ {x : x� 0}. It

follows from the Composition Rule that the function g � f : x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

is

For instance, the function

p xð Þ ¼ x17�34x5þ5x� 8

is continuous on R .

For instance, the function

r xð Þ ¼ 1� 2x

x2 � 4;

x 2 R � 2f g; is continuouson its domain.

Functions obtained in thisway are called compositefunctions.

136 4: Continuity

continuous at all points c of R (the domain of f) whose image f(c) lies in I. But

all such points f(c) lie in I; so it follows that in fact the composite g � f is

continuous on R .

Problem 6 Prove that the function f given by f xð Þ ¼ x32; x � 0; is

continuous.

Finally, just as we had a Squeeze Rule for convergent sequences, we have a

corresponding Squeeze Rule for continuous functions.


Let the functions f, g and h be defined on an open interval I, and c2 I. If:

1. g(x) f(x) h(x), for all x2 I,

2. g(c)¼ f(c)¼ h(c),

3. g and h are continuous at c,

then f is also continuous at c.

Example 5 Prove that the function f given by f xð Þ ¼ x2 sin 1x

� �

; x 6¼ 0,

0; x ¼ 0,

�

is continuous at 0.

Solution The diagram in the margin suggests that we should find functions

g and h that squeeze f near 0.

So, we define g(x)¼�x2, x2R , and h(x)¼ x2, x2R . With these two

chosen, we check the conditions of the Squeeze Rule.

The first condition is of course vital! Now we know that

�1 sin1

x

� �

1; for any x 6¼ 0:

It follows that

�x2 x2 sin1

x

� �

x2; for any x 6¼ 0;

so that

g xð Þ ¼ �x2 �

f xð Þ x2 ¼ �

h xð Þ; for any x 2 R :

So condition 1 of the Squeeze Rule is satisfied.

Next, the functions f, g and h all take the value 0 at the point 0. Thus

condition 2 of the Squeeze Rule is satisfied.

Finally, the functions g and h are polynomials, and so in particular they are

continuous at 0. So condition 3 of the Squeeze Rule is satisfied.

It follows then from the Squeeze Rule that f is continuous at 0, as required.&

To test your understanding of these techniques, try the following problems.

Problem 7 Prove that the following function is continuous on R ,

stating each rule or fact about continuity that you are using

f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ xþ 1p

� 5x

1þ x2; x 2 R :

We state this rule only in thecase of an interior point of anopen interval. However moregeneral versions also exist!

y

Ic x

y = h(x)

y = f (x)y = g (x)

We shall study thetrigonometric functions indetail in Sub-section 4.1.3.The only property that youneed here is that, for any realnumber x, jsin xj 1.

y = x2 sin 1x

x

y

0

These inequalities areobviously true for x¼ 0, aswell as for x 6¼ 0.

Recall that all polynomials arecontinuous on R .


Problem 8 Using the elementary properties of the trigonometric func-

tions, determine whether the following functions are continuous at 0:

(a) f ðxÞ ¼ x sinð1xÞ; x 6¼ 0,

0; x = 0;

�

(b) f ðxÞ ¼ sinð1xÞ; x 6¼ 0,

0; x = 0.

�

Proofs

We now give the proofs of the Combination, Composition and Squeeze Rules.

First, recall the Combination Rules.

Combination Rules for continuous functions

If f and g are functions that are continuous at a point c, then so are:

Sum Rule fþ g;

Multiple Rule lf, for l2R ;

Product Rule fg;

Quotient Rule fg, provided that f(c) 6¼ 0.

The proofs of all these are very similar, and depend on the corresponding

results for sequences. We prove only the Sum Rule.

Proof of the Sum Rule We want to prove that fþ g is continuous at c.

Suppose that f and g have domains S1 and S2, respectively. Then the domain

of fþ g is S1 \ S2, and this set contains the point c.

Thus, we have to show that:

for each sequence {xn} in S1 \ S2 such that xn! c, then

f xnð Þ þ g xnð Þ ! f cð Þ þ g cð Þ:

We know that the sequence {xn} lies in S1 and in S2, and that both functions

f and g are continuous at c. Hence

f xnð Þ ! f cð Þ and g xnð Þ ! g cð Þ;

and it follows from the Sum Rule for Sequences that f(xn)þ g(xn)!f(c)þ g(c), as required. &

Next, recall the Composition Rule.

Composition Rule Let f be continuous on a set S1 that contains a point c,

and g be continuous on a set S2 that contains the point f(c). Then g � f is

continuous at c.

Proof We want to prove that g � f is continuous at c.

Now, we know that g � f is certainly defined on the set S¼ {x : x2 S1 and

f(x)2 S2}, and this set contains the point c.

Thus, we have to show that:

for each sequence {xn} in S such that xn! c, then g f xnð Þð Þ ! g f cð Þð Þ.We know that the sequence {xn} lies in S1, and that f is continuous at c.

Hence, we have that f(xn)! f(c).

You may omit the rest of thisSub-section at a first reading.

138 4: Continuity

We also know that {f(xn)} lies in S2, and that g is continuous at f(c). It

follows that g f xnð Þð Þ ! g f cð Þð Þ, as required. &

Finally, recall the Squeeze Rule.


Let the functions f, g and h be defined on an open interval I, and c2 I. If:

1. g(x) f(x) h(x), for all x2 I,

2. g(c)¼ f(c)¼ h(c),

3. g and h are continuous at c,

then f is also continuous at c.

Proof We have to show that f is continuous at c.

Thus, we have to prove that:

for each sequence {xn} in the domain of f such that xn! c, thenf (xn)! f (c).

Now, since xn! c there is some number X such that xn2 I, for all n>X.

Hence, by condition 1 we have that

g xnð Þ f xnð Þ h xnð Þ; for all n > X: (1)

So, if we now let n!1 and use conditions 2 and 3, we get that

limn!1

g xnð Þ ¼ g cð Þ ¼ f cð Þ and limn!1

h xnð Þ ¼ h cð Þ ¼ f cð Þ: (2)

It follows, from (1), (2) and the Squeeze Rule for sequences, that

limn!1

f xnð Þ ¼ f cð Þ;

as required. &

4.1.3 Trigonometric functions and the exponentialfunction

Trigonometric functions

For the moment we are assuming that you have a knowledge of the trigono-

metric functions arising from your study of trigonometry.

We will prove that the trigonometric functions are continuous on the

whole of their domains. But, first, we need a basic inequality for the sine

function.

Lemma 1 0 sin x x, for 0 x p2:

Proof If x¼ 0, then sin 0¼ 0; so there is an equality.

Suppose next that 0 < x p2, and consider the following diagram, which

represents a quarter circle, centred at the origin, with radius 1.

This follows from thedefinition of a sequenceconverging to c.

Sub-section 2.3.3.

Since the circle has radius 1, the arc AB has length x and the perpendicular AC

has length sin x. Hence

0 < sin x x; for 0 < x p2: &

We can now extend the inequality in Lemma 1 to obtain a more general result.

Theorem 2 The Sine Inequality

sin xj j xj j; for x 2 R :

Proof We saw in Lemma 1 that the inequality holds for 0 x p2:

For x > p2, we have

sin xj j 1 <1

2p < x ¼ xj j;

and so the desired inequality is also true in this case.

Finally, the inequality also holds for x< 0, since

sin �xð Þj j ¼ sin xj j and �xj j ¼ xj j: &

This is the key tool that we need to prove the continuity of the trigonometric

functions.

Theorem 3 The trigonometric functions (sine, cosine and tangent) are

continuous.

Proof To prove that the sine function is continuous at each point c2R , we

need to show that:

for each sequence fxng in R such that xn ! c; then sin xn ! sin c: (3)

We use the formula

sin xn � sin c ¼ 2 cos1

2xn þ cð Þ

� �

sin1

2xn � cð Þ

� �

;

it follows that

For the shortest distance fromthe point A to the line BC isthe perpendicular from Ato BC.

y

1

0–π

y = ⎜x ⎜

y = ⎜sin x ⎜

⎜sin x ⎜≤ ⎜x ⎜, for x ∈π x

Recall that this means they arecontinuous at each point oftheir domains.

This is a standardtrigonometric formula.

140 4: Continuity

sin xn � sin cj j ¼ 2 cos1

2xn þ cð Þ

� ��

�

�

�

�

�

�

�

� sin1

2xn � cð Þ

� ��

�

�

�

�

�

�

�

2 sin1

2xn � cð Þ

� ��

�

�

�

�

�

�

�

21

2xn � cð Þ

�

�

�

�

�

�

�

�

¼ xn � cj j:

Hence, if {xn� c} is null, then {sin xn� sin c} is also null, by the Squeeze

Rule for null sequences. In other words, the result (3) holds.

The continuity of the cosine and tangent functions now follows from the

formulas

cos x ¼ sin xþ 1

2p

� �

and tan x ¼ sin x

cos x;

using the Combination Rules and the Composition Rule. &



f xð Þ ¼ x2 þ 1þ 3 sin x2 þ 1� �

; for x 2 R :

Problem 10 Prove that the function f xð Þ ¼ sin p2

cos x� �

is continuous

on R .

The exponential function x 7! ex

We now prove that the exponential function is continuous on R . But first we

start with some inequalities that we will need to do this.

Theorem 4 The Exponential Inequalities

(a) ex> 1þ x, for x> 0; and ex� 1þ x, for x� 0;

(b) ex 11�x

; for 0 x < 1;

(c) 1þ x ex 11�x

; for xj j < 1:

Proof We prove inequalities (a) and (b) using the exponential series

ex ¼ 1þ xþ x2

2!þ x3

3!þ � � �; for x � 0:

(a) For x> 0, we have x2

2! > 0, x3

3! > 0, and so on. Hence

ex > 1þ x; for x > 0:

It follows from this that ex� 1þ x, for x� 0, since e0¼ 1

(b) For x� 0, we have x2

2! x2, x3

3! x3, and so on. Hence

ex 1þ xþ x2 þ x3 þ � � �:The series on the right is a geometric series; it converges to the sum 1

1�x,

for 0 x 1. Hence

ex 1

1� x; for 0 x < 1:

By taking moduli.

For cos 12

xn þ cð Þ�

�

�

� 1:

Here we use the SineInequality, Theorem 2.

You met this series inSub-section 3.4.1.

Section 4.4.

For example

f xð Þ ¼ 3x7 � 4x2 þ 5;

f xð Þ ¼ 3x

x4 � 6;

f xð Þ ¼ xj j;f xð Þ ¼ x

12; x � 0;

f xð Þ ¼ x13; x 2 R ;

f xð Þ ¼ sin x; cos x; tan x;

f xð Þ ¼ ex:

(c) We have just seen that these inequalities hold for 0 x< 1.

For �1< x< 0, we have that 0<�x< 1, so that, by parts (a) and (b)

1þ �xð Þ e�x 1

1� �xð Þ ; for �1 < x < 0:

By taking reciprocals and reversing the inequalities, we may reformulate

this in the form

1þ x ex 1

1� x; for �1 < x < 0:

This completes the proof of the desired result. &

Remark

Notice that, when x 6¼ 0, the results of Theorem 4 hold with strict inequalities.

We are now able to prove the continuity of the exponential function.

Theorem 5 The exponential function x 7! ex, x2R , is continuous.

Proof To prove that the exponential function is continuous at each point

c2R , we need to show that:

for each sequence fxng in R such that xn ! c; then exn ! ec: (4)

We use the formula exn � ec ¼ ec exn�c � 1ð Þ. If we apply Theorem 4, part (c),

with xn� c in place of x, we obtain

1þ xn � cð Þ exn�c 1

1� xn � cð Þ ; for xn � cj j < 1:

Thus, if {xn� c} is null, then jxn� cj< 1 eventually, and so exn�c ! 1 as

n!1, by the Squeeze Rule for sequences. Hence exn ! ec, so that the desired

result (4) holds. &

Remark

Later we shall give a rigorous definition of the general exponential function

x 7! ax, for x2R and a> 0, and we shall show that all exponential functions are

continuous on R .



f xð Þ ¼ cos x5 � 5x2� �

þ 7e�x2

:

We end this section by listing the various types of functions that we have

found to be continuous on their domains.

Basic continuous functions The following functions are continuous:

� polynomials and rational functions;

� modulus function;

� nth root function;

� trigonometric functions (sine, cosine and tangent);

� the exponential function.

142 4: Continuity

4.2 Properties of continuous functions

4.2.1 The Intermediate Value Theorem

In Section 4.3 we shall prove that the function f(x)¼ x5þ x� 1, x2R , is a

one-function on R . From the graph of f it certainly looks as though f maps R

onto R ; how can we prove this?

For example, is there a value of x for which f(x)¼ 0? In other words,

is there a root of the equation x5þ x� 1¼ 0? The shape of the graph

y¼ x5þ x� 1 certainly suggests that such a number x exists; since f(0)¼�1

and f(1)¼ 1, we would expect there to be some number x in the interval (0, 1)

such that f(x)¼ 0. However we do not have a formula for solving the equation

to find x.

Now, we introduced the concept of continuity on the grounds that it would

enable us to pin down precisely the idea that the graph of a ‘well-behaved’

function does not have ‘gaps’ or ‘jumps’, and this is the key to proving that

such a number x exists.

Theorem 1 Intermediate Value Theorem

Let f be a continuous function on [a, b], and let k be any number lying

(strictly) between f(a) and f(b). Then there exists a number c in (a, b) such

that f(c)¼ k.

This result is illustrated below in the two possible cases:

As the graph above on the right shows, there may be more than one possible value

of c such that f(c)¼ k. All that we claim is that there is at least one such point c.

The requirement in Theorem 1 that f be continuous at each point of [a, b] is

essential. For example, the function

f ðxÞ ¼1x; � 1 x < 0,

0; x ¼ 0,1x; 0 < x 1,

8

<

:

is continuous on [�1, 1] except at 0 – where it is discontinuous. For this

function, f(�1)¼�1 and f(1)¼ 1, but there is no number c in (�1, 1) such that

f ðcÞ ¼ 12

(for example).

The following example shows a typical application of the Intermediate

Value Theorem.

This is one of the mainexistence theorems inAnalysis.

Note that f(a) 6¼ f(b) anda< c< b.

Example 1 Prove that there is a number c in (0, 1) such that c5þ c� 1¼ 0.

Solution Consider the function f(x)¼ x5þ x� 1 on the interval [0, 1]. Then

f is continuous on [0, 1] since it is a basic continuous function, and f(0)¼�1

and f(1)¼ 1.

Since f(0)< 0< f(1), it then follows from the Intermediate Value Theorem

that there is a number c in (0, 1) such that f(c)¼ 0; that is, such that

c5þ c� 1¼ 0. &

If f is a function and c is a real number such that f(c)¼ 0, then c is called a

zero of the function f. We often show that an equation has a solution by proving

that a related continuous function has a zero (by using the Intermediate Value

Theorem with k¼ 0).

Problem 1 Prove that there is a real number c in (0, 1) such that

cos c¼ c.

Problem 2 Let the function f: [0, 1]! [0, 1] be continuous. Prove that

there is a real number c in [0, 1] such that f(c)¼ c.

Proof of Theorem 1 We use the method of repeated bisection.

We shall assume that f(a)< f(b). If, in fact, f(a)> f(b), the proof is very

similar.

First, denote the closed interval [a, b] as [A1, B1]. Then, denote by p the

midpoint of the interval [A1, B1]. Notice that, if f(p)¼ k, then the proof is

complete, since we can take c¼ p.

Otherwise, we define one of the two intervals [A1, p] and [p, B1] to be

[A2, B2] in the following way

A2;B2½ � ¼ A1; p½ �; if f ( p) > 0,

p;B1½ �; if f ( p) < 0.

�

In either case, we obtain:

1. [A2, B2]� [A1, B1];

2. B2 � A2 ¼ 12

B1 � A1ð Þ;3. f(A2)< k< f(B2).

We now repeat this process indefinitely often, bisecting [A2, B2] to obtain

[A3, B3], and so on. If, at any stage, we encounter a bisection point p such

that f(p)¼ k, then the proof is complete.

Otherwise, we obtain a sequence of closed intervals {[An, Bn]} with the


1. [Anþ1, Bnþ1]� [An, Bn], for each n2N;

2. Bn � An ¼ 12

� �n�1B1 � A1ð Þ, for each n2N;

3. f(An)< k< f(Bn), for each n2N .

Property 1 implies that the sequence {An} is increasing and bounded above by

B1¼ b. Hence by the Monotone Convergence Theorem, {An} is convergent;

denote by A its limit. By the Limit Inequality Rule for sequences, we must have

that A b.

Similarly, Property 1 implies that the sequence {Bn} is decreasing and

bounded below by A1¼ a. Hence by the Monotone Convergence Theorem,

For f is a polynomial.

The function f is strictlyincreasing on [0, 1], and so thenumber c must be unique inthis case.

We do not consider thepossibility of complex zerosin this book.

For examplef xð Þ ¼ 1

3þ 1

2x sin p

2x

� �

:

You should at least skim thisproof on a first reading, as themethod is an important one.

p ¼ 12

A1 þ B1ð Þ:


144 4: Continuity

{Bn} is convergent; denote by B its limit. By the Limit Inequality Rule for

sequences, we must have that B� a.

We may then deduce, by letting n!1 in Property 2 and using the

Combination Rules for sequences, that

B� A ¼ limn!1

Bn � limn!1

An ¼ limn!1

Bn � Anð Þ

¼ limn!1

12

� �n�1B1 � A1ð Þ

¼ B1 � A1ð Þ limn!1

12

� �n�1¼ 0:

In other words, the sequences {An} and {Bn} both converge to a common limit.

Denote this limit by c; then we must have a c b.

Now we use the fact that f is continuous at c. It follows that

limn!1

f Anð Þ ¼ f cð Þ and limn!1

f Bnð Þ ¼ f cð Þ:

Then, by Property 3, f(An)< k, for n¼ 1, 2, . . .; so that, by letting n!1, we

obtain f(c) k by the Limit Inequality Rule for sequences. Similarly, we can

deduce from the fact that k< f(Bn) that f(c)� k. It follows that f(c)¼ k.

Finally, notice that, since f(a)< k< f(b), we cannot have either c¼ a or

c¼ b; so that, in fact, a< c< b as required. &

The method of repeated bisection is very powerful, and of wide application

in Mathematics.

Now, we saw above that the continuous function f(x)¼ x5þ x� 1, x2 [0, 1],

has a zero in (0, 1). Now, since f 12

� �

¼ 132þ 1

2� 1 ¼ � 15

32< 0 and f(1)¼ 1 > 0,

we can make the stronger statement that f must have a zero in 12; 1

� �

, by the

Intermediate Value Theorem.

Problem 3 Use the method of repeated bisection to find an interval of

length 18

that contains a zero of the function f(x)¼ x5þ x� 1, x2 [0, 1].

Antipodal points

Two points on the surface of the Earth are called antipodal points if the line

between them passes through the centre of the Earth. (In this sense, antipodal

points are ‘opposite each other’.)

The following result is a rather interesting application of the Intermediate

Value Theorem! It makes the (physically reasonable) assumption that tem-

perature is a continuous function of position on the Earth’s surface.

Theorem 2 Antipodal Points Theorem

There is always a pair of antipodal points on the Equator of the Earth at

which the temperature is the same.

To prove this, we must set up the situation as a mathematical problem.

Let f(�) denote the temperature at a point on the Equator at an angle � radians

East of Greenwich, for 0 �< 2p, and extend f to be defined on [0, 2p] by

That is, c2 [a, b].

Recall that taking limits‘flattens inequalities’.

That is, c2 (a, b).

In fact the same result holdsfor any ‘Great Circle’ on theEarth’s surface – that is, theintersection of a plane throughthe Earth’s centre with theEarth’s surface.

This process is often calledMathematical modelling.

requiring that f(2p)¼ f(0). Then Theorem 2 can be rephrased in the following

equivalent way:

Theorem 20 Antipodal Points Theorem

Let f : [0, 2p]!R be continuous, with f(0)¼ f(2p). Then there exists a

number c in [0, p] such that f(c)¼ f(cþ p).

Proof of Theorem 20 Notice, first, that if f(0)¼ f(p) then we can take c¼ 0.

We now prove the result under the assumption that f(0)< f(p). (The proof in

the case that f(0)> f(p) is very similar.)

Next, define the function g as follows

g �ð Þ ¼ f �ð Þ � f �þ pð Þ; for � 2 0; p½ �:Then, since f is continuous on [0, 2p], it follows that g is continuous on [0, p],


But

g 0ð Þ ¼ f 0ð Þ � f pð Þ < 0

and

g pð Þ ¼ f pð Þ � f 2pð Þ ¼ f pð Þ � f 0ð Þ> 0:

It then follows from the Intermediate Value Theorem, with k¼ 0, that there

exists some number c in (0, p) such that g(c)¼ 0; in other words, such that

f(c)� f(cþ p)¼ 0. This completes the proof. &

4.2.2 Zeros of polynomials

You will have already met a standard method for solving a polynomial

equation of degree 2, and possibly equations of degrees 3 and 4 too. How-

ever there exists no method of solving a general polynomial equation of

degree 5 or higher by means of formulas. How many zeros can a polynomial

equation have?

In fact, a polynomial equation of degree n can have at most n roots in R .

Theorem 3 Fundamental Theorem of Algebra

Let p(x)¼ anxnþ an�1xn�1þ � � � þ a1xþ a0, x2R , where an 6¼ 0. Then

the equation p(x)¼ 0 has at most n roots in R .

Remark

In fact, in its most general form the Fundamental Theorem of Algebra states

that, if p(x)¼ anxnþ an�1xn�1þ � � � þ a1xþ a0, where the coefficients ak may

For, if f(c)¼ f(cþ p), then cand cþ p are antipodal pointswith the same temperature.

Quite often in mathematicswe obtain our results byapplying a standard result tosome cunningly chosenauxiliary function!

These are called quadratic,cubic and quartic equations,respectively.

This is a quite difficult resultto prove!

We do not prove this result,which requires methodsbeyond the scope of this book.

146 4: Continuity

be real or complex numbers and an 6¼ 0, then the equation p(x)¼ 0 has exactly n

roots in C, the set of all complex numbers.

Now, it is sometimes straight-forward to locate zeros of a polynomial if we

have some idea of where to look for them in the first place.

Problem 4 Let p xð Þ ¼ x6 � 4x4 þ xþ 1, x 2 R . Prove that p has a

zero in each of the intervals (�1, 0), (0, 1) and (1, 2).

However to follow this approach for finding zeros (if any) of a given

polynomial, we need first to have an inkling where to look for them. We

usually start by applying the following result, which gives an interval in

which the zeros must lie:

Theorem 4 Zeros Localisation Theorem

Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , be a polynomial. Then

all the zeros of p (if there are any) lie in the open interval (�M, M), where

M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g:

Example 2 Prove that the polynomial p xð Þ ¼ x4 � 2x2 � xþ 1, x 2 R , has

at least two zeros in R .

Solution We will apply the Zeros Localisation Theorem to p, since its

leading coefficient is 1. Since

M ¼ 1þmax �2j j; �1j j; 1j jf g¼ 3;

it follows that all the zeros of p lie in (�3, 3).

We now compile a table of values of p(x), for x¼�3, �2, �1, 0, 1, 2, 3:

x �3 �2 �1 0 1 2 3

p(x) 67 11 1 1 �1 7 61

We find that p(0) and p(1) have opposite signs, as do p(1) and p(2), so p

must have a zero in each of the intervals (0, 1) and (1, 2), by the Intermediate

Value Theorem.

Thus we have proved that p has at least two zeros in R . &

Problem 5 Prove that the polynomial p xð Þ ¼ x5 þ 3x4 � x� 1, x 2 R ,

has at least three zeros in R .

Although we cannot at this stage prove the Fundamental Theorem of

Algebra, nevertheless we can use the Zeros Localisation Theorem and the

Intermediate Value Theorem to prove the following:

Theorem 5 Every real polynomial of odd degree has at least one zero in R .

Thus, for example, for sufficiently large x the polynomial p xð Þ ¼xn þ an�1xn�1 þ � � � þ a1xþ a0, x 2 R is essentially dominated by its leading

term xn, and so is positive for large positive values of x and is negative for large

negative values of x.

Notice that the coefficient ofxn, the leading coefficient of p,has been set as 1.

The proof appears at the endof the sub-section.

Often, it is not necessary tocompute the values of p(x) forall integers x in [�M, M].

In fact this polynomial hasexactly two zeros in R .

For example,p xð Þ ¼ x5 þ 3x4 � x� 1:

Since n is odd.


No corresponding result holds for polynomials of even degree. Thus, for

instance, the polynomial p xð Þ ¼ x2 þ 1 has no zeros in R .

Proofs of Theorems 4 and 5

We supply these proofs which were omitted earlier, so as not to disturb the flow

of the text.

Theorem 4 Zeros Localisation Theorem

Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , be a polynomial. Then

all the zeros of p (if there are any) lie in the open interval (�M, M), where

M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g:

Proof In order to concentrate on the dominating term xn, we define the

function r as follows

r xð Þ ¼ p xð Þxn� 1 ¼ an�1

xþ � � � þ a1

xn�1þ a0

xn; x 2 R � 0f g:

Then, by using the Triangle Inequality, we obtain that, for jxj> 1

r xð Þj j ¼ an�1

xþ � � � þ a1

xn�1þ a0

xn

�

�

�

�

�

�

an�1

x

�

�

�

�

�

�þ � � � þ a1

xn�1

�

�

�

�

�

�þ a0

xn

�

�

�

�

�

�

max an�1j j; . . .; a1j j; a0j jf g � 1

xj j þ � � � þ1

xj jn�1þ 1

xj jn

!

< M � 1

xj j þ � � � þ1

xj jn�1þ 1

xj jn þ � � � !

¼ M �1xj j

1� 1xj j¼ M

xj j � 1:

It follows that, if xj j � M ¼ 1þmax an�1j j; . . .; a1j j; a0j jf g, then r xð Þj j < 1:Now, from the definition of r(x) we see that

p xð Þ ¼ xn 1þ r xð Þð Þ; for xj j � M:

But since r xð Þj j < 1, we certainly have that 1þ r xð Þ > 0. It follows from the

above expression for p(x) in terms of r(x) that p(x) must have the same sign as

xn, for xj j � M:It follows that any zero of p must lie in (�M, M). &

Theorem 5 Every polynomial of odd degree has at least one zero in R .

Proof By dividing the polynomial by its leading coefficient, we may assume

that the polynomial is of the form

p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R , where n is odd.

We can then define M and r(x) as in Theorem 4, and all the statements in the

proof of Theorem 4 then hold here too.

Since n is odd, xn is positive for x�M and negative for x�M. It follows

from the arguments in the proof of Theorem 4 that p(x) must be positive for

x�M and negative for x�M. In particular, p(M)> 0 and p(�M)< 0.

Then, by the Intermediate Value Theorem, p must have a zero in (�M, M).&

You may omit these at a firstreading.

Recall that the coefficient ofxn, the leading coefficientof p, is 1.

Here the modulus of eachcoefficient is at most themaximum value over allcoefficients.

jxj> 1, so that 1xj j < 1:

By summing the geometricseries.

r xð Þj j51, �15r xð Þ51:

So we have certainly got goodvalue from the arguments inthe previous proof!

148 4: Continuity

4.2.3 The Extreme Values Theorem

We now look at whether continuous functions on closed intervals are bounded

or unbounded. It turns out that they are bounded, and we use this fact a great

deal throughout Analysis!

Theorem 6 The Extreme Values Theorem

A continuous function on a closed interval possesses a maximum value and

a minimum value on the interval. In other words, if f is a function that is

continuous on a closed interval [a, b], then there exist points c and d in [a, b]

such that

f cð Þ f xð Þ f dð Þ; for x 2 a; b½ �:

If f is continuous on an interval that is not closed, then it may not be

bounded. For example, the function f xð Þ ¼ 1x, x 2 0; 1ð �, is continuous on (0,1]

but is not bounded above (and so it is not bounded), but it is bounded below

(by 1).

Similarly, if a function is not continuous on a closed interval, then it may not

be bounded. For example, the function

f ðxÞ ¼ 1; x ¼ 0;1x; 0 < x 1;

�

takes the values 1 when x¼ 0 and 1x

when 0< x 1. f is bounded below, by 0,

but is not bounded above (and so it is not bounded).

Problem 6

(a) Determine the maximum and the minimum of the function

f xð Þ ¼ x2, x 2 �1; 2½ �, on [�1, 2]. Specify all points in [�1, 2]

where these are attained.

(b) Determine the maximum and the minimum of the function

g xð Þ ¼ sin x, x 2 0, 2p½ �, on [0, 2p]. Specify all points in [0, 2p]

where these are attained.

Since a function is bounded if and only it is both bounded above and

bounded below, we sometimes use the following version of Theorem 6 in

applications:

Corollary 1 The Boundedness Theorem

A continuous function on a closed interval is bounded. In other words, if f is

a function that is continuous on the closed interval [a, b], then there exist a

number M such that

f xð Þj j M; for x 2 a; b½ �:

On other occasions, we shall find the following consequence of Theorem 6

and the Intermediate Value Theorem useful:

Corollary 2 The Interval Image Theorem

The image of a closed interval under a continuous function is a closed

interval.

We discussed bounds,suprema, infima, maxima andminima of functions earlier, inSub-section 1.4.2.

We prove this at the end of thesub-section.

For it will often be sufficientfor our purposes.

Proof of Theorem 6 Without some systematic approach to studying con-

tinuous functions, we would never be able to prove this theorem, and would be

reduced to much hand-waving arguments and loose assertions. In this case we

shall use our earlier work on sequences, and the following related result:

Lemma 1 Let f be a function defined on an interval I.

(a) If f is bounded above on I and sup f xð Þ: x 2 If g ¼ M, then there exists

some sequence {xn} in I such that f xnð Þ ! M as n!1.

(b) If f is not bounded above on I, then there exists some sequence {xn} in I

such that f xnð Þ!1 as n!1.

Proof

(a) If there is some point, c say, in I for which f(c)¼M, then the constant

sequence {c} has the desired property.

Now suppose that no such point c exists. Then, since sup f xð Þ:fx 2 Ig ¼ M, it follows from the definition of supremum that there is

some point, x1 say, in I for which

f x1ð Þ > M � 1:

Next, since f(x1)<M (from our assumption that there is no point where

f takes the value M), choose a point, x2 say, in I for which

f x2ð Þ > max f x1ð Þ; M � 1

2

�

;

notice, in particular, that this last inequality ensures that x2 6¼ x1.

Continuing this process indefinitely, we obtain a sequence {xn} of

distinct points in I for which

f xnð Þ > max f x1ð Þ; f x2ð Þ; . . .; f xn�1ð Þ; M � 1

n

�

:

Since the sequence M � 1n

� �

converges to M, it follows, by the Squeeze

Rule for sequences, that f xnð Þ ! M as n!1.

(b) It follows from the definition of ‘unbounded above’ that there is some

point, x1 say, in I for which f(x1)> 1. Then, for each n> 1, we can

construct a sequence {xn} of distinct points in I for which

f xnð Þ > max f x1ð Þ; f x2ð Þ; . . .; f xn�1ð Þ; nf g:Since the sequence {n} tends to 1, it follows, by the Squeeze Rule for

sequences, that f(xn)!1 as n!1. &

We are now in a position to prove Theorem 6.

Theorem 6 The Extreme Values Theorem

A continuous function on a closed interval possesses a maximum value and

a minimum value on the interval. In other words, if f is a function that is

continuous on the closed interval [a, b], then there exist points c and d in

[a, b] such that

f cð Þ f xð Þ f dð Þ; for x 2 a; b½ �:

Notice that this is a resultabout sup for any function f onan interval; no assumption ofcontinuity is involved.

A similar result holds forfunctions that are boundedbelow on I or are unboundedbelow on I.


So f x2ð Þ > f x1ð Þ andf x2ð Þ > M � 1

2. Also,

f x2ð Þ < M:

For M � 1n< f xnð Þ < M:

150 4: Continuity

Proof First, we shall assume that f is not bounded above, and verify that this

assumption leads to a contradiction with known facts about f.

It follows from part (b) of Lemma 1 that there exists some sequence {xn} in

[a, b] such that f(xn)!1 as n!1. Then, by the Bolzano–Weierstrass

Theorem, {xn} must contain a convergent subsequence xnkf g; denote by d

the limit of xnkf g. Since all the xnk

lie in [a, b], it follows, by the Limit

Inequality Rule for sequences, that d2 [a, b].

Since f is continuous at d and the sequence xnkf g converges to d, it follows

that f xnkð Þ ! f dð Þ. However, f xnk

ð Þf g is a subsequence of the sequence {f(xn)}

which tends to1, so that f xnkð Þ ! 1:

This is a contradiction. So f must be bounded above on [a, b] after all.

Next, denote by M the number sup f xð Þ: x 2 a; b½ �f g. It follows, from part (a)

of Lemma 1, that there exists some sequence {xn} in [a, b] such that f(xn)!M as

n!1. Then, by the Bolzano–Weierstrass Theorem, {xn} must contain a con-

vergent subsequence xnkf g; denote by d the limit of xnk

f g. Since all the xnklie in

[a, b], it follows, by the Limit Inequality Rule for sequences, that d2 [a, b].

Since f is continuous at d and the sequence xnkf g converges to d, it follows

that f xnkð Þ ! f dð Þ. Therefore, since f xnk

ð Þf g is a subsequence of the sequence

{f(xn)} which tends to M, we have M¼ f(d).

This complete the proof that there exists a point d in [a, b] such that

f xð Þ f dð Þ; for all x 2 a; b½ �:The proof of the existence of a point c in [a, b] such that f(c) f(x), for all

x2 [a, b], is similar; we omit it. &

4.3 Inverse functions

4.3.1 Existence of an inverse function

Let f be the function f(x)¼ 2x, x2R . Then, given any number y in R , we can

find a unique number x ¼ 12

y in the domain of f such that y¼ f(x)¼ 2x.

The inverse function f �1, defined by f�1 yð Þ ¼ 12

y; y 2 R , undoes the

‘effect’ of f; that is, f �1( f(x))¼ x, x2R .

Also, f undoes the effect of f �1; that is, f ( f �1(y))¼ y, y2R .

Not every function has an inverse function! For example, consider the

function

g xð Þ ¼ x2; x 2 R :

Since g(2)¼ 4¼ g(�2), we cannot define g�1(4) uniquely. Thus g fails to

have an inverse function, because it is not one–one. However the function

h xð Þ ¼ x2; x 2 ½0;1Þ;is one–one and has an inverse function

h�1 yð Þ ¼ ffiffiffi

yp; y 2 ½0;1Þ:

In general, if f : A! R is one–one, then, for each point y in the image f(A),

there is a unique point x in A such that f(x)¼ y. Thus f is a one–one correspon-

dence between A and f(A); and so we can define the inverse function f �1 by

f �1( y)¼ x, where y¼ f(x).

It is also possible to proveTheorem 6 by the method ofrepeated bisection.


xnk2 a; b½ � , a xnk

b:

Inheritance Property ofSubsequences, Theorem 5,Sub-section 2.4.4.

M must exist, since we haveshown that the setf xð Þ: x 2 a; b½ �f g is bounded.


Definition Let f : A! R be a one–one function. Then the inverse functionf �1 has domain f(A) and is specified by

f�1 yð Þ ¼ x; where y ¼ f xð Þ; x 2 A:

For some functions f, we can find the inverse function f �1 directly, by

solving the equation y¼ f(x) algebraically to obtain x in terms of y.

Example 1 Prove that the function f defined by f xð Þ ¼ 11�x

; x 2 ð�1; 1Þ,has an inverse function defined on (0,1).

Solution First, we solve the equation y ¼ 11�x

to give x in terms of y. By

simple manipulation, we obtain

y ¼ 1

1� x, x ¼ 1� 1

y:

Now, for each x2 (�1, 1), we have x< 1, and so f xð Þ ¼ 11�x

> 0; thus

f �1,1ð Þð Þ � 0,1ð Þ. Also, for each y2 (0,1), we have

x ¼ 1� 1

y2 �1; 1ð Þ;

and so f is a one–one correspondence between (�1, 1) and (0,1). Hence

f�1 yð Þ ¼ 1� 1

y; y 2 ð0;1Þ: &

Remark

Usually, when defining a function we write x for the domain variable. To

conform with this practice, we may rewrite the inverse function f �1 in

Example 1 as follows

f�1 xð Þ ¼ 1� 1

x; x 2 ð0;1Þ:

The graph y¼ f �1 (x) is obtained by reflecting the graph y¼ f(x) in the line

y¼ x. This reflection interchanges the x- and y-axes.

Proving that a function f is one–one

We have seen that, if f : A! R is one–one, then f has an inverse function f �1

with domain f(A). For the function f considered in Example 1, it is possible to

determine f �1 explicitly by solving the equation y¼ f (x) to obtain x in terms

of y. Unfortunately, it is generally not possible to solve the equation y¼ f (x) in

this way.

Nevertheless, it may still be possible to prove that f has an inverse function

f �1 by showing that f is one–one in some other way. For example, f is one–one

if it is either strictly increasing or strictly decreasing; that is, if f is strictly

monotonic.

Definitions A function f defined on an interval I is:

� increasing on I if x15x2 ) f x1ð Þ f x2ð Þ; for x1; x2 2 I;

� strictly increasing on I if x15x2 ) f x1ð Þ5 f x2ð Þ; for x1; x2 2 I;

� decreasing on I if x15x2 ) f x1ð Þ � f x2ð Þ; for x1; x2 2 I;

y

y = 1 1

0 1 x(– ∞, 1)

1 – x

152 4: Continuity

� strictly decreasing on I if x15x2 ) f x1ð Þ > f x2ð Þ; for x1; x2 2 I;

� monotonic on I if f is either increasing on I or decreasing on I;

� strictly monotonic on I if f is either strictly increasing on I or strictly

decreasing on I.

The most powerful technique for proving that a function f is strictly mono-

tonic is to compute the derivative f 0 of f and examine the sign of f 0(x). We shall

not study differentiation in detail until Chapter 6, so here we consider only

functions which can be proved to be strictly monotonic by manipulating

inequalities rather than by Calculus.

For example, if n2N , then the function f(x)¼ xn, x2 [0, 1), is strictly

increasing; and, if n is odd, the function f(x)¼ xn, x2R , is strictly increasing.

Similarly, if n2N , then the function f(x)¼ x�n, x2 (0, 1), is strictly

decreasing.

Example 2 Prove that the function f xð Þ ¼ x5 þ x� 1; x 2 R ; is one–one.

Solution If x1< x2, then x51< x5

2. Hence

x51 þ x1 � 1 < x5

2 þ x2 � 1;

so f is strictly increasing, and thus one–one. &

Problem 1 Prove that the following functions are one–one:

(a) f xð Þ ¼ x4 þ 2xþ 3; x 2 0;1½ Þ;(b) f xð Þ ¼ x2 � 1

x; x 2 0;1ð Þ:

4.3.2 The Inverse Function Rule

If the function f: A!R is strictly monotonic, then f is one–one, and so f has

an inverse function f �1 with domain f(A). However, it is not always easy to

determine f(A). However, if f is known to be continuous on A, then the

following result simplifies the problem immediately.

Theorem 1 Inverse Function Rule

Let f : I! J, where I is an interval and J is the image f(I), be a function such that:

1. f is strictly increasing on I;

2. f is continuous on I.

Then J is an interval, and f has an inverse function f� 1: J! I such that:

10. f �1 is strictly increasing on J;

20. f �1 is continuous on J.

Remarks

1. The interval I may be any type of interval: open or closed, half-open,

bounded or unbounded.

2. There is a similar version of the Inverse Function Rule with ‘strictly

increasing’ replaced by ‘strictly decreasing’.

We prove the InverseFunction Rule inSub-section 4.3.4.

Example 3 Prove that the function f xð Þ ¼ x5 þ x� 1; x 2 R ; has a contin-

uous inverse function, with domain R .

Solution The domain of f is R , which is an interval.

We have already seen, in Example 2, that f is strictly increasing on R and so

has an inverse function f�1. Since f is a polynomial, it is a basic continuous

function on R . Thus f satisfies the hypotheses of the Inverse Function Rule.

Then it follows from the Rule that the image J¼ f(R) is an interval, and that

the inverse function f�1: J!R is strictly increasing and continuous on J.

It remains to check that the image J is the whole of R . Notice that J¼ f(R)

contains each of the numbers

f nð Þ ¼ n5 þ n� 1; n 2 Z:

Now, J contains each of the intervals [f (�n), f (n)]; and f �nð Þ ¼ �n5�n�1! �1 as n!1, while f nð Þ ¼ n5 þ n� 1!1 as n!1. It follows

that, in fact, J¼ f (R) must be (�1,1)¼R .

Thus f has a continuous inverse function f �1: R!R . &

As the above example shows, when we apply the Inverse Function Rule the

hardest step is to determine the image J¼ f (I). Since J is an interval, it is

sufficient to determine the end-points of J, which may be real numbers or one

of the symbols1 and�1. We must also determine whether or not these end-

points belong to J.

The following diagrams illustrate two examples:

y

d = f (b)

c = f (a)

a x

y = f (x)y = f (x) J = (c, d ]

I = (a, b]I = [a, ∞)a b x

c

y

d

J = [c, d )

Notice that, if a is an end-point of I and a2 I, then c¼ f(a) is the correspond-

ing end-point of J and c2 J.

On the other hand, if a is an end-point of I and a =2 I (this includes the

possibility that a may be 1 or �1), then it is a little harder to find the

corresponding end-point of J. However, it can be shown that, if {xn} is a

monotonic sequence in I and xn! a, then f(xn)! c, and c =2 J: (We prove this

in Sub-section 4.3.4.)

Example 4 Prove that the function f xð Þ ¼ x4 þ 2xþ 3; x 2 0; 1½ Þ, has a

continuous inverse function with domain [3,1).

Solution The domain of f is [0,1), which is an interval.

Also, we know that f is strictly increasing and continuous on [0,1), and so

conditions 1 and 2 of the Inverse Function Rule hold. It follows that the

image J¼ f([0,1)) is an interval, and f has a continuous inverse function f �1:

J! [0,1) which is strictly increasing on J.

It remains to check that the image J is [3,1).

For the end-point 0 of I, we have 02 I, so the corresponding end-point of J is

f(0)¼ 3, and 32 J.

For example:

0; 1ð � has end-points 0

and 1;

1;1½ Þ has end-points 1

and1:Do not let this use of thesymbol1 tempt you to thinkthat1 is a real number.

y

d

J = [c, d )

c = f (a)

ac ∈ J

a ∈ II = [a, ∞)

x

y = f (x)

d = f (b)

{ f (an)}

{an}a

c

b x

y = f (x)

I = (a, b]

y

J = (c, d ]

c ∉ J

a ∉ I

You saw this in Problem 1(a ),above.

154 4: Continuity

The other end-point of I is1, so to find the corresponding end-point of J we

choose the monotonic sequence {n}, which lies in I and tends to infinity. Then

f nð Þ ¼ n4 þ 2nþ 3!1 as n!1. It follows that the corresponding end-

point of J is1. Thus, J¼ [3,1), as required. &

We now summarise the strategy for establishing that a given continuous

function f has a continuous inverse function.

Strategy To prove that f : I! J, where I is an interval with end-points a

and b, has a continuous inverse f �1: J! I:

1. show that f is strictly increasing on I;

2. show that f is continuous on I;

3. determine the end-point c of J corresponding to the end-point a of I as

follows:

� if a2 I, then f(a)¼ c (and c2 J);

� if a =2 I, then f(xn)! c (and c =2 J),

where {xn} is a monotonic sequence in I such that xn! a;

4. determine the end-point d of J corresponding to the end-point b of I,

similarly.

Problem 2 Use the above strategy to prove that the function

f xð Þ ¼ x2 � 1x; x 2 0;1ð Þ, has a continuous inverse function with

domain R .

Hint: Use the result of Problem 1(b) in Sub-section 4.3.1.

4.3.3 Inverses of standard functions

We now use the Inverse Function Rule and the above strategy to define

continuous inverse functions for certain standard functions. Although you

will be familiar with these inverse functions already, we can now prove that

they exist and are continuous. We also remind you of some properties of these

inverse functions.

For each function, we give brief remarks on the four steps of the strategy. In

each case, the continuity of the function f follows directly from the results of

Section 4.1.

The nth root function

We asserted the existence of the nth root function in Sub-section 1.5.2,

Theorem 1. We can at last provide the proof of that assertion!

The nth root function For any positive integer n� 2, the function

f xð Þ ¼ xn; x 2 0;1½ Þ;has a strictly increasing continuous inverse function f�1 xð Þ ¼

ffiffiffi

xnp

with

domain 0;1½ Þ, called the nth root function.

There is a correspondingversion of this strategy for thecase that f is decreasing on I.


In this case, the strategy is easy to apply:

1. f is strictly increasing on [0,1);

2. f is continuous on [0,1);

3. f(0)¼ 0;

4. f(k)¼ kn!1 as k!1.

It follows, from the Inverse Function Rule, that f has a strictly increasing

continuous inverse function f �1: [0,1)! [0,1).

Remark

If n is odd, then the nth root function can be extended to a continuous function

whose domain is the whole of R .

Inverse trigonometric functions

The function sin�1 The function

f xð Þ ¼ sin x; x 2 � 1

2p;

1

2p

� �

;

has a strictly increasing continuous inverse function with domain [�1, 1],

called sin�1.

In this case:

1. the geometric definition of f(x)¼ sin x shows that f is strictly increasing on

�12p; 1

2p

�

;

2. f is continuous on �12p; 1

2p

�

;

3. sin �12p

� �

¼ �1;

4. sin 12p

� �

¼ 1.

It follows that f �12p; 1

2p

��

¼ �1; 1½ �. Hence, by the Inverse Function

Rule, f has a strictly increasing continuous inverse function f�1:�1; 1½ � ! � 1

2p; 1

2p

�

:

The decreasing version of the strategy can be applied similarly to prove that

the cosine function has an inverse, if we restrict its domain suitably.

The function cos�1 The function

f xð Þ ¼ cos x; x 2 0; p½ �;has a strictly decreasing continuous inverse function with domain [�1, 1],

called cos�1.

The domain [0, p] of f is chosen here by convention, so that f is a strictly

monotonic restriction of the cosine function.

Similarly, to form an inverse of the tangent function we must restrict its

domain to � 12p; 1

2p

� �

, since the tangent function is strictly increasing and

continuous on this interval.

f is a basic continuousfunction.

We use {k} rather than {n}here, to avoid using n for twodifferent purposes in the sameexpression.

156 4: Continuity

The function tan�1 The function

f xð Þ ¼ tan x; x 2 � 1

2p;

1

2p

� �

;

has a strictly increasing continuous inverse function with domain R , called

tan�1.

In this case, the image set f � 12p; 1

2p

� ��

is R , because (for example) if {xn}

is a monotonic sequence in � 12p; 1

2p

� �

and xn ! 12p as n!1, then

f xnð Þ ¼ tan xn ¼sin xn

cos xn

!1 as n!1:

Remark

Some texts use arc sin, arc cos and arc tan instead of sin�1, cos�1 and tan�1,

respectively.

Problem 3

(a) Determine the values of sin�1ð 1ffiffi

2p Þ, cos�1 �1

2

� �

and tan�1ffiffiffi

3p� �

.

(b) Prove that cos 2 sin�1 x� �

¼ 1� 2x2, for x 2 ½�1; 1�.Hint: Let y ¼ sin�1 x:

The function loge

We now discuss one of the most important inverse functions.

The function loge The function

f xð Þ ¼ ex; x 2 R ;

has a strictly increasing continuous inverse function f �1 with domain

(0,1), called loge.

In this case:

1. f is strictly increasing on R , since

x1 < x2 ) x2 � x1 > 0

) ex2�x1 > 1 ðsince ex � 1þ x > 1; for x > 0Þ) ex2 > ex1 ;

2. f is continuous on R ;

3. f nð Þ ¼ en !1 as n!1;

4. f �nð Þ ¼ e�n ! 0 as n!1:

It follows that the image of R under f is f Rð Þ ¼ 0;1ð Þ. Hence, by the

Inverse Function Rule, f has a strictly increasing continuous inverse function

f�1: 0;1ð Þ ! R .

Problem 4 Prove that loge xþ loge y ¼ loge xyð Þ, for x; y 2 0;1ð Þ.Hint: Let a ¼ loge x and b ¼ loge y.

Inverse hyperbolic functions

The function sinh�1 The function

f xð Þ ¼ sinh x ¼ 1

2ex � e�xð Þ; x 2 R ;

has a strictly increasing continuous inverse function f �1 with domain R ,

called sinh�1.

In this case:

1. f is strictly increasing on R , since

x1 < x2 ) ex1<ex2

) �e�x1<�e�x2

) ex1 � e�x1< ex2 � e�x2

) sinh x1 < sinh x2;

2. f is continuous on R , by the Combination Rules;

3. f nð Þ ¼ 12

en � e�nð Þ ! 1 as n!1;

4. f �nð Þ ¼ 12

e�n � enð Þ ! �1 as n!1.

It follows that the image of R under f is f (R)¼R . Hence, by the Inverse

Function Rule, f has a strictly increasing continuous inverse function

f �1: R!R .

The function cosh�1 The function

f xð Þ ¼ cosh x ¼ 1

2ex þ e�xð Þ; x 2 0;1½ Þ;

has a strictly increasing continuous inverse function f �1 with domain

[1,1), called cosh�1.

In this case:

1. f is strictly increasing on [0,1), since

x1 < x2 ) sinh x1 < sinh x2

) 1þ sinh2 x1

� �

12< 1þ sinh2 x2

� �

12

) cosh x1 < cosh x2; since cosh2 x ¼ 1þ sinh2 x;

2. f is continuous on [0,1), by the Combination Rules;

3. f 0ð Þ ¼ 1;

4. f nð Þ ¼ 12

en þ e�nð Þ ! 1 as n!1.

It follows that the image of [0,1) under f is f ([0,1))¼ [1,1). Hence, by

the Inverse Function Rule, f has a strictly increasing continuous inverse func-

tion f �1: [1,1)! [0,1).

The strategy can be applied in a similar way to show that f(x)¼ tanh x is

strictly increasing and continuous on R , with f (R)¼ (�1, 1). We omit the

details.

158 4: Continuity

The function tanh�1 The function

f xð Þ ¼ tanh x ¼ sinh x

cosh x; x 2 R ;

has a strictly increasing continuous inverse function f�1 with domain

(�1, 1), called tanh�1.

The inverse hyperbolic functions can be expressed in terms of loge, as the

following example shows.

Example 5 Prove that sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

� �

, for x 2 R .

Solution Let y ¼ sinh�1 x, for x 2 R . Then

x ¼ sinh y ¼ 1

2ey � e�yð Þ;

and so

e2y � 2xey � 1 ¼ 0:

This is a quadratic equation in ey, so that

ey ¼ xffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

:

Since ey> 0, we must choose theþ sign here, so that

y ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

� �

: &

Problem 5 Prove that cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p� �

, for x 2 1;1½ Þ.

4.3.4 Proof of the Inverse Function Rule

We now prove the Inverse Function Rule, and justify our strategy for determin-

ing the domains of inverse functions.


Let f: I! J, where I is an interval and J is the image f (I), be a function

such that:

1. f is strictly increasing on I;

2. f is continuous on I.

Then J is an interval, and f has an inverse function f �1: J! I such that:

10. f �1 is strictly increasing on J;

20. f �1 is continuous on J.

Proof The proof is in four parts:

J¼ f(I) is an interval.

Let y1, y22 f(I), with y1< y2, and let y2 (y1, y2). Now y1¼ f (x1) and

y2¼ f(x2), for some x1, x22 I, with x1< x2 since f is strictly increasing on I.

It follows, by the Intermediate Value Theorem, that there is some number

x2 (x1, x2) for which f(x)¼ y. Hence y2 f(I).

It follows that f(I) is an interval.

You may omit the proofs inthis sub-section at a firstreading, but you should atleast read the statement ofTheorem 2 below.

The inverse function f �1: J! I exists.

The function f is strictly increasing and is therefore one–one; also, f maps

I onto J, so the function f �1: J! I exists, by definition.

f �1 is strictly increasing on J.

We have to show that

y1 < y2 ) f�1 y1ð Þ < f�1 y2ð Þ; for y1; y2 2 J:

Notice that

f�1 y1ð Þ � f�1 y2ð Þ ) f f�1 y1ð Þ� �

� f f�1 y2ð Þ� �

) y1 � y2:

Hence y1 < y2 ) f�1 y1ð Þ < f�1 y2ð Þ, as required.

f �1 is continuous on J.

Let y2 J, and (for simplicity) assume that y is not an end-point of J. Then

y¼ f(x), for some x2 I, and we want to prove that

yn ! y) f�1 ynð Þ ! f�1 yð Þ ¼ x:

Thus, we want to deduce that:

for each "> 0, there is some number X such that

x� " < f�1 ynð Þ < xþ "; for all n > X: (1)

Since f is strictly increasing, we know that

f x� "ð Þ < f xð Þ < f xþ "ð Þ;also, since yn! y¼ f(x), there is some number X such that

f x� "ð Þ < yn < f xþ "ð Þ; for all n > X:

If we then apply the strictly increasing function f �1 to these inequalities, we

obtain (1), as required.

This completes the proof of the Inverse Function Rule. &

In fact, a careful check of the first part of the above proof shows that a slight

change enables us to prove the following result that is of interest in its own right.

Theorem 2 The image of an interval under a continuous function is also

an interval.

Proof Let f be a continuous function on an interval I. We shall assume that f

is non-constant on I, since otherwise the result is trivial.

Let y1, y22 f(I), with y1< y2, and let y2 (y1, y2). Now y1¼ f(x1) and

y2¼ f(x2), for some x1, x22 I, and x1 6¼ x2. Let I0 denote the interval with end-

points x1 and x2. It follows, by applying the Intermediate Value Theorem to the

function f on I0, that there is some number x2 I0 for which f(x)¼ y. Hence

y2 f(I).

It follows that f(I) is an interval. &

Finally, we justify our strategy for determining the end-points of J¼ f(I).

In Corollary 2,Sub-section 4.2.3, we showedthat the image of a closedinterval under a continuousfunction is a closed interval.

We must introduce thesymbol I0 since we do notknow in general whetherx1< x2 or x2< x1.

160 4: Continuity

Strategy for finding the end-points of J If a is an end-point of the

interval I, then we can find the corresponding end-point c of J as follows:

� if a2 I, then f (a)¼ c (and c2 J);

� if a =2 I, then f(xn)! c (and c =2 J),

where {xn} is a monotonic sequence in I such that xn! a.

Proof For simplicity, we shall suppose that a is the left end-point of I.

y

c = f (a)

d = f (b)

{ f(xn)}

{xn}

a b x

y = f (x)y = f (x)

J = [c, d )

I = [a, ∞)I = (a, b]

a x

d

y

J = (c, d ]

c ∉ J

a ∉ I

c ∈ J

a ∈ I

If a2 I, then c¼ f (a)2 J and

f xð Þ � f að Þ ¼ c; for x 2 I;

and so c is the corresponding left end-point of J.

On the other hand, if a =2 I, then we select any decreasing sequence {xn} in I

such that xn! a as n!1. Then {f(xn)} is also decreasing; it therefore

follows, by the Monotonic Sequence Theorem for sequences, that

f xnð Þ ! c as n!1; (2)

where c is a real number or �1.

Now, f(xn)2 J for n¼ 1, 2, . . .; therefore, since J is an interval, it follows that

f xnð Þ; f x1ð Þ½ � � J; for n ¼ 1; 2; . . .:

Hence, by (2), we have that

c; f x1ð Þð � ¼[

1

n¼1

f xnð Þ; f x1ð Þ½ � � J:

To deduce, finally, that c is the left end-point of J, we need to show that c =2 J.

Suppose that in fact c2 J. Then c¼ f(x), for some x2 I; it follows that

f xð Þ ¼ c < f xnð Þ; for n ¼ 1; 2; . . .¼) x < xn; for n ¼ 1; 2; . . .¼) x a:

Thus x =2 I. This contradiction completes the proof. &

4.4 Defining exponential functions

4.4.1 The definition of ax

Earlier, we looked at the definition of the irrational numberffiffiffi

2p¼ 1:4142 . . ..

We have also defined ax for a> 0 when x is rational, but we have not yet

defined ax when x is irrational.

Here we consider only theend-point a of I; a similarresult holds for the otherend-point b of I.

This was Theorem 2 in Sub-section 2.5.1: if the sequence{an} is monotonic, then either{an} is convergent oran!1.

Taking limits ‘flattens’inequalities.

Sub-section 1.1.1.

Sub-section 1.5.3.

One possible method for defining the irrational power 2ffiffi

2p

involves the

decimal representation offfiffiffi

2p

. Each of the truncations offfiffiffi

2p¼ 1:4142 . . . is

a rational number, and the corresponding rational numbers

21; 21:4; 21:41; 21:414; 21:4142; . . . (1)

are defined, and form an increasing sequence which is bounded above by

22¼ 4 (for example). Hence, by the Monotone Convergence Theorem for

sequences, the sequence (1) is convergent, and the limit of this sequence can

be taken as the definition of 2ffiffi

2p:

We can define ax for a> 0 and x2R similarly. However, with this definition

it is difficult to establish the properties of ax, such as the Exponential Laws. It is

more convenient to define ax by using the exponential function x 7! ex, whose

properties we have already discussed.

Recall that

ex ¼ limn!1

1þ x

n

� �n

¼X

1

n¼0

xn

n!; for x 2 R ;

ex ¼ e�xð Þ�1; for x 2 R ;

and

exþy ¼ ex � ey; for x; y 2 R : (2)

Recall too that the function x 7! ex is strictly increasing and continuous; and

hence, by the Inverse Function Rule, it has a strictly increasing continuous

inverse function x 7! loge x, x2 (0,1). Thus we have

loge exð Þ ¼ x; for x 2 R ;(3)

and

eloge x ¼ x; for x 2 0;1ð Þ:

Now let a¼ ex and b¼ ey. Then, by equation (2), we have

ab ¼ exey ¼ exþy;

so that, by equation (3), we obtain loge(ab)¼ logeaþ logeb, for a, b2 (0,1).

We deduce that, if a> 0 and n2N , then

loge anð Þ ¼ n loge a;

and so

an ¼ en loge a:

With a little more manipulation, we can show that the equation

ax ¼ ex loge a

is true for each rational number x. This suggests that we define ax, for a> 0 and

x irrational, by means of this equation.

Definition If a> 0, then ax ¼ ex loge a; for x 2 R :

For example, 2x ¼ ex loge 2 for x2R , so that the graph y¼ 2x is obtained from

the graph y¼ ex by a scaling in the x-direction with scale factor 1loge 2

.

Sub-section 2.5.1.

Section 3.4.

Sub-section 3.4.3.

162 4: Continuity

This relationship between the graphs y¼ ex and y¼ 2x suggests that the

function x 7! 2x must also be continuous. In fact, we can deduce the continuity

of the function x 7! 2x ¼ ex loge 2; x 2 R , from the continuity of the function

x 7! ex, by using the Multiple Rule and the Composition Rule for continuity.

Remark

Since x 7! 2x is continuous, we deduce that the sequence

21; 21:4; 21:41; 21:414; 21:4142; . . .;

where 1, 1.4, 1.41, 1.414, 1.4142, . . . are truncations offfiffiffi

2p¼ 1:4142 . . .,

does converge to 2ffiffi

2p

, and so both definitions of 2ffiffi

2p

agree.

In general, we have the following result.

Theorem 1 If a> 0, then the function x 7! ax ¼ ex loge a; x 2 R , is

continuous.

Problem 1 Prove that the following functions are continuous:

(a) f(x)¼ x�, where x2 (0,1) and �2R ;

(b) f(x)¼ xx, where x2 (0,1).

4.4.2 Further properties of exponentials

Our definition of ax enables us to give straight-forward proofs of the following

Exponent Laws.

Exponent Laws

� If a, b> 0 and x2R , then axbx ¼ (ab)x.

� If a> 0 and x, y2R , then axay¼ axþ y.

� If a> 0 and x, y2R , then axð Þy ¼ axy.

You first met these laws inSub-section 1.5.3, but therex, y2Q .


For exampl e, to prove the final Exponent Law, notice that, from the defini tion

of ax, we have loge( ax) ¼ x log e a; so that

axð Þy ¼ e y log e a xð Þ

¼ e xy loge a ¼ axy :

Thus manipul ations such as

ffiffiffi

2p ffiffi

2p� �

ffiffi

2p

¼ffiffiffi

2p ffiffi

2p�ffiffi

2pð Þ ¼

ffiffiffi

2p� �2

¼ 2

are inde ed justi fied.

Problem 2 Prove that, if a > 0 and x, y 2R , then axay ¼ a x þ y.

Finally, our defini tion of a x enables us to prove the rule for rearra nging

inequal ities by taking powers.

Rule 5 For any non-ne gative a, b 2R , and any p > 0, a 0ð Þ

, e p log e a < e p log e b

, ap < bp :

The followi ng exampl e illus trates the use of Rule 5.

Exampl e 1 Determine which of the numbers e p and pe is g reater.

Solut ion We use the inequal ity

e x > 1 þ x ; for x > 0 :

Applying this ineq uality with x ¼ pe

� �

� 1, we obta in

epeð Þ� 1 > 1 þ p

e

� �

� 1� �

¼ pe¼ pe � 1 ;

then, by multipl ying thr ough by the positive factor e , we obtain that epe > p.

Finally, by applying Rule 5 with p¼ e, we obtain ep> pe. &

Problem 3 Prove that ex> xe, for x> e.

4.5 Exercises

Section 4.1

1. Use the appropriate rules, together with the list of basic continuous

functions, to prove that the following functions are continuous:

You met this Rule in Sub-section 1.2.1.

This was one of theExponential Inequalities:part (a) of Theorem 4 inSub-section 4.1.3.

164 4: Continuity

(a) f xð Þ ¼ exp sin x2 þ 1ð Þð Þ; x 2 R ;

(b) f xð Þ ¼ effiffi

xpþ x5; x 2 0;1½ Þ:

2. Determine whether the following functions are continuous at 0:

(a) f xð Þ ¼ sin x sinð1xÞ; x 6¼ 0,

0; x ¼ 0;

�

(b) f xð Þ ¼1xsin 1

x

� �

; x 6¼ 0,

0; x ¼ 0:

�

3. Prove that each of the following sequences is convergent, and determine its

limit:

(a) sin e1n � 1

� �n o

; (b) cos 12n

� �� 12

n o

:

4. Let f be defined on an open interval I, and c2 I. Prove that, if f is continuous

at c and f(c) 6¼ 0, then there is an open interval J� I such that c2 J and

f(x) 6¼ 0, for any x2 J.

5. Determine the points where the function f xð Þ¼ 1; x ¼ 0; 1;xþ 2x½ �; 0 < x < 1;

�

is

(a) continuous on the left; (b) continuous on the right;

(c) continuous.

6. Prove that the following function is continuous on R

f xð Þ ¼�1; x � 1

2p;

sin x; � 12p < x < 1

2p;

1; x � 12p:

8

<

:

7. Determine at which points the following function is continuous

f xð Þ ¼ x; �1 x < 0;ex; 0 x 1:

�

8. Write down examples of functions with the following properties:

(a) f and g are discontinuous on R , but g� f is continuous on R ;

(b) f and g are discontinuous on R , but fþ g and fg are continuous on R ;

(c) f is continuous on R � 1; 12; 1

3; 1

4; . . .

� �

but discontinuous at

1; 12; 1

3; 1

4; . . .

� �

:

Section 4.2

1. The function f is continuous on (0, 1), and takes every real value at most

once. Use the Intermediate Value Theorem to prove that f is strictly mono-

tonic on (0, 1).

2. Give examples of functions f continuous on the half-open interval [0, 1) in

R , to show that f ([0,1)) can be open, closed or half-open.

3. Prove that each of the following polynomials has the stated number of (real)

zeros:

(a) p xð Þ ¼ x4 � 4x3 þ 3x2 þ 2x� 1; 4 zeros;

(b) p xð Þ ¼ 3x3 � 8x2 þ xþ 3; 3 zeros:

4. Prove that the function f xð Þ ¼ x� sin x� 23p; x 2 R , has a zero in

23p; 5

6p

� �

.

We shall use this resultin Chapter 6.

4.5 Exercises 165

5. Using the Zeros Localisation Theorem and the Extreme Values Theorem,

prove that every polynomial of even degree n

p xð Þ ¼ anxn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R ;

where an 6¼ 0, has a minimum value on R .

6. Write down an example of a function f(x), x2 (0, 1), that is continuous on

(0, 1) and for which f ((0, 1))¼ (0, 1), but such that there is no point c in

(0, 1) for which f (c)¼ c.

Section 4.3

1. For each of the following functions, prove that it has a continuous inverse

function and determine the domain of that inverse function:

(a) f xð Þ ¼ x3 þ 1� 1x2 ; x 2 0;1ð Þ;

(b) f xð Þ ¼ 1

1þx3ð Þ2 ; x 2 �1;1ð Þ:2. Determine whether each of the following statements is true:

(a) sin sin�1 x� �

¼ x, for x 2 ½�1; 1�;(b) sin�1 sin xð Þ ¼ x, for x 2 R .

3. (a) Prove that tan�1 xþ tan�1 y ¼ tan�1 xþy1�xy

� �

, provided that

tan�1 xþ tan�1 y lies in � 12p; 1

2p

� �

.

(b) Use the result in part (a) to evaluate tan�1 12

� �

þ tan�1 13

� �

.

This result is closely related toProblem 2 in Sub-section 4.2.1, so you mightlike to look back at thatproblem and compare the two.

This is known as the AdditionFormula for tan�1.

166 4: Continuity

5 Limits and continuity

In Chapter 4 we made some progress in pinning down the idea of ‘a well-

behaved function’ in precise terms. Using our earlier work on sequences we

defined what was meant by a function being continuous: roughly speaking, its

graph has no jumps or gaps.

However, a number of functions arise quite naturally in mathematics where

we need to handle functions that are already defined near some particular point,

but that are not defined at the point itself.

For example, the function

f xð Þ ¼ sin x

x; x 6¼ 0;

which arises when we examine the question of whether the sine function is

differentiable. Clearly from their graphs, the behaviour of f differs significantly

from the behaviour of the function

g xð Þ ¼ sin1

x

� �

; x 6¼ 0;

that you have already met. It is possible to assign a value (namely, 1) to f at 0 so

that this extension of the domain of f makes f continuous on a domain that is an

interval. On the other hand, it is not possible to assign a value to g at 0 so that this

extension of the domain of g makes g continuous on a domain that is an interval.

In Section 5.1, we discuss limits of functions, and show that the existence of

this helpful value 1 for f can be stated as limx!0

sin xx¼ 1. The concept of a limit of a

function is closely related to that of a continuous function, and many of the

rules for calculating limits are similar to those for continuity. We also discuss

one-sided limits.

In Section 5.2, we discuss the behaviour of functions near asymptotes of

their graphs. In particular, we prove that, if n2Z, then limx!1

xne�x ¼ 0:

Next, in Section 5.3, we introduce a slightly different definition of the limit

limx!c

f xð Þ of a function f at a point c. Instead of using sequences tending to c to

define the limit of f at c, we define limx!c

f xð Þ directly in terms of inequalities

involving x and f (x), and we verify that this new definition is completely

equivalent to the earlier definition. We also illustrate the changes to proofs of

results about limits that the new definition involves.

In Section 5.4, we introduce a definition for the continuity of a function f at a

point c in terms of inequalities involving x and f (x), rather than in terms of the

behaviour of f on sequences that tend to c. We verify that this new definition is

completely equivalent to the earlier definition of continuity, and illustrate the

changes to proofs of results about continuity that the new definition involves.

Finally, in Section 5.5, we introduce a concept that will be extremely

powerful in your further study of Analysis; namely, that of uniform continuity.

1

–1

y = sin 1x

At first sight this newdefinition will seem morecomplicated. Howeverthroughout the rest of thebook we shall see just howpowerful this new definitionturns out to be!

167

Before starting on Sections 5.3 and 5.4, you may find it useful to quickly

revise the material in Sections 1.2 and 1.3 on inequalities.

5.1 Limits of functions

5.1.1 What is a limit of a function?

The graph of the function

f xð Þ ¼ sin x

x; x 6¼ 0;

shows that, if x takes values which are ‘close to’ but distinct from 0, then f(x)

takes values which are ‘close to’ 1. The closer that x gets to 0, the closer f(x)

gets to 1. We now pin down this idea precisely.

First, we need the following fact.

Theorem 1 If {xn} is a null sequence whose terms are non-zero, then

sin xn

xn

!1 as n!1:

Proof First we deduce from the inequality

sin x � x; for 0 < x <p2;

that

sin x

x� 1; for 0 < x <

p2:

Next, we need to use the formula for the area of a sector of a disc of radius 1

in Figure (a), below. Compare the area of this sector with the area of the

triangle in Figure (b), below.

We find that

x � tan x; for 0 < x <p2:

Since tan x ¼ sin xcos x

and cos x > 0 for 0 < x < p2, we obtain

cos x � sin x

x; for 0 < x <

p2:

Combining this inequality with our earlier upper estimate for sin xx

, we find that

cos x � sin x

x� 1; for 0 < x <

p2:

‘may’ means ‘will’!

That is, limn!1

sin xn

xn¼ 1.

Lemma 1, Sub-section 4.1.3.

We discussed the area andperimeter of a disc of radius 1in Sub-section 2.5.4 and inExercise 4 on Section 2.5 inSection 2.6.

168 5: Limits and continuity

In fact, this inequality holds for 0 < jxj < p2, since

cos �xð Þ ¼ cos x andsin �xð Þ�xð Þ ¼

sin x

x:

Now, if {xn} is any null sequence with non-zero terms, then the terms xn

must eventually satisfy the inequality xnj j < p2, and so there is some number X

such that

cos xn �sin xn

xn

� 1; for n > X:

But cos xn! 1 as n!1, since the cosine function is continuous at 0 and

cos 0¼ 1. Hence, by the Squeeze Rule for sequences

sin xn

xn

! 1: &

This behaviour of sin xcos x

near 0 is an example of a function f tending to a limit as x

tends to a point c.

To define this concept, we need to ensure that the function f is defined near

the point c, but not necessarily at the point c itself. We first introduce the idea

of a punctured neighbourhood of c.

Definitions A neighbourhood of a point c of R is an open interval that

contains the point c, and a punctured neighbourhood of a point c of R is a

neighbourhood of c from which the point c itself has been deleted.

For example, the sets (0, 9), (1,1) and R are neighbourhoods of the point 2,

and (1, 2)[ (2, 5) and (�1, 2)[ (2, 4) are punctured neighbourhoods of the

point 2. In general, a neighbourhood of c is an interval of the form (c� r, cþ s),

for some r, s> 0, and a punctured neighbourhood of c is the union of a pair of

intervals (c� r, c)[ (c, cþ s), for some r, s> 0. In practice, we often choose as

a neighbourhood of c an open interval (c� r, cþ r), r > 0, with centre at c; and

as a punctured neighbourhood of c the union (c� r, c)[ (c, cþ r) of two open

intervals of equal length.

We now define the limit of a function in terms of limits of sequences.

Definition Let the function f be defined on a punctured neighbourhood N

of a point c. Then f (x) tends to the limit ł as x tends to c if:

for each sequence xnf g in N such that xn ! c; then f xnð Þ ! ‘: (1)

We write this as either ‘limx!c

f xð Þ ¼ ‘’ or ‘f (x)! ‘ as x! c’.

Let us check that this definition holds for f ðxÞ ¼ sin xx

at 0. This function f is

defined on the domain R � {0}, and so in particular on every punctured

neighbourhood of 0. We have just seen that the statement (1) holds; it follows

then that

limx!0

sin x

x¼ 1:

Since the definition of the limit of a function involves the limit of sequences,

we can use our various Combination Rules for sequences to determine the

limits of many functions.

Take " ¼ p2

in the definition ofnull sequence.

So the ‘puncture’ is at c.

These choices are simplymatters of convenience!

Note that xn 6¼ c, for any n.

For example, f is defined on(�1, 0)[ (0, 1).

Example 1 Prove that each of the following functions tends to a limit as x

tends to 2, and determine these limits:

(a) f xð Þ¼ x2 � 4x� 2

; (b) f xð Þ¼ x3� 3x� 2x2� 3xþ 2

:

Solution

(a) First, notice that f is defined on every punctured neighbourhood of 2.

Next, notice that

f xð Þ ¼ x2 � 4

x� 2¼ xþ 2; for x 6¼ 2:

Thus, if {xn} is any sequence that lies in some punctured neighbourhood

N of 2 and xn! 2, then

f xnð Þ ¼ xn þ 2! 4 as n!1;

by the Sum Rule for sequences. It follows that

limx!2

x2 � 4

x� 2¼ 4:

(b) The function

f xð Þ ¼ x3 � 3x� 2

x2 � 3xþ 2¼ x� 2ð Þ x2 þ 2xþ 1ð Þ

x� 2ð Þ x� 1ð Þ

has domain R � {1, 2}, and so f is defined on the punctured neighbourhood

N¼ (1, 2)[ (2, 3) of 2.

Next, notice that

f xð Þ ¼ x3 � 3x� 2

x2 � 3xþ 2¼ x2 þ 2xþ 1

x� 1;

for x 2 N ¼ ð1; 2Þ [ ð2; 3Þ:Thus, if {xn} is any sequence that lies in N¼ (1, 2)[ (2, 3) and xn! 2, then

f xnð Þ ¼x2

n þ 2xn þ 1

xn � 1! 4þ 4þ 1

2� 1¼ 9 as n!1;

by the Combination Rules for sequences. It follows that

limx!2

x3 � 3x� 2

x2 � 3xþ 2¼ 9: &

Later in this section we give several further techniques for calculating limits.

Our next example illustrates how to prove that a limit does not exist.

Example 2 Prove that each of the following functions does not tend to a limit

as x tends to 0:

(a) f ðxÞ ¼ 1x; x 6¼ 0; (b) f ðxÞ ¼ sin 1

x

� �

; x 6¼ 0; (c) f xð Þ ¼ffiffiffi

xp; x� 0:

Solution

(a) The function f is defined on the punctured neighbourhood N¼ (�2, 0)[ (0, 2)

of 0. The null sequence 1n

� �

lies in N and tends to 0, but f 1n

� �

¼ n!1.

Hence f does not tend to a limit as x tends to 0.

(b) The function f is defined on the punctured neighbourhood N¼ (�2, 0)[ (0, 2)

of 0.

We can cancel x� 2, sincex 6¼ 2.

For example,N¼ (1, 2)[ (2, 3); any otherpunctured neighbourhood of2 would serve our purposeequally well.

Of course any smallerpunctured neighbourhood of 2would serve our purposeequally well.

Sub-sections 5.1.2 and 5.1.3.

Any other puncturedneighbourhood of 0 wouldserve equally well here, ofcourse.


To prove that f(x) does not tend to a limit as x tends to 0, we choose two null

sequences {xn} and {x0n} that lie in N, such that

f xnð Þ ! 1 and f x0n� �

! �1:

Since sin 2npþ 12p

� �

¼ 1 and sin 2npþ 32p

� �

¼ �1 for n2Z, we can choose

xn ¼1

2npþ 12p

and x0n ¼1

2npþ 32p; for n ¼ 0; 1; 2; . . .:

Hence f does not tend to a limit as x tends to 0.

(c) The function f xð Þ¼ffiffiffi

xp

has domain [0,1), and so f is not defined on any

punctured neighbourhood of zero. Hence f does not tend to a limit as x

tends to zero. &

We collect these techniques together in the form of a strategy.

Strategy To show that limx!c

f xð Þ does not exist:

1. Show that there is no punctured neighbourhood N of c on which f is

defined;

OR

2. Find two sequences {xn} and {x0n} (in some punctured neighbourhood N

of c) which tend to c, such that { f (xn)} and { f (x0n)} have different limits;

OR

3. Find a sequence {xn} (in some punctured neighbourhood N of c) which

tends to c, such that f (xn)!1 or f (xn)!�1.

Problem 1 Determine whether the following limits exist:

(a) limx!0

x2þxx

; (b) limx!0

xj jx:

5.1.2 Limits and continuity

Consider the function

f xð Þ ¼ 1; x 6¼ 0;0; x ¼ 0:

Does this function tend to a limit as x tends to zero; and, if it does, what is the

limit?

Certainly, f is defined on any punctured neighbourhood of zero, since the

domain of f is R . Also, if {xn} is any null sequence with non-zero terms, then

f (xn)¼ 1, for n¼ 1, 2, . . ., so that f (xn)! 1 as n!1. It follows that limx!0

f xð Þ ¼ 1.

This example serves to emphasise that the value of a limit limx!c

f xð Þ has nothing

to do with the value of f (c) – even if f happens to be defined at the point c.

However, if f is defined at c, and f is also continuous at c, then the only

possible value for limx!c

f xð Þ is f (c).

We put these observations together in the following result.

Theorem 2 Let the function f be defined on an open interval I, with c 2 I.

Then

f is continuous at c, limx!c

f xð Þ ¼ f cð Þ:

In particular, the terms xn and

x 0n will be non-zero.

Notice that I is aneighbourhood of c.


Using Theorem 2 and our knowledge of continuous functions, we can eval-

uate many limits rather easily.

For example, to determine limx!2

3x5 � 5x2 þ 1� �

notice that the function

f (x)¼ 3x5� 5x2þ 1 is defined on R and is continuous at 2, since f is a poly-

nomial. Hence, by Theorem 2

limx!2

3x5 � 5x2 þ 1� �

¼ f 2ð Þ¼ 77:

We saw earlier that the following functions are continuous at 0

f xð Þ ¼ x sin 1x

� �

; x 6¼ 0;0; x ¼ 0;

and f xð Þ ¼ x2 sin 1x

� �

; x 6¼ 0;0; x ¼ 0:

It therefore follows from Theorem 2 that

limx!0

x sin1

x

� �

¼ 0 and limx!0

x2 sin1

x

� �

¼ 0:

On the other hand, we saw in part (b) of Example 2 that

limx!0

sin1

x

� �

¼ 0 does not exist:

It follows from Theorem 2 that, no matter how we try to extend the domain

of f ðxÞ ¼ sin 1x

� �

to include x¼ 0, we can never obtain a continuous function.

Remark

If f is defined on an open interval I, with c 2 I, and limx!c

f xð Þ exists but

limx!c

f xð Þ 6¼ f cð Þ, then f is said to have a removable discontinuity at c. For

example, the function f xð Þ ¼ x sin 1x

� �

; x 6¼ 0;3; x ¼ 0;

has a removable discontinuity

at 0.

Problem 2 Use Theorem 1 to determine the following limits:

(a) limx!2

ffiffiffi

xp

; (b) limx!p

2

ffiffiffiffiffiffiffiffiffi

sin xp

; (c) limx!1

ex

1þx

In the remainder of this chapter we shall frequently use Theorem 2. When the

function f is one of our basic continuous functions, however, we shall not

always refer to the theorem explicitly. For example, f (x)¼ x2þ 1 and g(x)¼sin x are basic continuous functions, and so we can write lim

x!2x2 þ 1ð Þ¼ 5 and

limx!0

sin x ¼ 0 without further explanation.

5.1.3 Rules for limits

As you might expect from your experience with sequences, series and con-

tinuous functions, we often find limits by using various rules. First, we state the

Combination Rules.


If limx!c

f xð Þ¼ ‘ and limx!c

g xð Þ¼ m, then:

Sum Rule limx!c

f xð Þ þ g xð Þð Þ ¼ ‘þ m;

Multiple Rule limx!c

lf xð Þ¼l‘; for l 2 R ;

Polynomials are basiccontinuous functions: seeSub-section 4.1.3.

Problem 8 and Example 5,Sub-section 4.1.2,respectively.

Sub-section 5.1.1.

This means that, if weredefine the value of f just at citself, the resulting function isthen continuous at c.

Basic continuous functions:

� polynomials andrational functions;

� modulus function;� nth root function;� trigonometric functions

(sine, cosine and tangent);� the exponential function.


Product Rule limx!c

f xð Þg xð Þ ¼ ‘m;

Quotient Rule limx!c

f xð Þg xð Þ ¼ ‘

m; provided that m 6¼ 0.

For example, since limx!0

sin xx¼ 1 and lim

x!0x2 þ 1ð Þ ¼ 1; we have

limx!0

sin x

xþ 2 x2 þ 1� �

� �

¼ 1þ 2� 1 ¼ 3:

The proofs of these rules are all simple consequences of the corresponding

results for sequences. We illustrate this by proving just one rule.

Proof of the Sum Rule Let the functions f and g be defined on a punctured

neighbourhood N of a point c. We want to show that:

for each sequence xnf g in N such that xn ! c, then f xnð Þ þ g xnð Þ ! ‘þ m:

Now, since limx!c

f xð Þ ¼ ‘, we know that, for any sequence {xn} in N for which

xn! c, then f (xn)! ‘. Also, since limx!c

g xð Þ¼m, we know that, as xn! c, then

g(xn)!m.

It follows by the Sum Rule for sequences that, as xn! c, then

f (xn)þ g(xn)! ‘þm. This completes the proof. &

We can also use the following Composition Rule.

Theorem 4 Composition Rule

If limx!c

f xð Þ¼ ‘ and limx!‘

g xð Þ ¼ L, then limx!c

g f xð Þð Þ ¼ L, provided that:

EITHER f (x) 6¼ ‘ in some punctured neighbourhood of c;

OR g is defined at ‘ and is continuous at ‘.

Remarks on the Composition Rule

(a) In any particular case, the Composition Rule may be FALSE if we do not

ensure that one or other of the two provisos holds! For example, if

f xð Þ ¼ 0; x 2 R ; and g xð Þ ¼sin x

x; x 6¼ 0;

0; x ¼ 0;

then

limx!0

f xð Þ ¼ 0 and limx!0

g xð Þ ¼ 1; but limx!0

g f xð Þð Þ ¼ limx!0

0ð Þ ¼ 0:

(b) Suppose the first proviso ‘f (x) 6¼ ‘ in some punctured neighbourhood of c’

in Theorem 4 holds. Then we know that, for each sequence {xn} in a

punctured neighbourhood N of c for which xn! c, f (xn) does not take the

value ‘ but must lie in a punctured neighbourhood of ‘. In this case, the

desired result will follow from the facts that (i) the sequence {f (xn)}

converges to ‘, (ii) f (xn) lies in a punctured neighbourhood of ‘, and

(iii) limx!‘

g xð Þ ¼ L:

(c) Suppose the second proviso ‘g is defined at ‘ and is continuous at ‘’in Theorem 4 holds. In this case, the desired result will follow from

There are no new ideas in theproof. All that we have to do isto set up things so that we canuse the Sum Rule forsequences.

Note that the second limit is alimit as x! ‘, not as x! c.

Here we take c¼ 0, ‘¼ 0 andL¼ 1.

So, in this case,limx!0

g f xð Þð Þ 6¼ L:

We omit the details of theproof in this case, which arenow straight-forward to writedown.


the facts that (i) the sequence {f (xn)} converges to ‘, and (ii) g is contin-

uous at ‘.The following example illustrates the use of the Composition Rule.

Example 3 Determine the following limits:

(a) limx!0

sin sin xð Þsin x

; (b) limx!0

1þ sin xx

� �2 �

:

Solution

(a) Let f (x)¼ sin x, x2R , and gðxÞ ¼ sin xx

, x 6¼ 0: Then

limx!0

f xð Þ ¼ limx!0

sin x ¼ 0 and limx!0

g xð Þ ¼ limx!0

sin x

x¼ 1:

Also, f (x)¼ sin x 6¼ 0 in the punctured neighbourhood �p; 0ð Þ [ 0; pð Þ of 0

(for example). It follows, by the Composition Rule, that

limx!0


sin sin xð Þsin x

¼ 1:

(b) Let f ðxÞ ¼ sin xx

, x 6¼ 0, and g xð Þ ¼ 1þ x2, x 2 R : Then

limx!0

sin x

x¼1 and lim

x!11þ x2� �

¼2:

Also, g is defined and continuous at 1. It follows, by the Composition Rule,

that

limx!0


1þ sin x

x

� �2 !

¼ 2: &

Problem 3 Use the Combination Rules and the Composition Rule to

determine the following limits:

(a) limx!0

sin x2xþx2 ; (b) lim

x!0

sin x2ð Þx2 ; (c) lim

x!0

xsin x

� �12:

Problem 4 For the functions

f xð Þ ¼0; x ¼ 0;�2 x ¼ 1;2þ x; x 6¼ 0; 1;

8

<

:

and g xð Þ ¼ 0; x ¼ 0;1þ x; x 6¼ 0;

determine f � gð Þ xð Þ; f � gð Þ 0ð Þ; f

limx!0

g xð Þ�

and limx!0

f g xð Þð Þ:

There is also a Squeeze Rule for limits.


Let the functions f, g and h be defined on a punctured neighbourhood N of a

point c. If:

1. g xð Þ � f xð Þ � h xð Þ; for x 2 N; and

2. limx!c

g xð Þ¼ limx!c

h xð Þ ¼ ‘;then lim

x!cf xð Þ¼‘:

Problem 5 Use the Squeeze Rule for limits to prove that:

(a) limx!0

x2 sin 1x

� �

¼ 0; (b) limx!0

x cos 1x

� �

¼ 0:

Here we have c¼ 0, ‘¼ 0 andL¼ 1.

Here we have c¼ 0, ‘¼ 1 andL¼ 2.

The proof of Theorem 5 is astraight-forward applicationof the Squeeze Rule forsequences.

y

cN

x

y = h(x)y = f (x)y = g(x)

The Limit Inequality Rule is another result for limits of functions that is an

analogue of the corresponding result for sequences.

Theorem 6 Limit Inequality Rule

If limx!c

f xð Þ ¼ ‘ and limx!c

g xð Þ ¼ m, and also

f xð Þ � g xð Þ, on some punctured neighbourhood of c,

then

‘ � m:

5.1.4 One-sided limits

In Example 2(c) above, we saw that limx!0

ffiffiffi

xp

does not exist, because the

function f xð Þ ¼ffiffiffi

xp

, x � 0, is not defined on any punctured neighbourhood

of zero. Nevertheless, if {xn} is any sequence in the domain [0, 1) of f for

which xn! 0, thenffiffiffiffiffi

xnp ! 0. Thus f xð Þ ¼

ffiffiffi

xp

tends to f (0)¼ 0 as x tends to 0

from the right.

Definition Let f be defined on (c, cþ r), for some r > 0. Then f(x) tendsto the limit ł as x tends to c from the right if:

for each sequence {xn} in (c, cþ r) such that xn! c, then f (xn)! ‘.

We write this either as ‘ limx!cþ

f xð Þ ¼ ‘’ or as ‘f (x)! ‘ as x! cþ ’.

There is a corresponding definition for limits as x tends to c from the left.

Definition Let f be defined on (c� r, c), for some r > 0. Then f(x) tends

to the limit ł as x tends to c from the left if:

for each sequence {xn} in (c� r, c) such that xn! c, then f(xn)! ‘.

We write this either as ‘ limx!c�

f xð Þ ¼ ‘’ or as ‘f xð Þ ! ‘ as x! c�’.

Sometimes both left and right limits exist. Even when this happens, the two

values need not be equal, as the following example shows.

Example 4 Prove that the function f xð Þ ¼ xj jx

, x 6¼ 0, tends to different limits

as x tends to 0 from the right and from the left.

Solution The function f is defined on (0, 1), and f (x)¼ 1 on this interval.

Thus, if {xn} is a null sequence in (0, 1), then

limn!1

f xnð Þ ¼ limn!1

1ð Þ ¼ 1:

So limx!0þ

f xð Þ¼1:

Similarly, f is defined on (�1, 0), and f(x)¼�1 on this interval. Thus, if {xn}

is a null sequence in (�1, 0), then

limn!1

f xnð Þ¼ limn!1

�1ð Þ¼�1:

So limx!0�

f xð Þ¼�1: &


Note that f need not be definedat the point c.

For example, limx!0þ

ffiffiffi

xp¼ 0:

Again, f need not be defined atthe point c.


Problem 6 Write down a function f defined on the interval [�1, 3] for

which limx!0�

f xð Þ does not exist but limx!0þ

f xð Þ ¼ 1. Verify that f has the

specified properties.

The relationship between one-sided limits and ‘ordinary’ limits is given by

the following result.

Theorem 7 Let f be defined on a punctured neighbourhood of the point c.

Thenlimx!c

f xð Þ ¼ ‘

, limx!cþ

f xð Þ and

limx!c�

f xð Þ both exist and equal ‘:

Remark

If f is defined on an open interval I that contains the point c, and

limx!cþ

f xð Þ and limx!c�

f xð Þ both exist; but limx!cþ

f xð Þ 6¼ limx!c�

f xð Þ;then f is said to have a jump discontinuity at c.

Analogues of the Combination Rules, Composition Rule and Squeeze Rule can

also be used to determine one-sided limits. In the statements of all these rules,

we simply replace limx!c

by limx!cþ

or limx!c�

, and replace the open interval I contain-

ing c by (c, cþ r) or (c� r, c), as appropriate.

Problem 7 Prove that:

(a) limx!0þ

sin xxþ

ffiffiffi

xp� �

¼ 1; (b) limx!0þ

sinffiffi

xpð Þffiffi

xp ¼ 1:

Problem 8 Use the inequalities 1þ x � ex � 11�x

; for jxj< 1, to

prove that

limx!0þ

ex�1

x¼ lim

x!0�

ex�1

x¼1:

Deduce that

limx!0

ex�1

x¼1:

5.2 Asymptotic behaviour of functions

In this section we define formally a number of statements that you will have

already met in your study of Calculus, such as

1

x!1 as x!0þ and ex!1 as x!1;

and describe the relationship between them.

5.2.1 Functions which tend to infinity

Just as we defined the statement limx!c

f xð Þ ¼ ‘ in terms of the convergence of

sequences, so we can define the statement f (x)!1 as x! c.

We omit a proof of this result.

We verified these inequalitiesas part (c) of Theorem 4 (‘TheExponential Inequalities’) inSub-section 4.1.3.

Definition Let f be defined on a punctured neighbourhood N of the point c.

Then f (x)!1 as x! c if:

for each sequence {xn} in N such that xn! c, then f (xn)!1.

The statement ‘f (x)!�1 as x! c’ is defined similarly, with 1 replaced

by �1.

As with sequences, there is a version of the Reciprocal Rule which re-

lates the behaviour of functions which tend to infinity and functions which

tend to 0.

Theorem 1 Reciprocal Rule

(a) If the function f satisfies the following two conditions:

1. f (x)> 0 for all x in some punctured neighbourhood of c, and

2. f (x)! 0 as x! c,

then 1f xð Þ ! 1 as x! c.

(b) If f (x)!1 as x! c, then 1f xð Þ ! 0 as x! c.

For example, 1x2 !1 as x! 0, since f (x)¼ x2> 0, for x2R � {0}, and

limx!0

x2¼0.

The statements

‘f xð Þ ! 1 or�1 as x! cþ’ and ‘f xð Þ ! 1 or�1 as x! c�’

are defined similarly, with the punctured neighbourhood of c being replaced by

open intervals (c, cþ r) or (c� r, c), as appropriate, for some r> 0.

The Reciprocal Rule can also be applied with ‘x! c’ replaced by ‘x! cþ’

or ‘x! c�’, and with the punctured neighbourhood of c being replaced by open

intervals (c, cþ r) or (c� r, c), as appropriate, for some r > 0.

For example, 1x!1 as x! 0þ, since f(x)¼ x> 0, for x 2 0;1ð Þ, and

limx!0þ

x ¼ 0:

Problem 1 Prove the following:

(a) 1jxj ! 1 as x! 0; (b) 1

1� x3 ! �1 as x! 1þ;

(c) sin xx3 !1 as x! 0:

Remark

There are also versions of the Combination Rules and Squeeze Rule for

functions which tend to 1 or �1 as x tends to c, cþ or c�; here we state

only the Combination Rules for functions which tend to1 as x tends to c.


If f(x)!1 and g(x)!1 as x! c, then:

Sum Rule f(x)þ g(x)!1;

Multiple Rule l f(x)!1, for l> 0;

Product Rule f(x) g(x)!1.


These are similar to resultsstated for sequences inTheorems 3 and 4,Sub-section 2.4.3.


Problem 2 Prove that the following statements are false:

(a) If f(x)!1 and g(x)!1 as x! 0, then f(x)� g(x)!1;

(b) If f(x)!1 and g(x)! 0 as x! 0, then f(x) g(x)! 0 or

f(x) g(x)!1.

5.2.2 Behaviour of f(x) as x tends to1 or �1

Finally, we define various types of behaviour of real functions f (x) as x tends to

1 or �1. To avoid repetition, here we allow ‘ to denote either a real number

or one of the symbols1 and �1.

Definition Let f be defined on an interval (R,1), for some real number R.

Then f (x)! ‘ as x!1 if:

for each sequence {xn} in (R,1) such that xn!1, then f (xn)! ‘.

The statement ‘f (x)! ‘ as x!�1’ is defined similarly, with1 replaced

by�1, and (R,1) by (�1, R). In practice, we usually prove that f (x)! ‘ as

x!�1 by showing that f (�x)! ‘ as x!1.

When ‘ is a real number, we also use the notation

limx!1

f xð Þ ¼ ‘ and limx!�1

f xð Þ ¼ ‘:

Once again, we can use versions of the Combination Rules and Reciprocal

Rule to obtain statements about the behaviour of given functions as x tends to1or �1. In the statements of these rules, we need only replace c by1 or �1,

and the punctured neighbourhood of c by (R,1) or (�1, R), as appropriate.

For example, for any positive integer n

xn !1 as x!1 and limx!1

x�n ¼ 0:

More generally, we have the following result for the behaviour of poly-

nomials as x!1.

Theorem 3 If a0, a1, . . ., an�1 are real numbers and

p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0; x 2 R ;

then

p xð Þ ! 1 as x!1 and1

p xð Þ ! 0 as x!1:

There are also versions of the Squeeze Rule for functions as x tends to1,

which have some important applications.


Let the functions f, g and h be defined on some interval (R,1).

(a) If f, g and h satisfy the following two conditions:

1. g(x)� f(x)� h(x), for all x in (R,1), and

2. limx!1

g xð Þ ¼ limx!1

h xð Þ ¼ ‘;then lim

x!1f xð Þ ¼ ‘:

We adopt this convention forsimplicity ONLY in thissub-section!

We ask you to prove the firstpart of this result inSection 5.6. The second partthen follows at once.

We omit the proof of thistheorem, which is straight-forward.


(b) If f and g satisfy the following two conditions:

1. f(x)� g(x), for all x in (R,1), and

2. g(x)!1 as x!1,

then f (x)!1 as x!1.

An important application of this Squeeze Rule is to show that ex tends to1faster than any power of x, as x tends to1. We prove this in the next example.

Example 1 Prove that for each n¼ 0, 1, 2, . . .,ex

xn !1 as x!1.

Solution We use the power series expansion

ex ¼ 1þ xþ x2

2!þ � � � þ xn

n!þ xnþ1

nþ 1ð Þ!þ � � �

for x� 0. Since x� 0, all the terms on the right here are non-negative, and so

ex � xnþ1

nþ 1ð Þ! ; for x � 0:

It follows that

ex

xn� x

nþ 1ð Þ! ; for x > 0:

Since xðnþ1Þ!!1 as x!1, it follows from part (b) of the Squeeze Rule that

ex

xn !1 as x!1. &

Remark

We deduce from Example 1 and the Product Rule that, for any integer n,

xnex!1 as x!1. Thus, by the Reciprocal Rule, for any integer n,

xne�x! 0 as x!1.

Problem 3 Determine the behaviour of the following functions as

x!1:

(a) f xð Þ ¼ 2x3þxx3 ; (b) f xð Þ ¼ sin x

x.

In our next example, we describe the behaviour of loge x as x!1.

Example 2 Prove that loge x!1 as x!1.

Solution We prove this from first principles.

First, note that the function f (x)¼ loge x is defined on (0,1).

Next, we have to prove that:

for each sequence {xn} in (0,1) such that xn!1, then loge xn!1.

To prove that loge xn!1, we need to show that:


loge xn > K; for all n > X: (1)

However, since we know that xn!1, we can choose X such that

xn > eK ; for all n > X:

Since the function loge is strictly increasing, the statement (1) is therefore

true. It follows that loge x!1 as x!1, as required. &

y y = h(x)

y = f (x)l

y = g(x)

R

yy = f (x)

y = g(x)

(R, ∞) x

R (R, ∞) x

See Section 3.4.

There is no easy way to provethis result by using theSqueeze Rule, since loge xtends to1 ‘ratherreluctantly’; that is, moreslowly than any function thatwe have considered so far.


5.2.3 Composing asymptotic behaviour

Earlier we gave a Composition Rule for limits. We now describe a more

general Composition Rule which permits the composition of the different

types of asymptotic behaviour that we have now met.

For example, since

f ðxÞ ¼ 1

x!1 as x! 0þ and gðxÞ ¼ ex !1 as x!1;

we should expect that

g f xð Þð Þ ¼ e1x !1 as x! 0þ:

This result is true, as the following Composition Rule shows. To avoid

repetition, we again allow ‘ and L to denote either a real number or one of

the symbols1 and �1.


If:

1. f (x)! ‘ as x! c (or cþ, c�,1 or �1), and

2. g (x)! L as x! ‘,

then

g f xð Þð Þ ! L as x! c ðor cþ; c�;1 or �1; respectivelyÞ;provided that:

EITHER f (x) 6¼ ‘ in some punctured neighbourhood of c (or in (c, cþ r),

(c� r, c), (R,1) or (�1, R), respectively, for some r> 0)

OR ‘ is finite, and g is defined at ‘ and is continuous at ‘.

Example 3 Prove that

(a) ex2

x!1 as x!1; (b) xe

1x !1 as x! 0þ;

(c) x sin 1x

� �

! 1 as x!1:

Solution

(a) Let f xð Þ ¼ x2

, x 2 R , and g xð Þ ¼ ex

x, x 2 R � 0f g, so that

g f xð Þð Þ ¼ ex2

x2

¼ 2ex2

x, for x 2 R � 0f g, and so in

particular for x 2 0;1ð Þ:Now, by the Multiple Rule, f(x)!1 as x!1; and, by Example 1,

g(x)!1 as x!1. It follows, by the Composition Rule, that

g f xð Þð Þ ¼ 2ex2

x!1 as x!1,

so that; by the Multiple Rule;we have ex2

x!1 as x!1:

(b) Let f xð Þ ¼ 1x, x 2 R � 0f g, and g xð Þ ¼ ex

x, x 2 R � 0f g, so that

g f xð Þð Þ ¼ e1x

1x

¼ xe1x, for x 2 R � 0f g, and so in

particular for x 2 0;1ð Þ:


We again adopt thisconvention for simplicityONLY in this sub-section!

We omit the proof ofTheorem 5.

If ‘ denotes1 or �1,then the first proviso isautomatically satisfied. Notethat we must have conditions1 and 2 and the first provisosatisfied or conditions 1 and 2and the second provisosatisfied if we are to make anyapplication of this result.


Now, f (x)!1, as x! 0þ; and, by Example 1, g(x)!1 as x!1. It

follows, by the Composition Rule, that

g f xð Þð Þ ¼ xe1x !1 as x! 0þ:

(c) Let f xð Þ ¼ 1x

, x 2 R � 0f g; and g xð Þ ¼ sin xx; x 2 R � 0f g, so that

g f xð Þð Þ ¼sin 1

x

� �

1x

¼ x sin1

x

� �

; for x 2 R � 0f g, and so

for x 2 0;1ð Þ:

Now

f xð Þ ! 0 as x!1;g xð Þ ! 1 as x! 0; and

f xð Þ 6¼ 0; for x 2 0;1ð Þ:

It follows, by the Composition Rule, that

g f xð Þð Þ ¼ x sin1

x

� �

! 1 as x!1: &

Problem 4 Prove that:

(a) loge (loge x)!1 as x!1; (b) xe1x ! 0 as x! 0�.

Hint for part (b): Use the fact that 1x! �1 as x! 0�.

Problem 5 Give examples of functions f and g (and a specific value for

each of ‘ and m) for which

1. f(x)! ‘ as x!1 and

2. g(x)!m as x! ‘,

but for which g f xð Þð Þ6! m as x!1.

5.3 Limits of functions – using e and d

Earlier we gave definitions of the limit of a sequence, continuity of a function

and limit of a function, and strategies for using these definitions. In each case,

the strategy was in two parts:

GUESS that the definition HOLDS: prove that it holds for all cases.

GUESS that the definition FAILS: find ONE counter-example.

Each definition is particularly convenient if we wish to prove that the

definition FAILS, but it is not always easy to work with it when we wish to

prove that the definition HOLDS.

For example, when we wish to prove that a given function f is discontinuous

at a point c, then we have to find ONE sequence in a punctured neighbourhood of

c such that

xn ! c BUT f xnð Þ 6! f cð Þ:

In Theorem 5, we have ‘¼ 0,L¼ 1.

This is condition 1.

This is condition 2.

So the first proviso issatisfied.


On the other hand, if we wish to use the definition to prove that f is

continuous at c, then we have to show that:

for each sequence {xn} in a punctured neighbourhood of c such thatxn ! c; then f xnð Þ ! f cð Þ :

As it is sometimes tricky to determine the behaviour of every such sequence

{xn}, this definition can be very inconvenient to use.

In this section we introduce a definition of the limit of a function that is

equivalent to our earlier definition, but which does not use sequences.

5.3.1 The e – d definition of limit of a sequence

We motivate our discussion by returning to the definition of a convergent

sequence.

Definition The sequence {an} is convergent with limit ł, or converges to

the limit ł, if {an� ‘} is a null sequence. In other words, if:


an � ‘j j < "; for all n > X: (1)

l + ε{an}

X n

l – ε < an < l + ε

|an – l| < εl

l − ε⇔

The condition (1) means that, from some point on, the terms of the sequence

all lie in the shaded strip between ‘� " and ‘ þ ", and thus lie ‘close to’ ‘. If we

choose a smaller number ", then the shaded strip in the diagram becomes

narrower, and we may need to choose a larger number X in order to ensure that

the inequality (1) still holds. But, whatever positive number " we choose, we

can always find a number X for which (1) holds.

Problem 1 Let an ¼ �1ð Þnn2 , n ¼ 1, 2, . . .. How large must we take X in

order that:

(a) |an� 0|< 0.1, for all n>X?

(b) |an� 0|< 0.01, for all n>X?

(c) |an� 0|<", for all n>X, where " is a given positive number?

Earlier we described a ‘game’, based on the above definition, in which

player A chooses a positive number " and challenges player B to find a number

X such that (1) holds. If the sequence in question converges, then such a

number X always exists, and player B can always win. If the sequence does

not converge, then, for SOME choices of ", NO number X exists such that (1)

holds, and so player A can always win.

In Section 5.4 we shallintroduce a definition ofcontinuity which does notuse sequences.

Sub-section 2.3.1.

Sub-section 2.2.1.

For example, consider the null sequence an ¼ 1ffiffi

np , n ¼ 1, 2, . . .. In this case,

the game might proceed as follows:

I'll playε = 0.1

I'll playX = 100

A B

B wins

Here player B wins, since, for n> 100, we have

1ffiffiffi

np � 0

�

�

�

�

�

�

�

�

¼ 1ffiffiffi

np <

1ffiffiffiffiffiffiffiffi

100p ¼ 0:1:

Player A tries again:

I'll playε = 0.01

I'll playX = 10 000

A B

B wins

Player B again wins, since, for n> 10 000, we have

1ffiffiffi

np � 0

�

�

�

�

�

�

�

�

¼ 1ffiffiffi

np <

1ffiffiffiffiffiffiffiffiffiffiffiffiffi

10 000p ¼ 0:01:

But now player B has figured out a winning strategy, and challenges player A

to do his worst!


ε

I'll playX = 1/ε

2

A B

B wins

This is indeed a winning strategy, since

for n >1

"2, we have

ffiffiffi

np

>1

", and so

1ffiffiffi

np � 0

�

�

�

�

�

�

�

�

¼ 1ffiffiffi

np < " :

The reason for introducing " in the definition of convergent sequence is to

formalise the idea of ‘closeness’. The statement (1) means that we can make

the terms an of the sequence as close as we please to ‘ by choosing X large

enough.

Here X¼ 100.

Here X¼ 10 000.

Here X ¼ 1"2.

Put another way, we can think of a sequence {an} as a function with domain R

n 7! n; anð Þ:

The number " formalises the idea of closeness of an to ‘ (that is, in the

codomain). The condition ‘for all n>X’ restricts the values n in the domain

to values for which condition (1) holds.

an close to l {

{an}

for all n > X

X n

l + ε

l – εl

|an – l| < ε for all n > X

5.3.2 The e – d definition of limit of a function

The concept of the limit of a function as x! c also involves the idea of

closeness. To define

f xð Þ ! ‘ as x! c; (2)

we require the statement

f xð Þ � ‘j j < ";

this means that the values of f (x) and ‘ are within a distance " of each other.

Earlier, we formalised the statement (2) by defining the limit of a function in

terms of limits of sequences.



for each sequence xnf g in N such that xn ! c, then f xnð Þ ! ‘: (3)

Intuitively, this means that, as xn gets close to c, so f (xn) gets close to ‘. More

precisely, we can make f (xn) as close as we please to ‘ by choosing xn

sufficiently close to c; we can ensure this by considering only xns for suffi-

ciently large n. But the sequence {xn} can be any sequence of points in N that

converges to c, so what we are really saying is that we can make f (x) as close as

we please to ‘ by choosing x sufficiently close to c (but not equal to c).

Thus we now need to specify not only closeness in the codomain

f xð Þ � ‘j j < ";

but also closeness in the domain. To do this, we introduce another small

positive number � (which depends on ") and specify closeness in the domain

by a statement of the form

0 < x� cj j < �:

In general, the smaller thevalue of ", the larger the X thatwe have to choose.

Sub-section 5.1.1.

Note that xn 6¼ c, for any n.

Actually it is the requirementj x� c j<� that specifies thecloseness in the domain. Therequirement 0< j x� c j ismade simply to ensure thatx 6¼ c.

f (xn) close to l

y = f (x) y = f (x)

{ f (xn)}

{xn}

c x

xn close to c

c x

x close to c

y y

l{ f (x) close to l

l{

Problem 2 Let f(x)¼ 2xþ 3, x 2 R . How small must we choose � in

order that:

(a) j f (x)� 5j< 0.1, for 0< jx� 1j<�?(b) j f (x)� 5j< 0.01, for 0< jx� 1j<�?(c) j f (x)� 5j<", for 0< jx� 1j<�, where " is a given positive

number?

In general, the statement ‘f (x)! ‘ as x! c’ means the following:

for EACH positive number ", there is a CORRESPONDING positive number� such that

f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �: (4)

The following diagram illustrates how the numbers " and � are used to

formalise the idea of closeness.

f (x) close to l

y = f (x) y = f (x)

c y

x close to c

c – δ

l + εl

l – ε

c + δc x

y y

l{

This leads us to the following definition of the limit of a function.



for each positive number ", there is a positive number � such that

f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < � : (5)

The same number �may servefor various different values of"; but, in general, the value of� depends on the value of ".

Condition (4) means that, forall x with 0< jx� cj<�, thepoints (x, f (x)) of the graphlie in the heavily shadedrectangle, and thus lie ‘closeto’ (c, ‘).

Thus

f ðxÞ is ‘close to’ ‘

wheneverx is ‘close to’ c:

Remarks

1. We assume that � is chosen sufficiently small in (5), so that the interval

(c� �, cþ �) lies within N. We also require that 0< jx� cj in (5), since we

are not concerned with the value of f at c, or even whether f is defined at c.

2. There are similar definitions of limits from the left and limits from the right

at c, where we simply replace ‘0< jx� cj<�’ by ‘c� � < x< c’ and

‘c< x< cþ �’, respectively; and similar definitions of limx!1

f xð Þ and

limx!�1

f xð Þ.

We now formally state the equivalence of the two definitions of ‘limit of a

function’.

Theorem 1 The ‘"� � definition’ and the ‘sequential definition’ of the

statement limx!c

f xð Þ ¼ ‘ are equivalent.

Proof Let the function f be defined on a punctured neighbourhood N of a

point c.

We have to show that:

for each sequence {xn} in N such that xn ! c, then f xnð Þ ! ‘ (6)

, for each positive number ", there is a positive number � such that

f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < � : (7)

First, let us assume that (6) holds. Then, for each sequence {xn} in N and


f xnð Þ � ‘j j < "; for all n > X: ()

We will now use a ‘proof by contradiction’. So, suppose that (7) does not

hold. Then there must exist some positive number " for which there is no

positive number � such that


In particular, we can always assume that � < r, so that {x : 0< jx� cj<�}N.

It follows that, for each positive integer n, there is some number xn (say) in

N such that

f xnð Þ � ‘j j � "; where 0 < xn � cj j < 1

n: (9)

But, from our assumption (6), the sequence {xn} must satisfy (*). Hence, in

addition, in particular, for the positive number " that we have chosen for (*),

there is a number X such that j f (xn)� ‘j<", for all n>X. But this is incon-

sistent with (9), so that we have a contradiction. It follows, therefore, that (7)

must hold after all!

Next, let us assume that (7) holds. That is, for each positive number ", there

is a positive number � such that


For otherwise the values f (x)may be undefined.The existence and value of alimit is a local property, butthe behaviour of the functionAT the point in question isirrelevant.


We shall assume, forconvenience, that

N ¼ c� r; cð Þ [ c; cþ rð Þ¼ x : 0 < x� cj j < rf g:

Here we simply restate (6).

Here we write down what ismeant by ‘(7) does not hold’.

Recall that

N ¼ x : 0 < x� cj j < rf g:This is possible since (8) doesnot hold for any �, and since wemay take as successive choicesof � the values 1; 1

2; 1

3; . . ..

For all n>X, f (xn) now has tosatisfy the two inconsistentinequalities

f xnð Þ � ‘j j � " and

f xnð Þ � ‘j j < ":

Here we simply restate (7).

Now let {xn} be any sequence in N such that xn! c. In particular, for the

positive number � that we have chosen for (10), there is a number X such

that (0< )jxn� cj<� for all n>X. It follows from (10) that jf(xn)� ‘j<"for all n>X.

Since this holds for each positive number ", we deduce that f (xn)! ‘, so that

(6) holds, as required. &

Remark

Depending on the particular circumstances, we can therefore use whichever

definition of limit is the more convenient for our purposes.

We can think of the choice of � in terms of " in the definition of limit of a

function as a game with players A and B, just as we did with the choice of X in

the definition of convergent sequence.

Thus, player A chooses some value for ", and then player B has to try to find a

value of � such that

f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �:

For example, consider how we might show that f (x)¼ x2! 0 as x! 0.

Here, for the function f (x)¼ x2, x 2 R , and c¼ 0, player B has to find a number

� such that

f xð Þ � 0j j ¼ 4x2�

�

�

� < "; for all x satisfying 0 < xj j < �:

The game might proceed as follows:

I'll playε = 0.2

I'll playδ = 0.1

A B

B wins

Here player B wins, since

for 0 < xj j < 0:1; we have that 4x2�

�

�

� � 0:04 < 0:2:

So player A tries again:

I'll playε = 0.02

I'll playδ = 0.1

A B

A wins

Here player A wins, since

x ¼ 0:09 satisfies 0 < xj j < 0:1 BUT 4� 0:092�

�

�

� ¼ 0:0324 6< 0:02:

Player B lost because he did not choose � sufficiently small; so he tries again.

We have 0< jxn� cj since xn

lies in a puncturedneighbourhood of c.

I'll playε = 0.02

I'll playδ = 0.01

A B

B wins

This time player B wins, since

for 0 < xj j < 0:01;we have that 4x2�

�

�

� � 0:0004 < 0:02:

But now player B has figured out a winning strategy! He challenges player A to

do his worst!


ε

I'll playδ =

12√ε

A B

B wins

This is indeed a winning strategy, since

for 0 < xj j < 1

2

ffiffiffi

"p

;we have that 4x2�

�

�

�< 4� 1

2

ffiffiffi

"p� �2

¼ ":

In general, if f (x)! ‘ as x! c, then such a number � always exists, and

player B can always win. On the other hand, if f ðxÞ 6! ‘ as x! c, then for SOME

choices of " no corresponding positive number � exists, and so player A can

always win.

Our strategy for using the ‘"� � definition’ of limx!c

f xð Þ in particular cases is

as follows.

Strategy for using the ‘e� d definition’ of limit of a function

1. To show that f (x)! ‘ as x! c, find an expression for � in terms of " such

that:

for each positive number "

f xð Þ � ‘j j < "; for all x satisfying 05 x� cj j5�: (11)

2. To show that f ðxÞ 6! ‘ as x! c, find:

ONE positive number " for which there is NO positive number �such that


Whatever value of " is chosenby player A, player B needonly specify � ¼ 1

2

ffiffiffi

"p

.

Of course the choice of � maydepend on c too.

Example 1 Using the strategy, prove that

(a) limx!1

2xþ 3ð Þ ¼ 5; (b) limx!0

2x3ð Þ ¼ 0;

(c) limx!0

x sin 1x

� �

¼ 0; (d) limx!0

sin 1x

� �

does not exist.

Solution

(a) The function f (x)¼ 2xþ 3 is defined on every punctured neighbour-

hood of 1.

We must prove that:


ð2xþ 3Þ � 5j j < "; for all x satisfying 0 < x� 1j j < � ;

that is

2 x� 1ð Þj j < "; for all x satisfying 0 < x� 1j j < �: (12)

We choose � ¼ 12"; then the statement (12) holds, since

for 0 < x� 1j j < 12", we have 2 x� 1ð Þj j < ":

Hence, limx!1

2xþ 3ð Þ ¼ 5.

(b) The function f (x)¼ 2x3 is defined on every punctured neighbourhood of 0.

We must prove that:


2x3 � 0�

�

�

� < "; for all x satisfying 0 < x� 0j j < � ;

that is

2x3�

�

�

� < "; for all x satisfying 0 < xj j < � : (13)

We choose � ¼ 12

ffiffiffi

"3p

; then the statement (13) holds, since:

for 0 < xj j < 12

ffiffiffi

"3p

, we have that

2x3�

�

�

� ¼ 2 xj j3< 2� 1

2

ffiffiffi

"3p� �3

¼ 1

4" < ":

Hence, limx!0

2x3ð Þ ¼ 0.

(c) The function f xð Þ ¼ x sin 1x

is not defined at 0, but it is defined on every

punctured neighbourhood of 0.

We must prove that:


x sin1

x� 0

�

�

�

�

�

�

�

�

< "; for all x satisfying 0< x� 0j j< � ;

that is

x sin1

x

�

�

�

�

�

�

�

�

< "; for all x satisfying 0 < xj j < �: (14)

Now, sin 1x

�

�

�

� � 1, for all non-zero x, so that

We could equally well choose� to be any positive numberless than 1

2", since the

statement (12) would stillhold.

Notice that many otherchoices of � would also haveserved our purpose.

x sin1

x

�

�

�

�

�

�

�

�

� xj j; for all non-zero x:

We therefore choose �¼ "; then the statement (14) holds.

Hence, limx!0

x sin 1x

� �

¼ 0.

(d) Suppose that the limit exists; call it ‘. Take " ¼ 12, say. Then there is some

positive number �, and points xn and yn of the form

xn ¼1

2nþ 12

� �

pand yn ¼

1

2nþ 32

� �

p; where n 2 N ;

such that

0 < xnj j < � and f xnð Þ � ‘j j < 1

2; and

0 < ynj j < � and f ynð Þ � ‘j j < 1

2:

:

Hence, for sufficiently large n

f xnð Þ � f ynð Þj j ¼ f xnð Þ � ‘ð Þ � f ynð Þ � ‘ð Þj j

� f xnð Þ � ‘j j þ f ynð Þ � ‘j j <1

2þ 1

2¼ 1:

But f (xn)� f (yn)¼ 1� (�1)¼ 2, which contradicts the previous inequal-

ity. It follows that no such numbers ‘ and � exist, so that in fact limx!0

sin 1x

� �

does not exist. &

Remarks

1. In Example 1, parts (a) and (b), we can find an appropriate choice of � by

‘working backwards’.

For example, in part (b) we have

2x3�

�

�

� < ", x3�

�

�

� <1

2"

, xj j < 1

213

ffiffiffi

"3p

;

so we can choose � ¼ 1

213

ffiffiffi

"3p

. The choice that we made in Example 1 was to

take � ¼ 12

ffiffiffi

"3p

, which was a somewhat smaller (but equally valid) value for �.Both possibilities are valid, since we only require to show that

0 < x� 0j j < � ) 2x3 � 0�

�

�

� < ";

rather than

0 < x� 0j j < � , 2x3 � 0�

�

�

� < ":

On the other hand, in part (c) it is not possible to ‘solve for �’ in this way –

that is, to solve the inequality x sin 1x

�

�

�

� < " to obtain a solution such that

x sin1

x

�

�

�

�

�

�

�

�

< "; for 0 < xj j < �:

For, if " is sufficiently small, the solution set of the inequality x sin 1x

�

�

�

� < " is

not even an interval but a union of disjoint intervals.

Of course any choice for �smaller than this particularvalue will also serve ourpurposes – such as 1

7" or 1

625".

2. In Example 1, part (d), the point is that, whatever the value of �, there are

always points in any punctured neighbourhood of 0 where f takes the values

1 and �1. Thus there cannot exist numbers ‘ and � such that

f xð Þ � ‘j j < 1

2; for all x satisfying 0 < xj j < �:

Problem 3 Using the Strategy (except in part (d)), prove that:

(a) limx!3

5x� 2ð Þ ¼ 13; (b) limx!0

1� 7x3ð Þ ¼ 1;

(c) limx!0

x2 cos 1x3

� �

¼ 0; (d) limx!0

xþ xj jx

does not exist.

Sometimes it is a little tricky to find an expression for � in terms of " such that


The following example illustrates one approach that is sometimes useful.

Example 2 Use the "� � definition to prove that limx!2

x2 ¼ 4.

Solution According to the definition, we must show that for x ‘close to 2’ the

following difference is small

x2 � 4�

�

�

� ¼ xþ 2j j � x� 2j j:For x close to 2, the term jxþ 2j is approximately 4, so that the product

jxþ 2j � jx� 2j is approximately equal to 4jx� 2j. Certainly, if we restrict x

to the punctured neighbourhood (1, 2)[ (2, 3) in which jxþ 2j< 5, we know that

x2 � 4�

�

�

� < 5 x� 2j j:

So, we proceed as follows.

The function f (x)¼ x2 is defined on every punctured neighbourhood of 2.

We must prove that


x2 � 4�

�

�

� < "; for all x satisfying 0 < x� 2j j < � ;

that is

xþ 2j j � x� 2j j < "; for all x satisfying 0 < x� 2j j < �:

We choose � ¼ min 1; 15"

� �

, so that for 0< jx� 2j<� we have

xþ 2j j < 5 and x� 2j j < 1

5" :

It follows that

xþ 2j j � x� 2j j < 5� 1

5" ¼ "; for all x satisfying 0 < x� 2j j < �:

Hence limx!2

x2 ¼ 4: &

Problem 4 Use the "� � definition to prove that limx!1

x3 ¼ 1.

Hint: Try � ¼ min 1, 17"

� �

.

The discussion so far has beenour motivation.

The formal proof now starts!

Proofs

Since the definition of limit of a function introduced in this section is equiva-

lent to the earlier sequential definition, the rules for combining limits can also

be proved using the "� � approach. To give you a flavour of what this

involves, we prove here just one of the Combination Rules.

Example 3 Prove the Sum Rule:

If limx!c


g xð Þ ¼ m; then limx!c

f xð Þ þ g xð Þð Þ ¼ ‘þ m:

Solution Since the limits for f and g exist, the functions f and g must

be defined on some punctured neighbourhoods (c� r1, c)[ (c, cþ r1) and

(c� r2, c)[ (c, cþ r2) of c. It follows that the function fþ g is certainly defined

on the punctured neighbourhood (c� r, c)[ (c, cþ r), where r¼min{r1, r2}.

We want to prove that:


f xð Þ þ g xð Þð Þ � ‘þ mð Þj j < "; for all x satisfying 0 < x� cj j < �: (15)

We know that, since limx!c

f xð Þ ¼ ‘, there is a positive number �1 such that

f xð Þ � ‘j j < 1

2"; for all x satisfying 0 < x� cj j < �1; (16)

similarly, since limx!c

g xð Þ ¼ m, there is a positive number �2 such that

g xð Þ � mj j < 1

2"; for all x satisfying 0 < x� cj j < �2: (17)

We now choose �¼min{�1, �2}. Then both statements (16) and (17) hold

for all x satisfying 0< |x� c|<�, so that

f xð Þ þ g xð Þð Þ � ‘þ mð Þj j ¼ f xð Þ � ‘ð Þ þ g xð Þ � mð Þj j� f xð Þ � ‘j j þ g xð Þ � mj j

<1

2"þ 1

2" ¼ ";

so that the statement (15) holds.

Hence limx!c

f xð Þ þ g xð Þð Þ ¼ ‘þ m. &

We now introduce an analogue of a useful tool for writing out proofs that

you met earlier in your work on sequences, the so-called ‘K" Lemma’.

In order to prove that a function f has a limit as x! c, using the definition of

limit, you need to prove that:



It can be tedious in the process of the proof to make a complicated choice for � in

order to end up with a final inequality that says that some expression is ‘<"’. While

this means that we end up with an inequality that shows at once that the desired

result holds, in fact it is not strictly necessary to end up with precisely ‘< "’.

Of course f and g may bedefined on some larger setstoo!

Sub-section 2.2.2.

Lemma The ‘Ke Lemma’ For the function f, suppose that:


f xð Þ � ‘j j < K"; for all x satisfying 0 < x� cj j < �;

where K is a positive real number that does not depend on " or X.

Then f (x)! ‘ as x! c.

We omit a proof of the Lemma, as it is essentially the same as the previous

proof of the K" Lemma for sequences. There are analogues of the K" Lemma

for the definition of continuity (in terms of " and �), differentiability and

integrability, but we shall not always mention them explicitly every time an

analogue could be stated.

From time to time we shall use this Lemma in order to avoid arithmetic

complexity in proofs.

Problem 5 Use the "� � definition to prove the Product Rule:

If limx!c


g xð Þ ¼ m; then limx!c

f xð Þg xð Þ ¼ ‘m:

Problem 6 Use the "� � definition to prove that, if limx!c

f xð Þ ¼ ‘ and

‘ 6¼ 0, then there exists some punctured neighbourhood of c on which f (x)

has the same sign as ‘.

Hint: Take " ¼ 12‘j j in the definition of limit.

5.4 Continuity – using e and d

In this section we introduce a definition of continuity of a function that is

equivalent to our earlier definition, but which does not use sequences.

5.4.1 The e–d definition of continuity

Recall our earlier definitions of continuity. The first version was for continuity

of a function at an interior point of an interval.

Definition A function f defined on an interval I that contains c as an

interior point is continuous at c if

for each sequence {xn} in I such that xn! c, then f (xn)! f (c).

We then saw how to extend this definition to include continuity of a function at

an end-point of an interval.

Definition A function f defined on a set S in R that contains a point c is

continuous at c if:

for each sequence {xn} in S such that xn! c, then f (xn)! f (c).

Loosely speaking, we mayexpress this result as ‘K" isjust as good as "’ in thedefinition of limit.

For example, K might be 2 or p7

or 259, but it could not be 2xor 1

".

For example, in Problem 5below.

You may omit Problem 5 ifyou are short of time.

Sub-section 4.1.1.

Sub-section 4.1.1.

Here c may be an interiorpoint of an interval, or it maybe an end-point of an interval;both possibilities are coveredby the phrase ‘each sequence{xn} in S’.

We start by reformulating the first version of this definition in terms of " and �.So, if we replace ‘ by f (c) in the definition of limit of a function, we

immediately obtain the following definition of continuity of a function at an

interior point of an interval.

Definition A function f defined on an interval I that contains c as an inte-

rior point is continuous at c if:


f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �: (1)

Remarks

1. Notice that in (1) we write jx� cj<� rather than 0< jx� cj<�, since in

fact (1) is always satisfied when x¼ c.

2. We may assume that � has been chosen sufficiently small so that (c� �,cþ �) lies in I.

3. It is sufficient in (1) to have

f xð Þ � f cð Þj j < K"; for all x satisfying x� cj j < �;

where K is some positive constant. This fact is often referred to as the ‘K"Lemma’ for continuity.

4. The condition (1) means that, for all x ‘sufficiently close to’ c, the values of f(x)

lie ‘close to’ f(c). If we choose a smaller number", then we may need to choose

a smaller number � in order to ensure that (1) still holds. But, whatever positive

number " we choose, we can always find a number � for which (1) holds.

5. We can describe the definition of continuity in terms of an ‘"� � game’ in

much the same way as we described the definition of limit. WHATEVER choice

of " player A makes, then player B can ALWAYS choose a value of � such that

the required condition (1) holds.

6. For a given value of the positive number ", we may need to choose different

values of � for different points c.

7. It follows immediately from our definition of continuity in terms of limits

that, if a function f is defined on an open interval containing c as an interior

point, then

f is continuous at c, limx!c

f xð Þ ¼ f cð Þ.

8. We can define one-sided continuity in a similar fashion. Thus a function f is

continuous on the left at c if:


f xð Þ � f cð Þj j < "; for all x satisfying c� � < x < c;

and is continuous on the right at c if:


f xð Þ � f cð Þj j < "; for all x satisfying c < x < cþ �:It follows from these definitions and the previous remark that, if a function f

is defined on an open interval containing c as an interior point, then

f is continuous at c, f is continuous on the left and on the right at c.

Sub-section 5.3.2.

Note that the inequalityj f (x)� f (c)j<" specifiescloseness in the codomain,whereas the inequalityjx� cj<� specifies closenessin the domain.

c – δ c + δ xc

y = f (x)

f (c)f (c) + ε

f (c) – ε

y

We shall return to this point inSection 5.5.

We shall not spend muchtime looking at one-sidedcontinuity; however there is awide range of results similarto the results for ‘ordinary’continuity.

We can use the definitions in the last Remark to extend the definition of

continuity of a function f from points of an open interval to all points of a

general interval, in a natural way. We say that f is continuous on an interval I if

f is continuous at all interior points of I, continuous on the right at the left end-

point of I if it belongs to I, and continuous on the left at the right end-point of

I if it belongs to I.

This leads us to extend our initial definition of continuity, as follows.

Definition A function f defined on an interval I that contains a point c is

continuous at c if:


f xð Þ � f cð Þj j < "; for all x in I satisfying x� cj j < �:

f is said to be continuous on I if it is continuous at each point c of I.

Since the ‘"� � definition’ and the ‘sequential definition’ of limits are

equivalent, and the two definitions of continuity are simply reformulations

of those definitions, it is obvious that the two definitions of continuity must

also be equivalent. Depending on the particular circumstances, we can there-

fore use whichever definition of continuity is the more convenient for our

purposes.

Theorem 1 The ‘"� � definition’ and the ‘sequential definition’ of the

statement ‘f is continuous at c’ are equivalent.

Our strategy for using the ‘"� � definition’ of continuity in particular cases

is also similar to the strategy for using the ‘"� � definition’ of limits.

Strategy for using the ‘e� d definition’ of continuity

1. To show that f is continuous at an interior point c of an interval, find an

expression for � in terms of " such that:

for each positive number "

f xð Þ � f cð Þj j5 "; for all x satisfying x� cj j < �: (2)

2. To show that f is discontinuous at an interior point c of an interval, find

ONE positive number " for which there is NO positive number � such that


Example 1 Use the Strategy to prove that f (x)¼ x2 is continuous at 2.

Solution The function f is defined on R .

We must prove that:


x2 � 4�

�

�

� < "; for all x satisfying x� 2j j < �;

that is

xþ 2j j x� 2j j < "; for all x satisfying x� 2j j < �:

This was Theorem 1 in Sub-section 5.3.2.

Note that, for x near to 2, xþ 2is near to 4 and x� 2 near to 0.

We choose � ¼ min 1; 15"

� �

; then, for jx� 2j<�, we have

xþ 2j j < 5 and x� 2j j < 1

5":

Hence

x2 � 4�

�

�

� < 5� 1

5" ¼ "; for all x satisfying x� 2j j < �:

It follows that f is continuous at 2. &

Problem 1 Use the Strategy to prove that f (x)¼ x3 is continuous at 1.

Hint: Try � ¼ min 1; 17"

� �

.

Problem 2 Use the Strategy to prove that f xð Þ ¼ffiffiffi

xp

, x� 0, is contin-

uous at 4.

Example 2 Use the Strategy to prove that the following function is discon-

tinuous at 0

f xð Þ ¼ x; if x � 0;1; if x < 0:

Solution The function f is defined on R .

The graph of f suggests that, while f is ‘well behaved’ to the right of 0, we

should examine the behaviour of f to the left of 0. If we choose " to be any

positive number less than 1, then there will always be points x (with x< 0) as

close as we please to 0, where f (x) is not within a distance " of f(0)¼ 0.

So, take " ¼ 12, say; and let xn ¼ � 1

n, for n¼ 1, 2, . . .. Then, for any positive

number �

xn ¼ �1

n

� �

2 ��; 0ð Þ; for all n > X; where X ¼ 1

�;

but

f xnð Þ � f 0ð Þj j ¼ f � 1

n

� �

� f 0ð Þ�

�

�

�

�

�

�

�

¼ 1� 0

¼ 1 6< 1

2¼ ":

Thus, with this choice of ", no value of � exists such that

f xð Þ � f 0ð Þj j < "; for all x satisfying x� 0j j < �:

This proves that f is discontinuous at 0. &

Problem 3 Use the Strategy to prove that the following function is

discontinuous at 2

f xð Þ ¼x� 1

2; if x > 2;

1; if x ¼ 2;x� 1; if x < 2:

8

<

:

Finally, we demonstrate a nice application of the "� � approach to continuity

in order to obtain a result that will be useful later on.

We could equally well choose� to be any positive numbersmaller than � ¼ min 1; 1

5"

� �

,since the rest of the argumentwould still apply. We choose� < 1 to get a bound on theterm xþ 2, and � < 1

5" as a

small multiple of " such thatthe final product is at most ".

y

0

1

x

For example, the points ofthe sequence {xn}, wherexn ¼ � 1

n, for all n� 1, since

then f (xn)¼ 1.

We make a suitable choice of" such that no corresponding �exists that satisfies therequirement (3).

This was requirement (3).

Theorem 2 Let f be continuous at an interior point c of an interval I, and

f cð Þ 6¼ 0. Then there exists a neighbourhood N¼ (c� r, cþ r) of c on

which:

(a) f(x) has the same sign as f(c), and

(b) f xð Þj j > 12

f cð Þj j.

Proof Since f(c) 6¼ 0, we may take 12

f cð Þj j as the positive number " in the

definition of continuity at c. It follows that there exists some positive number r

such that

f xð Þ � f cð Þj j < 1

2f cð Þj j; for all x satisfying x� cj j < r: (4)

We now rewrite (4) in the following equivalent form

� 1

2f cð Þj j < f xð Þ � f cð Þ < 1

2f cð Þj j; for all x 2 c� r; cþ rð Þ: (5)

Suppose, first, that f (c)> 0. It follows from the left inequality in (5) that, for

x2 (c� r, cþ r), we have �12

f cð Þ < f xð Þ � f cð Þ – in other words, that12

f cð Þ < f xð Þ. Thus, in this case, both (a) and (b) hold for x2 (c� r, cþ r).

Suppose, next, that f (c)< 0. It follows from the right inequality in (5) that,

for x2 (c� r, cþ r), we have f xð Þ � f cð Þ < 12

f cð Þj j ¼ �12

f cð Þ – in other

words, that f xð Þ < 12

f cð Þ. Thus, in this case too, both (a) and (b) hold for

x2 (c� r, cþ r).

This completes the proof of the theorem. &

5.4.2 The Dirichlet and Riemann functions

We can use our new "� � approach to continuity to tackle two very strange

functions that were devised in the second half of the nineteenth century. We

start with the Dirichlet function.

Definition Dirichlet’s function is defined on R by the formula

f xð Þ ¼ 1; if x is rational;0; if x is irrational:

The graph of f looks rather like two parallel lines, but each line has

‘infinitely many gaps in it’.

Theorem 3 The Dirichlet function is discontinuous at each point of R .

Proof Let c be any point of R . Then, for each n = 1, 2, . . ., by the Density

Property of R , the open interval c� 1n

, cþ 1n

� �

contains a rational number, xn

say, and an irrational number, yn say. Then xn! c and yn! c as n!1, but

f xnð Þ ¼ 1 and f ynð Þ ¼ 0:

Thus f (x) does not tend to a limit as x tends to c, whether c is rational or

irrational.

It follows that f is discontinuous at each point c of R . &

Here r> 0.

For convenience, here we usethe symbol r rather than � inthe definition of continuity –this makes no real difference.

y

1f (xn)

f (yn)

c xxn yn c + 1n

Recall that the DensityProperty of R asserts that,between any two unequal realnumbers, there exists at leastone rational and one irrationalnumber – Sub-section 1.1.4.

Our next function possesses even stranger behaviour.

Definition Riemann’s function is defined on R by the formula

f xð Þ ¼1q;

0;

if x is a rational numberpq; in lowest terms; with q > 0;

if x is irrational:

It is not clear from the graph of f whether f is continuous at any point of R .

Theorem 4 Riemann’s function is discontinuous at each rational point of

R , and continuous at each irrational point of R .

Proof First, let c ¼ pq

be any rational point of R , where q> 0 andpq

is in its

lowest terms.

Then, by the Density Property of R , for each n = 1, 2, . . ., the open inter-

val c� 1n

, cþ 1n

� �

contains an irrational number xn, for which

f xnð Þ ¼ 0 6¼ 1q¼ f cð Þ.

Hence as n!1xn ! c but f xnð Þ 6! f cð Þ:

Thus f is discontinuous at any rational point c.

Next, suppose that c is any irrational point of R . We must prove that:



Since c is irrational, f (c)¼ 0; also f (x)� 0 for all x. It follows that condition

(6) becomes

f xð Þ < "; for all x satisfying x� cj j < �: (7)

Our task is therefore to find a value of � such that (7) holds.

Now, let N be any positive integer such that N > 1", and let SN be the set of

rational numbers pq

(in lowest terms) in the interval (c� 1, cþ 1) for which

0< q<N – that is

SN ¼p

q: c� p

q

�

�

�

�

�

�

�

�

� 1; 0 < q < N

:

Since there are only a finite number of points in SN and c =2 SN, we can define a

positive number � as follows

The expression ‘in lowestterms’ means that p and qhave no common factor.

In this part of the proof, weuse a sequential approach.

Note that xn 6¼ c.

In this part of the proof, weuse an "� � approach.

Thus, in particular, 1N< ".

We choose (c� 1, cþ 1)simply so that we are onlylooking at points ‘near to c’.

c =2 SN since c is irrational.

� ¼ min x� cj j : x 2 SNf g:Thus there are NO rational numbers p

q, with 0< q<N, that lie in the interval

(c� �, cþ � ).

It follows that, if jx� cj<�, then:

EITHER x is irrational, so that f(x) (¼ 0)<";OR x is rational, so that x ¼ p

q, with q�N, and f xð Þð¼ 1

qÞ � 1

N< ".

In either case, f(x)<".Hence we have proved that (7) holds, and so that f is continuous at c. &

5.4.3 Proofs

Since the definition of continuity introduced in this section is equivalent to the

earlier sequential definition, the rules for continuity can also be proved using

the "� � approach. To give you a flavour of what this involves, we first prove

one of the Combination Rules.

Example 3 (Sum Rule) Let f and g be defined on an open interval I contain-

ing the point c. Then, if f and g are continuous at c, so is the sum function fþ g.

Solution The functions f, g and fþ g are certainly defined on some neigh-

bourhood (c� r, cþ r) I of c, where r> 0.



f xð Þþg xð Þð Þ� f cð Þþg cð Þð Þj j<"; for all x satisfying x� cj j<� (8)

We know that, since f is continuous at c, there is a positive number �1 (which

we may assume is� r) such that

f xð Þ � f cð Þj j < 1

2"; for all x satisfying x� cj j < �1; (9)

similarly, since g is continuous at c, there is a positive number �2 (which we

may assume is also� r) such that

g xð Þ � g cð Þj j < 1

2"; for all x satisfying x� cj j < �2: (10)

We now choose �¼min{�1, �2}. Then both statements (9) and (10) hold for all

x satisfying jx� cj<�, so that

f xð Þ þ g xð Þð Þ � f cð Þ þ g cð Þð Þj j ¼ f xð Þ � f cð Þð Þ þ g xð Þ � g cð Þð Þj j� f xð Þ � f cð Þj j þ g xð Þ � g cð Þj j

<1

2"þ 1

2" ¼ ";

so that the statement (8) holds.

Hence fþ g is continuous at c. &

Problem 4 Prove that if f is defined on an open interval I containing

the point c at which it is continuous, and f (c) 6¼ 0, then 1f

is defined on

some neighbourhood of c and is also continuous at c.

Hint: Use the result of Theorem 2 in Sub-section 5.4.1.

� is the distance from c to thenearest point of SN.

You might like to comparethis solution with that ofExample 3 in Sub-section 5.3.2.

Finally, we give a proof of the Composition Rule for continuity; this is a

particularly pleasing example of the "� � approach.

Example 4 (Composition Rule) Prove that if f is continuous at c and g is

continuous at f (c), then g� f is continuous at c.

Solution We must prove that:


g f xð Þð Þ � g f cð Þð Þj j < "; for all x satisfying x� cj j < �: (11)

We know that, since g is continuous at f(c), there is a positive number �1

such that:

g f xð Þð Þ � g f cð Þð Þj j < "; for all f xð Þ satisfying f xð Þ � f cð Þj j < �1: (12)

f (c) – δ1 f (c) + δ1 g(f (c)) – ε

g

g( f (c)) + εg( f (c)) g( f (x))f (c) f (x)

Also, since f is continuous at c, there is a positive number � such that

f xð Þ � f cð Þj j < �1; for all x satisfying x� cj j < �: (13)

c – δ c + δc x f (c) – δ1 f (c) + δ1

f

f (c) f (x)

Combining (12) and (13), we deduce that:

for all x satisfying jx� cj<�, then f xð Þ � f cð Þj j < �1,

so that

g f xð Þð Þ � g f cð Þð Þj j < ":

Hence g� f is continuous at c. &

5.5 Uniform continuity

We start by reminding you of the "� � definition of continuity at points of an

interval I.

Definition A function f defined on an interval I that contains a point c is

continuous at c if:


f xð Þ � f cð Þj j < "; for all x in I satisfying x� cj j < �:

f is said to be continuous on I if it is continuous at each point c of I.

We do not bother to mentionappropriate neighbourhoodsof c and f (c) on which f and gare defined, respectively, justto simplify the statement ofthe Rule.

By (13).

By (12).

Here the choice of � depends,in general, on both " and c.

Earlier we pointed out that, for a given value of the positive number ", we may

need to choose different values of � for different points c. This suggests the

following related definition.

Definition A function f defined on an interval I is uniformly continuouson I if:


f xð Þ � f cð Þj j < "; for all x and c in I satisfying x� cj j < �: (1)

Since the definition of uniform continuity is more restrictive than that of

continuity, it follows that uniform continuity implies continuity.

Theorem 1 If a function f is uniformly continuous on an interval I, then it

is continuous on I.

However the converse result is not true in general.

Example 1 Prove that the function f xð Þ ¼ 1x, 0 < x � 1; is not uniformly

continuous on (0, 1].

Solution To prove that f is NOT uniformly continuous on (0,1], we need to

find ONE positive number " for which there is NO positive number � such that

f xð Þ� f cð Þj j< "; for all x and c in ð0;1� satisfying x� cj j5�: (2)

We will take as our choice of " the number 1, and prove that there is NO

corresponding positive number � such that

f xð Þ� f cð Þj j< 1; for all x and c in ð0;1� satisfying x� cj j<�: (3)

Suppose on the contrary that some number � did exist such that (3) was valid.

We will show that this assumption leads us to a contradiction.

Choose a positive integer N with

1

N< �:

Now each point of (0,1] belongs to at least one interval of the form

0

2N;

2

2N

� �

;1

2N;

3

2N

� �

; . . .;k

2N;k þ 2

2N

� �

; . . .;2N � 2

2N;2N

2N

� �

;2N � 1

2N; 1

� �

;

(4)and, in addition, all points x and c in any given such interval satisfy the

inequality

x� cj j < 2

2N¼

� �

1

N< �;

Hence, by (3)

all points x and c in ð0; 1� in any interval in ð4Þ satisfy f xð Þ � f cð Þj j < 1:

(5)Now, from the definition of f, we have

f xð Þ < 2N

2N � 1¼ K; say; for all x in

2N � 1

2N; 1

� �

:

Sub-section 5.4.1, Remark 6.

Here the choice of � dependsONLY on ", the same choicewhatever c may be.

Since f is a rational function, itis a basic continuous function,and so is continuous on (0, 1].

For the statement (2) assertsthat statement (1) is not valid.

This is what we now set outto do.

For, successive intervals in (4)overlap.

In particular, f (x)< f (c)þ 1.

For, f is strictly decreasingand f 2N�1

2N

� �

¼ 2N2N�1

.

5.5 Uniform continuity 201

Then, since the interv als 2 N � 12N

; 1� �

and 2 N � 22 N

; 2 N2 N

� �

overlap, ther e is a point c in2N � 2

2 N; 2 N

2 N

� �

for which f ( c ) < K. It follow s from ( 5) that in the second last

interv al 2N � 22 N

; 2 N2 N

� �

in ( 4) we have

f xð Þ < K þ 1; for all x in2N � 2

2N;

2N

2N

� �

:

Work ing backwar ds through the 2 N intervals, we can check that

f xð Þ < K þ 2N � 1ð Þ; for all x in0

2N ;

2

2N

� �

:

It follow s then, that in fact f( x ) < K þ (2N � 1), for all x in (0,1]. In othe r words,

f is bounded in (0,1]. But this is not the case, since f ( x ) !1 as x ! 0þ .

We have thus reac hed the desired cont radiction . &

Problem 1 Let f be the funct ion f (x ) ¼ x 2, x 2 [2, 3]. Fo r any give n

positive number ", find a formula for � in terms of " such that (1) holds.

This proves directly from the defi nition that f is uniforml y cont inuous on

[2, 3].

Hint: Use the fact that x 2 � c 2 ¼ (x þ c)( x � c ).

Notice, thoug h, that the function in Probl em 1 is a continuo us funct ion on a

closed interv al, and reca ll that we saw earlier that continuo us functi ons on

closed intervals have ‘nice properties ’. It tur ns out that clos ed interv als are

important too for unif orm continuit y.

Theorem 2 If a funct ion f is cont inuou s on a closed interv al [ a, b], then it

is uniforml y continuo us on [a , b].

Proof of The orem 2 We use the Bolza no–We ierstras s Th eorem, and a proof

by contradict ion.

So, suppos e that, in fact, f is n ot uniform ly continuo us on [ a, b]. Then for

some choi ce of ", there is NO corr espond ing positiv e numb er � such that

f xð Þ � f yð Þj j < "; for all x and y in a, b½ � satisfying x � yj j < �: (6)

We can reform ulate this statemen t (6) in the followi ng conven ient way :

for any posi tive number � , there are two points x , y in [ a, b] such that

f xð Þ � f yð Þj j � " and x � yj j < �: (7)

By applying ( 7) for the choi ces �n ¼ 1n , n ¼ 1, 2, . . ., in turn, we can then define

two seque nces {xn} and { y n} in [a,b] such that

f xnð Þ � f ynð Þj j � " and xn � y nj j < 1

n ; for n 2 N : (8)

Now, the seque nce {xn} is bo unded, so that b y the Bolza no–We ierstras s

Theore m it cont ains a subsequenc e xnkf g that conver ges to some point c of

[ a, b] as k !1. Then, since xnk� ynk

j j < 1nk

, it follows that the correspo nding

subsequence ynkf g must also converge to c as k!1.

Since f is continuous at c, we must have

f xnkð Þ ! f cð Þ and f ynk

ð Þ ! f cð Þ; as k!1: (9)

For f is strictly decreasing.

Section 4.2.

Example 1 shows that, ingeneral, f may NOT beuniformly continuous if itsinterval of continuity is a half-closed interval.

You may omit this proof on afirst reading.

" is now fixed from here on inthis proof.

We use y rather than c in (6),since it fits in better with thenotation that we will use laterin the proof.

The choice of x and y willdepend, in general, on thechoice of �.


From (8).

Now, it follows from (8) that

f xnkð Þ � f ynk

ð Þj j � "; for each k 2 N : (10)

So, if we now let k!1 in (10) and use the limits in (9), we deduce that

0 � ";which is absurd.

This contradiction shows that (7) and so (6) cannot be valid. It follows

that f must be uniformly continuous on [a, b] after all. &

We shall use Theorem 2 later, to prove that a function continuous on a closed

interval [a, b] is integrable on [a, b].

Theorem 2 is one of the important theorems in Analysis beyond the scope of

this book.

5.6 Exercises

Section 5.1

1. Determine whether the following limits exist:

(a) limx!1

x3�1x�1

; (b) limx!1

x3�1x�1j j; (c) lim

x!0esin x.

2. Determine the following limits:

(a) limx!0

sin xþ ex�1x

� �

; (b) limx!0

ex�1sin x

;

(c) limx!0

e xj j�1xj j ; (d) lim

x!1þx3�1x�1j j.

3. Write out the proof of Theorem 4 (the Composition Rule) in Sub-

section 5.1.3.

4. For the functions

f xð Þ ¼0; x ¼ 0;1; x ¼ 1;2; x 6¼ 0; 1;

8

<

:

and g xð Þ ¼ 0; x ¼ 0;1þ xj j; x 6¼ 0;

determine f g 0ð Þð Þ; f

limx!0

g xð Þ�

and limx!0

f g xð Þð Þ.

Section 5.2

1. Prove that

(a) 1x4 !1 as x! 0; (b) cot x!1 as x! 0þ;

(c) exp ex � xð Þ ! 1 as x!1; (d) loge x! �1 as x! 0þ;

(e) xþ sin x!1 as x!1; (f) xx !1 as x!1.

2. Let p xð Þ ¼ xn þ an�1xn�1 þ � � � þ a1xþ a0, where a0; a1; . . .; an�1 2 R .

Prove that p xð Þ ! 1 as x!1.

Hint: Write p xð Þ ¼ xn 1þ r 1x

� ��

, for a suitable polynomial r.

3. For a> 0 and n2Z, prove that:

(a) eaxxn !1 as x!1; (b) eaxxn ! 0 as x! �1.

Here we use the LimitInequality Rule, Theorem 6of Sub-section 5.1.3.

In Theorem 3, Sub-section 7.2.2.

5.6 Exercises 203

4. For a< 0 and n2Z, prove that:

(a) eaxxn ! 0 as x!1;

(b) eaxxn ! �1; as x! �1; if n is odd;1; as x! �1; if n is even:

Section 5.3

1. Use the strategy based on the "� � definition of limit to prove the following

statements:

(a) limx!3

3x� 4ð Þ ¼ 5; (b) limx!2þ


x2 � 4p

¼ 0.

2. Use the strategy based on the "� � definition of limit to prove the following

statements:

(a) limx!0

sin xxj j does not exist; (b) lim

x!0cos 1

x

� �

does not exist.

3. Use the strategy based on the "� � definition of limit to prove that, if

limx!c

f xð Þ ¼ ‘, where ‘ 6¼ 0, then limx!c

1f xð Þ exists and equals 1

‘.

Section 5.4

1. Use the strategy based on the "� � definition of continuity to prove that:

(a) f xð Þ ¼ x5 is continuous at 0; (b) f xð Þ ¼ 1x

is continuous at 2.

2. Use the strategy based on the "� � definition of continuity to prove that the

function

f xð Þ ¼ sin 1x; if x 6¼ 0;

0; if x ¼ 0;

is discontinuous at 0.

3. Use the "� � definition of continuity to prove that, if f and g are continuous

at c, then so is the product fg.

Section 5.5

1. A function f is defined on a bounded interval I, on which it satisfies the

following inequality for some given number K2R :

f xð Þ � f yð Þj j � K x� yj j; for all x; y in I:

Prove that f is (a) bounded on I, and (b) uniformly continuous on I.

2. Give an example of a function f that is continuous and bounded on R but is

not uniformly continuous on R , and verify your assertions.

3. Prove that, if a function f is continuous on a half-closed interval (a, b], then

it is uniformly continuous on (a, b] if and only if limx!aþ

f xð Þ exists (and is

finite).

6 Differentiation

The family of all functions is so large that there is really no possibility of

finding many interesting properties that they all possess. In the last two

chapters we concentrated our attention on the class of all continuous

functions, and we found that continuous functions share some important

properties – for example, they satisfy the Intermediate Value Theorem, the

Extreme Values Theorem and the Boundedness Theorem. However, many

of the most interesting and powerful properties of functions are obtained

only when we further restrict our attention to the class of all differentiable

functions.

You will have already met the idea of differentiating a given function f; that

is, finding the slope of the tangent to the graph y¼ f(x) at those points of the

graph where a tangent exists. The slope of the tangent at the point (c, f(c)) is

called the derivative of f at c, and is written as f 0(c). But when does a function

have a derivative? Geometrically, the answer is: whenever the slope of the

chord through the point (c, f(c)) and an arbitrary point (x, f(x)) of the graph

approaches a limit as x! c. In this chapter we investigate which functions are

differentiable, and we discuss some of the important properties that all differ-

entiable functions possess.

In Section 6.1 we give a strategy for determining whether a given function

f is differentiable at a given point c. In particular, we prove that, if f is

differentiable at c, then it is continuous at c; whereas a function f may be

continuous at c, but not differentiable at c. We also consider functions that

possess higher derivatives; that is, functions which can be differentiated more

than once.

In Section 6.2 we obtain various standard derivatives and rules for differ-

entiation, including the Inverse Function Rule.

In Sections 6.3 and 6.4 we study the properties of functions that are differ-

entiable on an interval, and establish some powerful results about derivatives

which are easy to describe geometrically.

In Section 6.5 we meet an important and useful result, called l’Hopital’s

Rule, which enables us to find limits of the form

limx!c

f xð Þg xð Þ ;

in some of the awkward cases when f (c)¼ g(c)¼ 0.

In Section 6.6 we construct the Blancmange function, a function that is

continuous everywhere on R , but differentiable nowhere on R . It is related to

certain types of fractals.

In fact there exist functionsthat are continuouseverywhere on R , butdifferentiable nowhere on R .This discovery by KarlWeierstrass in 1872 caused asensation in Analysis.

In this case, the Quotient Rulefor limits of functions fails.

205

6.1 Differentiable functions

6.1.1 What is differentiability?

Differentiability arises from the geometric concept of the tangent to a graph.

The tangent to the graph y¼ f (x) at the point (c, f (c)) is the straight line through

the point (c, f (c)) whose direction is the limiting direction of chords joining the

points (c, f (c)) and (x, f (x)) on the graph, as x! c.

The following three examples illustrate some of the possibilities that can

occur when we try to find tangents in particular instances.

The function f (x)¼ x2, x2R , is continuous on R , and its graph has a tangent

at each point; for example, the line y¼ 2x� 1 is the tangent to the graph at the

point (1, 1).

On the other hand, although the function g(x)¼ jx� 1j, x2R , is continuous

on R , its graph does not have a tangent at the point (1, 0); no line through the

point (1, 0) is a tangent to the graph.

Finally, the signum function

k xð Þ ¼�1; x < 0,

0; x ¼ 0,

1; x > 0,

(

is discontinuous at 0, and no line through the point (0, 0) is a tangent to the

graph.

We now make these ideas more precise, using the concept of limit to pin

down what we meant above by ‘limiting direction’. We define the slope of the

graph at (c, f (c)) to be the limit, as x tends to c, of the slope of the chord through

the points (c, f (c)) and (x, f (x)). The slope of this chord is

f xð Þ � f cð Þx� c

;

and this expression is called the difference quotient for f at c. Thus the slope

of the graph of f at (c, f (c)) is

limx!c


; (1)

provided that this limit exists.

Sometimes it is more convenient to use an equivalent form of the differ-

ence quotient, particularly when we are examining a specific function for

206 6: Differentiation

differentiability at a specific point. If we replace x by cþ h, then ‘x! c’ in (1)

is equivalent to ‘h! 0’. Thus we can rewrite the difference quotient for f at c as

Q hð Þ ¼ f cþ hð Þ � f cð Þh

;

and the slope of the graph of f at (c, f (c)) is then

limh!0

Q hð Þ; (2)

provided that this limit exists.

We say that f is differentiable at c if the graph y¼ f (x) has a tangent at the

point (c, f (c)), and that the derivative of f at c is the limit of the difference

quotient given by expression (1) or expression (2). To formalise this concept,

we need to ensure that f is defined near the point c, and so we assume that c

belongs to some open interval I in the domain of f.

Definitions Let f be defined on an open interval I, and c2 I. Then the

derivative of f at c is

limx!c


or limh!0

Q hð Þ;

provided that this limit exists. In this case, we say that f is differentiable at c.

The derivative of f at c is denoted by f 0(c), and the function f 0: x 7! f 0 xð Þis called the derived function. The operation of obtaining f 0(x) from f (x)

is called differentiation.

For future reference, we reformulate the definition in terms of " and � as

follows.

Definition Let f be defined on an open interval I, and c2 I. Then f is

differentiable at c with derivative f 0(c) if:

for each positive number ", there is a positive number � such thatf xð Þ�f cð Þ

x�c� f 0 cð Þ

�

�

�

�

�

�< ", for all x satisfying 0< jx� cj<�.

Remarks

1. In ‘Leibniz notation’, f 0(x) is written as dydx

, where y¼ f (x). As we shall see

later, this notation is sometimes useful; however, it is important to recall

that the symbol dydx

is purely notation and does NOT mean some quantity dy

‘divided by’ another quantity dx.

2. The existence of the derivative f 0(c) is not quite equivalent to the existence

of a tangent to the graph y¼ f (x) at the point (c, f (c)). For, if limx!c

f xð Þ�f cð Þx�c

exists (and so is some real number), then the graph has a tangent at the point

(c, f (c)), and the slope of the tangent is the value of this limit. On the other

hand, the graph may have a vertical tangent at the point (c, f (c)). In this

case,f xð Þ�f cð Þ

x�c!1 or �1 as x! c, so that lim

x!c


does not exist; thus

f is not differentiable at c.

3. Since the concept of a derivative is defined in terms of a limit, we shall use

many results obtained for limits of functions to prove analogous results for

derivatives.

Note that the differencequotient depends on thechoice of c; for differencechoices of c, the differencequotient is generally different.

Sometimes f 0 is denoted by Dfand f 0(x) by Df (x).

This is a simple reformulation,with nothing else taking place.We shall use the definitionoccasionally in this form,especially when provinggeneral results as distinct fromlooking at specific functionsand specific points.

For instance, in the statementof the Composition Rule inSub-section 6.2.2.

Example 1 Prove that the function f (x)¼ x3, x2R , is differentiable at any

point c, and determine f 0(c).

Solution At the point c


¼ cþ hð Þ3� c3

h

¼ 3c2hþ 3ch2 þ h3

h

¼ 3c2 þ 3chþ h2:

Thus, Q(h)! 3c2 as h! 0. It follows that f is differentiable at c, and that

f 0(c)¼ 3c2. &

For comparison, we now prove the same result using the "� � definition of

differentiability; for simplicity, though, we shall assume that c¼ 2 and we shall

simply check that f 0(2)¼ 12. We think that you will see why the Q(h) method is

often preferred!

Solution We have to show that for each positive number ", there is a positive

number � such that

x3 � 8

x� 2� 12

�

�

�

�

�

�

�

�

< "; for all x satisfying 0 < x� 2j j < �;

that is

x2 þ 2x� 8�

�

�

� < "; for all x satisfying 0 < x� 2j j < �:

Now, x2þ 2x� 8¼ (xþ 4)(x� 2); so, for 0< jx� 2j< 1, we have x2 (1, 3)

and jxþ 4j< 7.

Next, choose � ¼ min 1; 17"

� �

. With this choice of �, it follows that, for all x

satisfying 0< jx� 2j<�, we have

x2 þ 2x� 8�

�

�

� ¼ xþ 4j j � x� 2j j

< 7� 1

7" ¼ ":

It follows that f is differentiable at 2, with derivative f 0(2)¼ 12. &

To prove that a function is not differentiable at a point, we use the strategy for

limits that applies in these situations.

Strategy To show that limh!0

Q hð Þ does not exist:

1. Show that there is no punctured neighbourhood of c on which Q is defined;

OR

2. Find two non-zero null sequences {hn} and h0n� �

such that {Q(hn)} and

Q h0n� ��

have different limits;

OR

3. Find a null sequence {hn} such that Q(hn)!1 or Q(hn)!�1.

Example 2 Prove that the modulus function f(x)¼ jxj, x2R , is not differ-

entiable at 0.

Recall that h 6¼ 0.

Thus, the derived function off is f 0(x)¼ 3x2, x2R .

Here we use the fact thatx3�8¼ x�2ð Þ x2þ2xþ4ð Þ:

We arrange for x2 (1, 3), inorder to concentrate simply onvalues of x near to 2. Thebound 7 for jxþ 4j is simplyused in order to keep somegiven bound (not necessarily asmall bound) for this term.

It is now clear why we madethat particular choice for �; itarose from trying variousvalues and then adjusting thechoice appropriately in orderto end up with a neat ‘"’.Alternatively, we could haveused a different value for � interms of ", and the K " Lemma.

For convenience, we rephrasethis strategy in terms oflimh!0

Q hð Þ rather than limx!c

f xð Þ.

Solution The graph suggests that the slopes of chords joining the origin to

points on the graph of f to the left and to the right do not tend to the same limit

as these points approach the origin. This suggests that we investigate whether

case 2 of the above strategy may be useful.

At the point 0

Q hð Þ ¼ f 0þ hð Þ � f 0ð Þh

¼ hj j � 0j jh

¼ hj jh:

Now let {hn} and h0n� �

be two null sequences, where hn ¼ 1n

and h0n ¼ � 1n.

Then

Q hnð Þ ¼ Q1

n

� �

¼1n

�

�

�

�

1n

¼1n1n

! 1 as n!1;

and

Q h0n� �

¼ Q � 1

n

� �

¼�1

n

�

�

�

�

�1n

¼1n

�1n

! �1 as n!1:

It follows that f is not differentiable at 0. &

Problem 1

(a) Prove that the function f (x)¼ xn, x2R , n2N , is differentiable at

any point c, and determine f 0(c).

(b) Prove that the constant function on R is differentiable at any point c,

with derivative zero.

Problem 2 Prove that the function f xð Þ ¼ 1x; x 2 R � 0f g, is differen-

tiable at any point c 6¼ 0, and determine f 0(c).

Problem 3 Use the "� � definition of differentiability to prove that the

function f (x)¼ x4 is differentiable at 1, with derivative f 0(1)¼ 4.

Problem 4 Prove that the following functions f are not differentiable

at the given point c:

(a) f xð Þ ¼ xj j12; x 2 R , c¼ 0; (b) f(x)¼ [x], x2R , c¼ 1.

Looking back at Example 2, it looks as though chords joining the origin to

points (h, f (h)) have slopes that tend to a limit 1 if h! 0þ, whereas they have

slopes that tend to a limit�1 if h! 0�. This suggests the concept of one-sided

derivatives that will be useful in our work later on.

This is motivation for theapproach to the solution ratherthan part of the solution itself.

This is the start of thesolution.

y

y = ⏐x⏐

x01n

1n

–

In particular, the derivative ofthe identity function f (x)¼ x,x2R , is the constant functionf 0(x)¼ 1.

Here [x] denotes the integerpart of x.


c

slopef ′(c)

slopef ′(c)

y

x

y = f (x)

c

y

x

y = f (x)L

R

Definitions Let f be defined on an interval I, and c2 I. Then the leftderivative of f at c is

f 0L cð Þ ¼ limx!c�


or f 0L cð Þ ¼ limh!0�

Q hð Þ;

provided that this limit exists. In this case, we say that f is differentiable on

the left at c.

Similarly, the right derivative of f at c is

f 0R cð Þ ¼ limx!cþ


or f 0R cð Þ ¼ limh!0þ

Q hð Þ;

provided that this limit exists. In this case, we say that f is differentiable on

the right at c.

A function f whose domain contains an interval I is differentiable on I if

it is differentiable at each interior point of I, differentiable on the right at the

left end-point of I (if this belongs to I ) and differentiable on the left at the

right end-point of I (if this belongs to I ).

With this notation, the function f in Example 2 is differentiable on the left at

0 and f 0L 0ð Þ ¼ �1; it is also differentiable on the right at 0, and f 0R 0ð Þ ¼ 1.

The connection between the definitions of differentiability and of one-sided

differentiability is rather obvious.

Theorem 1 A function f whose domain contains an interval I that contains

c as an interior point is differentiable at c if and only if f is both differenti-

able on the left at c and differentiable on the right at c AND

f 0L cð Þ ¼ f 0R cð Þ:

Example 3 Determine whether the function

f xð Þ ¼ xþ x2; �1 � x < 0,

sin x; 0 � x � 2p,

is differentiable at the points c¼�1, 0 and 2p, and determine the correspond-

ing derivatives when they exist.

Solution We investigate the behaviour of the difference quotient Q(h) at

each point in turn.

At �1, the function is not defined to the left of �1; and, for 0< h< 1,

we have

Q hð Þ ¼ f �1þ hð Þ � f �1ð Þh

All these definitions areanalogous to definitions forcontinuity that you met inSub-section 4.1.1.

We omit a proof of thisstraight-forward result.

The common value issimply f 0(c).

y

–1 π2π

x

y = f (x)

¼�1þ hð Þ þ �1þ hð Þ2

n o

� �1ð Þ þ �1ð Þ2n o

h

¼ �hþ h2

h

¼ �1þ h! �1 as h! 0þ:

It follows that f is differentiable on the right at �1, and f 0R �1ð Þ ¼ �1.

At 0, the function is defined on either side of 0, but by a different formula;

we therefore examine each side separately.

At 0, for 0< h< 2p, we have

Q hð Þ ¼ f hð Þ � f 0ð Þh

¼ sin h� sin 0

h

¼ sin h

h! 1 as h! 0þ:

Also at 0, for �1< h< 0, we have


¼hþ h2� �

� sin 0f gh

¼ hþ h2

h

¼ 1þ h! 1 as h! 0�:

It follows that f is differentiable on the right at 0 and f 0R 0ð Þ ¼ 1, and that f

is differentiable on the left at 0 and f 0L 0ð Þ ¼ 1. Since the left and right deriva-

tives at 0 are equal, it follows that f is differentiable at 0 and f 0 0ð Þ ¼ 1.

Finally, at 2p, the function is not defined to the right of 2p; and,

for �2p< h< 0, we have

Q hð Þ ¼ f 2pþ hð Þ � f 2pð Þh

¼ sin 2pþ hð Þ � sin 2ph

¼ sin h

h! 1 as h! 0�:

It follows that f is differentiable on the left at 2p, and f 0L 2pð Þ ¼ 1. &

Problem 5 Determine whether the function

f xð Þ ¼�x2; �2 � x <0,

x4; 0 � x < 1,

x3; 1 � x � 2,

0; x > 2,

8

>

>

<

>

>

:

is differentiable at the points c¼�2, 0, 1 and 2, and determine the

corresponding derivatives when they exist.

Hint: Sketch the graph y¼ f(x) first.

Remarks

Just as with continuity, the definition of differentiability involves a function

f defined on a set in R , the domain A (say), that maps A to another set in R ,

the codomain B (say).

These remarks are analogousto similar remarks forcontinuity inSub-section 4.1.1.

1. Let f and g be functions defined on open intervals I and J, respectively,

where I� J; and let f (x)¼ g(x) on J. Technically g is a different func-

tion from f. However, if f is differentiable at an interior point c of J, it

is a simple matter of some definition checking to verify that g too is

differentiable at c. Similarly, if f is non-differentiable at c, g too is non-

differentiable at c.

2. The underlying point here is that differentiability at a point is a local

property. It is only the behaviour of the function near that point that

determines whether it is differentiable at the point.

6.1.2 Differentiability and continuity

All the functions that we have met so far that are differentiable at a particular

point c are also continuous at that point. In fact, this property holds in general.

Theorem 2 Let f be defined on an open interval I, and c2 I. If f is differ-

entiable at c, then f is also continuous at c.

Proof If f is differentiable at c, then there is some number f 0(c) such that

limx!c

f xð Þ � fðcÞx� c

¼ f 0 cð Þ:

It follows that

limx!c

f xð Þ � f cð Þf g ¼ limx!c


� x� cð Þ

¼ limx!c


� limx!c

x� cð Þn o

¼ f 0 cð Þ � 0 ¼ 0:

Hence, by the Sum and Multiple Rules for limits

limx!c


Thus f is continuous at c. &

In fact this gives us a useful test for non-differentiability.

Corollary 1 If f is discontinuous at c, then f is not differentiable at c.

For example, the signum function

f xð Þ ¼�1; x < 0,

0; x ¼ 0,

1; x > 0,

(

is discontinuous at c, and so cannot be differentiable at c.

It is often worth checking whether a function is even continuous at a point

before setting out on a complicated investigation of difference quotients to

determine whether it is differentiable at that point.

differentiable) continuous

There is a similar result forone-sided derivatives.

By the Product Rule forLimits, Sub-section 5.1.3.

Sub-section 5.1.3

Problem 5, Sub-section 4.1.1.

Problem 6 Prove that the function

f xð Þ ¼ sin 1x; x 6¼ 0,

0; x ¼ 0,

is not differentiable at 0.

Example 4 Show that the function

f xð Þ ¼ x sin 1x; x 6¼ 0,

0; x ¼ 0,

is continuous at 0, but not differentiable at 0.

Solution We proved earlier that f is continuous at 0.

At 0


¼h sin 1

h

� �

� 0

h

¼ sin1

h

� �

:

Thus, if we define two null sequences xnf g ¼ 1np

� �

and x0n� �

¼n

1

2nþ12ð Þpo

;

n 2 N , we have

Q xnð Þ ¼ sin npð Þ ¼ 0 and Q x0n� �

¼ sin 2nþ 1

2

� �

p� �

¼ sin1

2p

� �

¼ 1;

so that {Q (xn)} and Q x0n� ��

tend to different limits as n!1. It follows that

f cannot be differentiable at 0. &


f xð Þ ¼ x2 sin 1x; x 6¼ 0,

0; x ¼ 0,

is differentiable at 0. Given that, for x 6¼ 0; f 0 xð Þ ¼ 2x sin 1x� cos 1

x,

is f 0 continuous at 0?

6.1.3 The sine, cosine and exponential functions

In order to study the differentiability of each of the sine, cosine and exponential

functions, we need to use the following three standard limits.

Theorem 3 Three standard limits

(a) limx!0

sin xx¼ 1; (b) lim

x!0

1�cos xx¼ 0; (c) lim

x!0

ex�1x¼ 1:

Proof The limits (a) and (c) have been verified earlier, so we have only to

verify (b) now.

Using the half-angle formula for cosine, we have

limx!0

1� cos x

x¼ lim

x!0

2 sin2 12

x� �

x

Sub-section 4.1.2,Problem 8 (a).

y

x

y = x sin 1x

y = x2 sin 1x

y

x

(a) was proved in Sub-section 5.1.1; (c) inSub-section 5.1.4, Problem 8.

For cos x ¼ 1� 2 sin2 12

x� �

.


¼ limx!0

sin 12

x� �

12

x� sin

1

2x

� ��

!

¼ limx!0

sin 12

x� �

12

x

!

� limx!0

sin1

2x

� ��

¼ 1� 0 ¼ 0: &

With this tool, we can now tackle the various functions.

Theorem 4 The function f (x)¼ sin x is differentiable on R , and f 0(x)¼ cos x.

Proof Let c be any point in R . At c, the difference quotient for f is

Q hð Þ ¼ sin cþ hð Þ � sin c

h

¼ sin c cos hþ cos c sin h� sin c

h

¼ cos c� sin h

h� sin c� 1� cos h

h;

so that

limh!0

Q hð Þ ¼ cos c� limh!0

sin h

h� sin c� lim

h!0

1� cos h

h

¼ cos c� 1� sin c� 0

¼ cos c:

It follows that f is differentiable at c, and f 0(c)¼ cos c. &

Problem 8 Prove that the function f (x)¼ cos x is differentiable on R ,

and f 0(x)¼�sin x.

Finally we find the derivative of the exponential function. This is an

extremely important result, as the function f (x)¼ lex, l an arbitrary constant,

is the only function f that satisfies the differential equation f 0(x)¼ f (x) on R .

Theorem 5 The function f (x)¼ ex is differentiable on R , and f 0(x)¼ ex.

Proof Let c be any point in R . At c, the difference quotient for f is

Q hð Þ ¼ ecþh � ec

h

¼ ec � eh � 1

h;

so that

limh!0

Q hð Þ ¼ ec � limh!0

eh � 1

h

¼ ec � 1

¼ ec:

It follows that f is differentiable at c, and f 0(c)¼ ec. &

By the Product Rule forlimits.

Since sine is continuous at 0.

That is, f is its own derivedfunction.


6.1.4 Higher-order derivatives

In the previous sub-section, we saw that sin0 ¼ cos and cos0 ¼�sin. An exactly

similar argument shows that (�sin)0 ¼�cos and (�cos)0 ¼ sin. Thus �sin

and �cos are all differentiable on R , and so clearly we can differentiate the

functions sin and cos as many times as we please on R .

In general, when we differentiate any given differentiable function f, we

obtain a new function f 0 (whose domain may be smaller than that of f ). The

notion of differentiability can then be applied to the function f 0, just as before,

yielding another function f 00 ¼ ( f 0)0, whose domain consists of those points

where f 0 is differentiable.

Definitions Let f be differentiable on an open interval I, and c2 I. If f 0 is

differentiable at c, then f is called twice differentiable at c, and the number

f 00(c) is called the second derivative of f at c. The function f 00 is called the

second derived function of f.

Provided that the derivatives exist, we can define f 000 or f (3), f (4), . . .,f (n), . . .. The functions f 00, f (3), f (4), . . ., f (n), . . . are called the higher-order

derived functions of f, whose values f 00(x), f (3)(x), f (4)(x), . . ., f (n)(x), . . .are called the higher-order derivatives of f.

However, not all derived functions are themselves differentiable. You have

already seen, for example, that the function

f xð Þ ¼ x2 sin 1x; x 6¼ 0,

0; x ¼ 0,

is differentiable at 0. In fact, it has as its derived function

f 0 xð Þ ¼ 2x sin 1x� cos 1

x; x 6¼ 0,

0; x ¼ 0,

which is not even continuous at 0.

Example 5 Prove that the function

f xð Þ ¼ �12

x2; x < 0,12

x2; x � 0,

is differentiable on R , but that its derived function is not differentiable at 0.

Solution For c> 0, the difference quotient for f at c is

Q hð Þ ¼12

cþ hð Þ2� 12

c2

h

¼ 1

22cþ hð Þ ! c as h! 0;

so that f is differentiable at c and f 0(c)¼ c.

When c¼ 0 and h> 0, a similar argument shows that f is differentiable on

the right at 0 and f 0R 0ð Þ ¼ 0.

For c< 0, the difference quotient for f at c is

Q hð Þ ¼�1

2cþ hð Þ2þ 1

2c2

h

¼ �1

22cþ hð Þ ! �c as h! 0;

f 0 0 ¼ (f 0)0 is sometimes writtenas f (2).

We assume that h issufficiently small that jhj< c,so that cþ h> 0; and hencethat we are using the correctvalue for f at cþ h.

We assume that h issufficiently small thatjhj<�c, so that cþ h< 0;and hence that we are usingthe correct value for f at cþ h.

so that f is differentiable at c and f 0(c)¼�c.

When c¼ 0 and h< 0, a similar argument shows that f is differentiable on

the left at 0 and f 0L 0ð Þ ¼ 0.

Since f has f 0L 0ð Þ ¼ f 0R 0ð Þ ¼ 0, it follows that f is differentiable at 0 and

f (0)¼ 0.

Thus the derived function for f is given by

f 0 xð Þ ¼ xj j; x 2 R :

This is the modulus function, which as we showed earlier is not differentiable

at 0. It follows that f 0 is not differentiable at 0. &


f xð Þ ¼ x2; x < 0;x3; x � 0;

is differentiable on R . Is f 0 differentiable at 0?

6.2 Rules for differentiation

In the last section we showed that the functions sin, cos and exp are differenti-

able on R , by appealing directly to the definition of differentiability.

However, it would be very tedious if, every time that we wished to prove that

a given function is differentiable and to determine its derived function, we had

to use the difference quotient method described in Section 6.1. Sometimes we

do need to use that method, but usually we can avoid the algebra involved in

the difference quotient method by using various rules for differentiation. In this

section, we introduce the Combination Rules, the Composition Rule and the

Inverse Function Rule for differentiable functions. These are similar to the

rules for continuous functions that you met earlier.

Note that each of the rules for differentiation supplies two pieces of

information:

1. a function of a certain type is differentiable;

2. an expression for the derivative.

6.2.1 The Combination Rules

The Combination Rules for differentiable functions are a consequence of the

Combination Rules for limits.


Let f and g be defined on an open interval I, and c2 I. Then, if f and g are

differentiable at c, so are:

Sum Rule fþ g, and ( fþ g)0(c)¼ f 0(c)þ g0(c);

Multiple Rule l f, for l2R , and (l f )0(c)¼ l f 0(c);

Product Rule fg, and ( fg)0(c)¼ f 0(c)g (c)þ f (c)g0(c);

Example 2, Sub-section 6.1.1.

In Sections 4.1 and 4.3.

We differentiate each termin turn.

We differentiate one term at atime.

Quotient Rule fg, provided that g(c) 6¼ 0; and

fg

� �0cð Þ ¼ g cð Þf 0 cð Þ�f cð Þg0 cð Þ

g cð Þð Þ2 :

Now, we saw in Section 6.1 that, for any n2N , the function

x 7! xn; x 2 R ;

is differentiable on R , and that its derived function is

x 7! nxn�1; x 2 R :

We can use this fact, together with the Combination Rules, to prove that any

polynomial function is differentiable on R , and that its derivative can be

obtained by differentiating the polynomial term-by-term.

Corollary 1 Let p(x)¼ a0þ a1xþ a2x2þ � � � þ anxn, x2R , where a0, a1,

a2, . . ., an2R . Then p is differentiable on R , and its derivative is

p0 xð Þ ¼ a1 þ 2a2xþ � � � þ nanxn�1; x 2 R :

Since a rational function is a quotient of two polynomials, it follows from

Corollary 1 and the Quotient Rule that a rational function is differentiable at

all points where the denominator does not vanish (that is, the denominator

does not take the value 0).

Example 1 Prove that the function f xð Þ ¼ x3

x2�1; x 2 R � �1f g, is differenti-

able on its domain, and find its derivative.

Solution The function f is a rational function whose denominator does not

vanish on R � {�1}; hence f is differentiable on this set (the whole of its

domain).

By Corollary 1, the derived function of x 7! x3 is x 7! 3x2, and the derived

function of x 7! x2 � 1 is x 7! 2x. It follows from the Quotient Rule that the

derivative of f is

f 0 xð Þ ¼ x2 � 1ð Þ � 3x2 � x3 � 2xð Þx2 � 1ð Þ2

¼ x4 � 3x2

x2 � 1ð Þ2: &

Problem 1 Find the derivative of each of the following functions:

(a) f (x)¼ x7� 2x4þ 3x3� 5xþ 1, x2R ;

(b) f xð Þ ¼ x2þ1x3�1

; x 2 R � 1f g;(c) f (x)¼ 2 sin x cos x, x2R ;

(d) f xð Þ ¼ ex

3þsin x�2 cos x; x 2 R :

Problem 2 Find the third order derivative of the function f (x)¼ xe2x,

x2R .

In the last section we found the derived functions for sin, cos and exp. We

now ask you to find the derived functions for the remaining trigonometric

functions and the three most common hyperbolic functions.

In particular,

1g

� �0cð Þ ¼ � g0 cð Þ

g cð Þð Þ2 :



(a) f xð Þ ¼ tan x; x 2 R � � 12p;� 3

2p;� 5

2p; . . .

� �

;

(b) f xð Þ ¼ cosec x; x 2 R � 0;�p;�2p; . . .f g;(c) f xð Þ ¼ sec x; x 2 R � � 1

2p;� 3

2p;� 5

2p; . . .

� �

;

(d) f xð Þ ¼ cot x; x 2 R � 0;�p;�2p; . . .f g:

Problem 4 Find the derivative of each of the following functions

(a) f (x)¼ sinh x, x2R ; (b) f(x)¼ cosh x, x2R ;

(c) f (x)¼ tanh x, x2R .

6.2.2 The Composition Rule

In the last sub-section we extended our stock of differentiable functions to

include all rational, trigonometric and hyperbolic functions. However, to differ-

entiate many other functions we need to differentiate composite functions,

such as the function f (x)¼ sin(cos x), x2R , which is the composite of two

differentiable functions – namely, f ¼ sin cos. The Composition Rule tells us

that the composite of two differentiable functions is itself differentiable.


Let g and f be defined on open intervals I and J, respectively, and let c2 I and

g Ið Þ J. If g is differentiable at c and f is differentiable at g(c), then f g is

differentiable at c and

ð f gÞ0 cð Þ ¼ f 0 g cð Þð Þg0 cð Þ:

Remarks

1. When written in Leibniz notation, the Composition Rule has a form that is

easy to remember: if we put

u ¼ g xð Þ and y ¼ f uð Þ ¼ f g xð Þð Þthen

dy

dx¼ dy

du� du

dx:

2. We can extend the Composition Rule to a composite of three (or more)

functions; for example

f g hð Þ0 xð Þ ¼ f 0 g h xð Þð Þ½ �g0 h xð Þð Þh0 xð Þ:In Leibniz notation, if we put

v ¼ h xð Þ; u ¼ g vð Þ and y ¼ f uð Þ ¼ f g h xð Þð Þ½ �;

This rule is sometimes knownas the Chain Rule.

We give the proof of the Rulein Sub-section 6.2.4.

We frequently use thisextension of the CompositionRule without mentioning itexplicitly.


then we obtain the chain

dy

dx¼ dy

du� du

dv� dv

dx:

Example 2 Prove that each of the following composite functions is differ-

entiable on its domain, and find its derivative:

(a) k(x)¼ sin (cos x), x2R ; (b) k(x)¼ cosh (e2x), x2R ;

(c) k xð Þ ¼ tan 14

ex2 þ sin px� �

; x 2 �1; 1ð Þ:

Solution

(a) Here k(x)¼ sin (cos x), so let

g xð Þ ¼ cos x and f xð Þ ¼ sin x; for x 2 R ;

then

g0 xð Þ ¼ �sin x and f 0 xð Þ ¼ cos x; for x 2 R :

By the Composition Rule, k ¼ f g is differentiable on R , and

k0 xð Þ ¼ f 0 g xð Þð Þg0 xð Þ¼ cos cos xð Þ � � sin xð Þ¼ �cos cos xð Þ � sin x:

(b) Here k(x)¼ cosh (e2x), so let

h xð Þ ¼ 2x; g xð Þ ¼ ex and f xð Þ ¼ cosh x; for x 2 R ;

then h, g and f are differentiable on R , and

h0 xð Þ ¼ 2; g0 xð Þ ¼ ex and f 0 xð Þ ¼ sinh x; for x 2 R :

By the (extended form of the) Composition Rule, k ¼ f g h is differ-

entiable on R , and

k0 xð Þ ¼ f 0 g h xð Þð Þ½ �g0 h xð Þð Þh0 xð Þ¼ sinh e2x� �

� e2x � 2

¼ 2e2x sinh e2x:

(c) Here k xð Þ ¼ tan 14

ex2 þ sin px� �

; x 2 �1; 1ð Þ, so let

g xð Þ¼ 1

4ex2 þ sin px� �

; x 2 �1; 1ð Þ; and

f xð Þ¼ tan x; x 2 � 1

2p;

1

2p

� �

:

Now, when x2 (�1, 1), we have

g xð Þj j � 1

4ex2 þ sin pxj j� �

� 1

4eþ 1ð Þ

< 1

<1

2p;

so that g(x) lies in � 12p; 1

2p

� �

, the domain of the differentiable func-

tion tan.

Now, by the Composition and Combination Rules, g is differenti-

able on (�1, 1), and

g0 xð Þ ¼ 1

42xex2 þ p cos px� �

; x 2 �1; 1ð Þ:

Also

f 0 xð Þ ¼ sec2 x; x 2 � 1

2p;

1

2p

� �

:

Finally, it follows, by the Composition Rule, that k ¼ f g is differentiable

on (�1, 1), and

k0 xð Þ¼ f 0 g xð Þð Þg0 xð Þ

¼ sec2 1

4ex2 þ sinpx� �

� �

�1

42xex2 þpcospx� �

; x2 �1;1ð Þ: &


(a) f(x)¼ sinh (x2), x2R ; (b) f (x)¼ sin (sinh 2x), x2R ;

(c) f xð Þ ¼ sin cos 2xx2

� �

; x 2 0;1ð Þ:

6.2.3 The Inverse Function Rule

Earlier we showed that, if a function f with domain some interval I and

image J is strictly monotonic on I, then f possesses a strictly monotonic and

continuous inverse function f�1 on J. In particular, the power function, the

trigonometric functions, the exponential function and the hyperbolic func-

tions all have inverse functions, provided that we restrict the domains, where

necessary.

These standard functions are all differentiable on their domains. Do their

inverse functions also have this property, as their graphs suggest?

As you saw earlier, we obtain the graph y¼ f�1(x) by reflecting the graph

y¼ f(x) in the line y¼ x. This reflection maps a typical point P(c, d) to the

point Q(d, c).

It follows that, if the slope of the tangent to the graph y¼ f (x) at the point P

is f 0(c)¼m, then the slope of the tangent to the graph y¼ f� 1(x) at Q is

f�1ð Þ0 dð Þ ¼ 1m:

However, if the graph of f has a horizontal tangent (m¼ 0) at a point P, then

the graph of f�1 has a vertical tangent at the corresponding point Q; in this case,

f�1 is not differentiable at Q, since 1m

is not defined when m¼ 0. We therefore

require the condition ‘f 0(x) is non-zero’ in our statement of the Rule.

Thus, in the Inverse Function Rule we require:

(a) f is strictly monotonic and continuous on an open interval I, so that f has

a strictly monotonic and continuous inverse function on the open interval

J¼ f(I);

(b) f is differentiable on I and f 0(x) 6¼ 0 on I, so that 1f 0ðxÞ is defined.

The Rule tells us that, under these conditions, f �1 is also differentiable, and

there is a simple formula for ( f �1)0.

Section 4.3.

Recall that strictly monotonicmeans that f is either strictlyincreasing or strictlydecreasing.

y

PQ

x

y = x

y = f –1(x)

y = f (x)

y

P

Q

x

y = x

y = f –1(x)y = f (x)



Let f : I! J, where I is an interval and J is the image f(I), be a function such

that:

1. f is strictly monotonic on I;

2. f is differentiable on I;

3. f 0(x) 6¼ 0 on I.

Then f�1 is differentiable on J. Further, if c2 I and d¼ f (c), then

f�1� �0

dð Þ ¼ 1

f 0 cð Þ :

Remark

The Leibniz notation for derivatives can be used to express the Inverse

Function Rule in a form that is easy to remember: if

y ¼ f xð Þ and x ¼ f�1 yð Þ;

and we write dydx

for f 0 xð Þ, and dxdy

for ( f�1)0(y), then

dx

dy¼ 1

dydx

:

Proof Let F¼ f�1, and let y¼ f (x), where x2 I, so that x¼F(y), where y2 J.

Then the difference quotient for F at d is

F yð Þ � F dð Þy� d

¼ x� c

f xð Þ � f cð Þ

¼ 1f xð Þ�f cð Þ

x�c:

Since f is one–one and differentiable on I, it is necessarily one–one and

continuous on I. Thus F is one–one and continuous on J. It follows that, for

y 6¼ d, we must have x 6¼ c, since x¼ f�1(y) and c¼ f�1(d).

Also, since f�1 is continuous, x! c as y! d. It follows that

limy!d

F yð Þ � F dð Þy� d

¼ limx!c

1f xð Þ�f cð Þ

x�c

!

¼ 1

limx!c


� �

¼ 1

f 0 cð Þ :

Thus F is differentiable at c, with derivative 1f 0 cð Þ. &

Example 3 For each of the following functions, show that f�1 is differenti-

able and determine its derivative:

(a) f (x)¼ xn, x> 0, n2N; (b) f xð Þ ¼ tan x; x 2 � 12p; 1

2p

� �

;

(c) f (x)¼ ex, x2R .

y

J

dy

x c I x


Solution

(a) The function

f xð Þ ¼ xn; x > 0;

is continuous and strictly increasing on (0,1), and f ((0, 1))¼ (0, 1).

Also, f is differentiable on (0,1), and its derivative f 0(x)¼ nxn�1 is non-

zero there. So f satisfies the conditions of the Inverse Function Rule.

Hence f�1 is differentiable on (0,1); and, if y¼ f (x), then

f�1� �0

yð Þ ¼ 1

f 0 xð Þ ¼1

nxn�1

¼ 1

nx1�n ¼ 1

ny

1�nn :

If we now replace the domain variable y by x, we obtain

f�1� �0

xð Þ ¼ 1

nx

1n�1; x 2 0;1ð Þ:

(b) The function

f xð Þ ¼ tan x; x 2 � 1

2p;

1

2p

� �

;

is continuous and strictly increasing on �12p; 1

2p

� �

, and f � 12p; 1

2p

� ��

¼ R .

Also, f is differentiable on �12p; 1

2p

� �

, and its derivative f 0 xð Þ ¼ sec2 x is

non-zero there. So f satisfies the conditions of the Inverse Function Rule.

Hence f�1¼ tan�1 is differentiable on R ; and, if y¼ f (x), then

f�1� �0

yð Þ ¼ 1

f 0 xð Þ ¼1

sec2 x

¼ 1

1þ tan2 x¼ 1

1þ y2:


tan�1� �0

xð Þ ¼ 1

1þ x2; x 2 R :

(c) The function

f xð Þ ¼ ex; x 2 R ;

is continuous and strictly increasing on R , and f (R)¼ (0, 1). Also, f is

differentiable on R , and its derivative f 0(x)¼ ex is non-zero there. So

f satisfies the conditions of the Inverse Function Rule.

Hence f� 1¼ loge is differentiable on (0,1); and, if y¼ f (x), then

f�1� �0

yð Þ ¼ 1

f 0 xð Þ ¼1

ex

¼ 1

y:


logeð Þ0 xð Þ ¼ 1

x; x 2 0;1ð Þ: &

Problem 6 For each of the following functions f, show that f�1 is

differentiable and determine its derivative:

(a) f(x)¼ cos x, x2 (0, p); (b) f(x)¼ sinh x, x2R .

y¼ xn, so that x ¼ y1n:


Most equations of the form y¼ f (x) cannot be solved explicitly to give x as

some formula involving y alone. The Inverse Function Rule is often used in

such situations to solve problems that would otherwise be intractable. The

following problem illustrates this type of application.

Problem 7 Prove that the function f (x)¼ x5þ x� 1, x2R , has an

inverse function f�1 which is differentiable on R . Find the values of

( f�1)0(d ) at those points d corresponding to the points c¼ 0, 1 and �1,

where d¼ f (c).

Exponential functions

Earlier we defined the number ax, for a> 0, by the formula

ax ¼ exp x loge að Þ:Since the functions exp and log are differentiable on R and (0,1), respectively,

it follows that we can use this formula to determine the derivatives of functions

such as x 7! x�; x 7! ax and x 7! xx.

Example 4 Prove that, for �2R , the power function f (x)¼ x�, x2 (0,1), is

differentiable on its domain, and that f 0(x)¼�x��1.

Solution By definition

f xð Þ ¼ exp � loge xð Þ:The function x 7!� loge x is differentiable on (0,1), with derivative �

x. By the

Composition Rule, f is differentiable on (0,1) with derivative

f 0 xð Þ ¼ exp � loge xð Þ � �

x

� �

¼ x� � �

x

� �

¼ �x��1; x 2 0;1ð Þ: &

Remark

In the case when � is a rational number, � ¼ mn

say, the result of Example 4 can

also be proved by applying the Composition Rule to the functions x 7! xm and

x 7! x1n. The function x 7! x

1n is differentiable on (0,1), by the Inverse Function

Rule, as it is the inverse of the function x 7! xn.

Example 5 Prove that, for a> 0, the function f (x)¼ ax, x2R , is differenti-

able on its domain, and that f 0(x)¼ ax loge a.

Solution By definition

f xð Þ ¼ exp x loge að Þ:The function x 7! x loge a is differentiable on R , with derivative loge a. By the

Composition Rule, f is differentiable on R , with derivative

f 0 xð Þ ¼ exp x loge að Þ � loge a

¼ ax loge a; x 2 R : &

Problem 8 Prove that the function f (x)¼ xx, x2 (0,1), is differenti-

able on its domain, and find its derivative.

Section 4.4.

Section 6.1.


We end this sub-section with a list of basic differentiable functions.

Basic differentiable functions The following functions are differentiable

on their domains:

� polynomials and rational functions

� nth root function

� trigonometric functions (sine, cosine and tangent)

� exponential function

� hyperbolic functions (sinh, cosh and tanh).

6.2.4 Proofs

We now supply the proofs of the Combination Rules and the Composition

Rule, which we omitted earlier. We also illustrate how to prove such results

using the "� � method.


Let f and g be defined on an open interval I, and c2 I. Then, if f and g are

differentiable at c, so are:

Sum Rule fþ g, and ( fþ g)0(c)¼ f 0(c)þ g0(c);

Multiple Rule l f, for l2R , and (l f )0(c)¼ l f 0(c);

Product Rule fg, and ( fg)0(c)¼ f 0(c) g(c)þ f (c) g0(c);

Quotient Rule fg, provided that g (c) 6¼ 0; and f

g

� �0cð Þ¼ g cð Þf 0 cð Þ�f cð Þg0 cð Þ

g cð Þð Þ2 :

Proof

Sum Rule

Let F¼ fþ g. Then

limx!c

F xð Þ � F cð Þx� c

¼ limx!c

f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf gx� c

¼ limx!c


þ limx!c

g xð Þ � g cð Þx� c

¼ f 0 cð Þ þ g0 cð Þ:Thus F is differentiable at c, with derivative f 0(c)þ g0(c).

Multiple Rule

This is just a special case of the Product Rule, with g(x)¼ l.

Product Rule

Let F¼ fg. Then

limx!c

F xð Þ�F cð Þx� c

¼ limx!c

f xð Þg xð Þ� f cð Þg cð Þx� c


In this proof we use theCombination Rules for limitsgiven in Sub-section 5.1.


¼ limx!c

f xð Þ � f cð Þf gg xð Þ þ f cð Þ g xð Þ � g cð Þf gx� c

¼ limx!c


g xð Þ þ limx!c

f cð Þ g xð Þ � g cð Þx� c

¼ limx!c


� limx!c

g xð Þ þ f cð Þ � limx!c

g xð Þ � g cð Þx� c

¼ f 0 cð Þg cð Þ þ f cð Þg0 cð Þ;since f and g are differentiable at c, and since g is continuous at c, so that

g(x)! g(c) as x! c.

Thus F is differentiable at c, with derivative f 0(c)g(c)þ f (c)g0(c).

Quotient Rule

Let F ¼ fg.

Recall, first, that, since g is continuous at c (as it is differentiable there) and

g (c) 6¼ 0, there exists an open interval J containing c on which g(x) 6¼ 0. Thus

the domain of F contains the open interval J, and c2 J.

Then

limx!c

F xð Þ�F cð Þx� c

¼ limx!c

f xð Þg xð Þ�

f cð Þg cð Þ

x� c

¼ limx!c

f xð Þg cð Þ� f cð Þg xð Þx� cð Þ�g xð Þg cð Þ

¼ limx!c

f xð Þ� f cð Þf gg cð Þ� f cð Þ g xð Þ�g cð Þf gx� cð Þ�g xð Þg cð Þ

¼ limx!c

f xð Þ� f cð Þx� cð Þ�g xð Þ� lim

x!c

f cð Þg cð Þ�

g xð Þ�g cð Þx� cð Þ�g xð Þ

¼ limx!c

f xð Þ� f cð Þx� c

� limx!c

1

g xð Þ�f cð Þg cð Þ� lim

x!c

g xð Þ�g cð Þx� cð Þ � lim

x!c

1

g xð Þ

¼ f 0 cð Þg cð Þ �

f cð Þg0 cð Þg2 cð Þ ¼

f 0 cð Þg cð Þ� f cð Þg0 cð Þg2 cð Þ ;

since f and g are differentiable at c, and g is continuous at c.

Thus F is differentiable at c, with derivativef 0 cð Þg cð Þ�f cð Þg0 cð Þ

g2 cð Þ : &

As an illustration of how such results may be proved using the "� � method

that we introduced in Section 5.4, we prove just the Sum Rule. Notice that in

our proofs we avoid many complications by judicious use of the ‘K" Lemma’.

Proof of the Sum Rule Let F¼ fþ g.

In view of the K" Lemma, we must prove that:



� f 0 cð Þ þ g0 cð Þf g�

�

�

�

�

�

�

�

< K" for all x satisfying

0 < x� cj j < �: (1)

First, we write the expression on the left-hand side of (1) in a convenient

form as

For, differentiable)continuous.

The ‘K" Lemma’ firstappeared in Sub-section 2.2.2.

As you work through thisproof, compare it with theearlier proof using limits.

Recall that K must NOT dependon x or ".

f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf gx� c

� f 0 cð Þ þ g0 cð Þf g

¼ f xð Þ � f cð Þf g þ g xð Þ � g cð Þf gx� c

� f 0 cð Þ þ g0 cð Þf g

¼ f xð Þ � f cð Þx� c

� f 0 cð Þ

þ g xð Þ � g cð Þx� c

� g0 cð Þ

: (2)

We now use the information that we already have to tackle each of the terms

in (2) in turn.

Since f is differentiable at c, there exists some positive number �1 such that

f xð Þ� f cð Þx� c

� f 0 cð Þ�

�

�

�

�

�

�

�

<"; for all x satisfying 0< x� cj j<�1: (3)

Also, since g is differentiable at c, there exists some positive number �2

such that

g xð Þ�g cð Þx� c

�g0 cð Þ�

�

�

�

�

�

�

�

<"; for all x satisfying 0< x� cj j<�2: (4)

Now let �¼min{�1, �2}. It follows that, for all x satisfying 0< jx� cj<�,both (3) and (4) hold. Hence, if we apply the Triangle Inequality to (2) and then

use both (3) and (4), we find that, for all x satisfying 0< jx� cj<�f xð Þ þ g xð Þf g � f cð Þ þ g cð Þf g

x� c� f 0 cð Þ þ g0 cð Þf g

�

�

�

�

�

�

�

�

� f xð Þ � f cð Þx� c

� f 0 cð Þ�

�

�

�

�

�

�

�

þ g xð Þ � g cð Þx� c

� g0 cð Þ�

�

�

�

�

�

�

�

< "þ " ¼ 2":

But this is just the statement (1) that we set out to prove, with K¼ 2. &

Remark

Note that our use of the ‘K" Lemma’ meant that we did not need to know at the

steps (3) and (4) to use expressions like 12" rather than " in order to end up with

a final conclusion that ‘some expression is<"’! For ‘some expression is <K"’is sufficient.

We end with the proof of the Composition Rule.


Let g and f be defined on open intervals I and J, respectively, and let c2 I

and g Ið Þ J. If g is differentiable at c and f is differentiable at g(c), then

f g is differentiable at c and

f gð Þ0 cð Þ ¼ f 0 g cð Þð Þg0 cð Þ:

This expression appears in (2).

By the Triangle Inequality.

By (3) and (4).

We would only discover theneed for such a choice oncewe had made a first attemptat the proof.

Proof Let F ¼ f g, and let y¼ g(x) and d¼ g(c).

The difference quotient for F at c is


¼ f g xð Þð Þ � f g cð Þð Þx� c

: (5)

Now, the right-hand side of (5) is equal to

f yð Þ � f dð Þy� d

� g xð Þ � g cð Þx� c

; (6)

provided that y 6¼ d.

Unfortunately, for some choices of g and c it is possible for y¼ d to occur

when y¼ g(x) and x is arbitrarily close to c but not actually equal to c. In such

situations the expression (6) will not exist.

To get round this problem, we introduce a carefully chosen auxiliary

function

h yð Þ ¼f yð Þ�f dð Þ

y�d; y 6¼ d,

f 0 dð Þ; y ¼ d.

Since f is differentiable at d, h(y)! f 0(d) as y! d; and, since h(d )¼ f 0(d ), it

follows that h is continuous at d.

We then apply the Composition Rule for continuous functions, to deduce

that the composite function

h g xð Þ ¼f g xð Þð Þ�f g cð Þð Þ

g xð Þ�g cð Þ ; g xð Þ 6¼ g cð Þ;f 0 g cð Þð Þ; g xð Þ ¼ g cð Þ;

(

is continuous at c.

Next, notice that, if g(x) 6¼ g(c), it follows from (5) that


¼ h g xð Þ � g xð Þ � g cð Þx� c

; (7)

and also that this last statement (7) is also valid when g(x)¼ g(c), since then

both sides of (7) are zero.

Hence, if we let x! c in (7) and use the continuity of the function h g, we

obtain

limx!c


¼ h g cð Þ � g0 cð Þ

¼ f 0 g cð Þð Þg0 cð Þ:

Thus F is differentiable at c, with derivative f 0(g(c))g0(c). &

Remark

Let

g xð Þ ¼ x2 sin 1x

� �

; x 6¼ 0,

0; x ¼ 0,

c ¼ 0; and d ¼ g 0ð Þ ¼ 0:

Some texts ignore thispossibility, and so giveincomplete proofs ofTheorem 2! In the Remarkbelow, we describe one suchsituation.


Since sin(np)¼ 0 for all n2N , it follows that y¼ g(x) takes the value d¼ 0

at the points 1np that are arbitrarily close to c¼ 0 but not equal to 0.

6.3 Rolle’s Theorem

In the next two sections we describe some of the fundamental properties of

functions that are differentiable, not just at a particular point but on an

interval. Our results are motivated by the geometric significance of differ-

entiability in terms of tangents, and explain why the graphs of differentiable

functions possess certain geometric properties.

6.3.1 The Local Extremum Theorem

Earlier we described some of the fundamental properties of functions which

are continuous on a closed interval. In particular, we proved the Extreme

Values Theorem, which states that, if a function ƒ is continuous on a closed

interval [a, b], then there are points c, d in [a, b] such that

f cð Þ � f xð Þ � f dð Þ; for all x 2 ½a; b�;

we called f(c) the minimum of ƒ on [a, b], and f(d) the maximum of ƒ on [a, b].

We used the term extremum to denote either a maximum or a minimum.

But how do we determine the points c, d where these extrema occur? In

general, unfortunately, this is not easy. However, if the function ƒ is differenti-

able, then we can, in principle, determine c and d by first finding any local

maxima or local minima of the function ƒ on the interval [a, b].

Roughly speaking, for a point c in (a, b), the value f(c) is a local maximum of

ƒ on [a, b] if

f ðxÞ � f ðcÞ; for all x in ½a; b� sufficiently close to c;

and a local minimum of ƒ on [a, b] if

f ðxÞ � f ðcÞ; for all x in ½a; b� sufficiently close to c:

To make this idea more precise, we use the concept of neighbourhood that you

met earlier.

Definitions Let ƒ be defined on an interval I, and let c2 I. Then:

� ƒ has a local maximum f(c) at c if there exists a neighbourhood N of c

such that

f ðxÞ � f ðcÞ; for x 2 N \ I;

� ƒ has a local minimum f(c) at c if there exists a neighbourhood N of c

such that

f ðxÞ � f ðcÞ; for x 2 N \ I;

� ƒ has a local extremum at c if f(c) is either a local maximum or a local

minimum.

This example shows that theparagraph ‘Unfortunately . . .’after equation (6) above doesdescribe a situation that canoccur.

Section 4.2.

Sub-section 4.2.3.

From this point onwards, weshall commonly use the letterc to denote an extremum – thatis EITHER a maximum OR aminimum.

Sub-section 5.1.1.

Notice that N \ I necessarilyincludes an interval of theform (c� r, c] or [c, cþ s),where r, s> 0.


When we wish to locate local extrema of a differentiable function ƒ, instead of

using the above definition we usually use the following result, which gives a

connection between local extrema of a function ƒ and the points where f 0 vanishes.

Theorem 1 Local Extremum Theorem

Let ƒ be defined on an interval [a, b]. If ƒ has a local extremum at c, where

a< c< b, and if ƒ is differentiable at c, then

f 0ðcÞ ¼ 0:

Proof Suppose that ƒ has a local maximum at c.

Since a< c< b, it follows from the definition of local maximum that there

exists a neighbourhood N of c with N ½a, b� such that

f ðxÞ � f ðcÞ; for x 2 N:

We now choose numbers r, s> 0 such that N¼ (c� r, cþ s).

First, looking to the left of c, we have

f ðxÞ � f ðcÞ � 0 and x� c < 0; for c� r < x < c;

so that

f ðxÞ � f ðcÞx� c

� 0; for c� r < x < c:

Thus

fL0ðcÞ ¼ lim

x!c�


� 0: (1)

Next, looking to the right of c, we have

f ðxÞ � f ðcÞ � 0 and x� c > 0; for c < x < cþ s;

so that


� 0; for c < x < cþ s:

Thus

fR0ðcÞ ¼ lim

x!cþ


� 0: (2)

Since ƒ is differentiable at c, the left and right derivatives at c exist and are

equal. Hence, in view of inequalities (1) and (2), their common value, f 0ðcÞ,must be 0. &

We prove only the localmaximum version: the proofof the local minimum versionis similar.

Remarks

1. The Local Extremum Theorem applies only if the function is differentiable

at a local extremum. For example, the function

f ðxÞ ¼ jxj; x 2 ½�1; 1�;has a local minimum 0 at 0, but ƒ is not differentiable at 0.

2. The Local Extremum Theorem does not assert that a point where the deriva-

tive vanishes is necessarily a local extremum. For example, the function

f ðxÞ ¼ x3; x 2 ½�1; 1�;does not have a local extremum at 0, although f 0(0)¼ 0.

3. The Local Extremum Theorem does not make any assertion about a

local extremum that occurs at a point c that is one of the end-points of the

interval [a, b].

Problem 1 Find the local extrema of the function

f xð Þ ¼ 1

4x4 � 1

3x3; x 2 �1; 2½ �:

Clearly any extremum of ƒ on [a, b] that occurs at a point other than a or b

must be a local extremum. It follows from Theorem 1 that such a point c must

be a point where f 0(c)¼ 0. Thus an immediate consequence of the Local

Extremum Theorem is the following criterion for finding all the extrema of

‘well-behaved functions’ on closed intervals.

Corollary 1 Let ƒ be continuous on the closed interval [a, b] and differ-

entiable on the open interval (a, b). Then the extrema of ƒ on [a, b] can occur

only at a, at b, or at points c in (a, b) where f 0(c)¼ 0.

We now reformulate Corollary 1 as a strategy for locating minima and

maxima.

Strategy To determine the maximum and the minimum of a function ƒwhich is continuous on [a, b] and differentiable on (a, b):

1. Determine the points c1, c2, . . . in (a, b) where f 0 vanishes;

2. Compare the values of f (a), f (b), f (c1), f (c2), . . .;

the least is the minimum, and the greatest is the maximum.

Problem 2 Use the above Strategy to determine the minimum and the

maximum of the function f xð Þ ¼ sin2 xþ cos x, for x 2 0; 12p

�

:

6.3.2 Rolle’s Theorem

In the previous sub-section we saw that, if a function ƒ is continuous on

the closed interval [a, b] and differentiable on the open interval (a, b), then

the extrema of ƒ can occur only at a, at b, or at points c in (a, b) where

f 0(c)¼ 0.


Now the function

f ðxÞ ¼ sin1

2px

� �

; x2 � 2

3;

2

3

� �

;

shows that it can happen that no interior point of (a, b) corresponds to a

maximum or a minimum of ƒ, and that f 0 need not vanish at any interior

point at all.

However, the situation is quite different for the function

f ðxÞ ¼ sin1

2px

� �

; x2½�2; 2�:

Here f(�2)¼ f(2)¼ 0; on [�2, 2] the function f has a maximum at 1 and a

minimum at �1, and at both these interior points f 0 vanishes. This is a special

case of Rolle’s Theorem which asserts that, if f(a)¼ f(b), then there is at least

one point c strictly between a and b at which f 0 vanishes.

Theorem 2 Rolle’s Theorem

Let ƒ be defined on the closed interval [a, b] and differentiable on the open

interval (a, b). If f(a)¼ f(b), then there exists some point c, with a< c< b,

for which f 0(c)¼ 0.

Remarks

1. This apparently simple result is one of the most important results in

Analysis.

2. In geometric terms, Rolle’s Theorem means that, if the line joining the

points (a, f(a)) and (b, f(b)) on the graph of ƒ is horizontal, then so is the

tangent to the graph at some point c in (a, b).

3. There may be more than one point c in (a, b) at which f 0 vanishes (as in the

diagram in the margin). Rolle’s Theorem simply asserts that at least one

such point c exists.

Proof If ƒ is constant on [a, b], then f 0(x)¼ 0 everywhere in (a, b); in this

case, we may take c to be any point of (a, b).

If ƒ is non-constant on [a, b], then either the maximum or the minimum (or

both) of ƒ on [a, b] is different from the common value f (a)¼ f (b). Since one

of the extrema occurs at some point c with a< c< b, the Local Extremum

Theorem applied to the point c shows that f 0(c) must be zero. &

Example 1 Verify that the conditions of Rolle’s Theorem are satisfied by the

function

f xð Þ ¼ 3x4 � 2x3 � 2x2 þ 2x; x 2 �1; 1½ �;and determine a value of c in (�1, 1) for which f 0(c)¼ 0.

Solution Since f is a polynomial function, f is continuous on [�1, 1] and

differentiable on (�1, 1). Also, f (�1)¼ f (1)¼ 1. Thus f satisfies the condi-

tions of Rolle’s Theorem on [�1, 1].

It follows that there exists a number c2 (�1, 1) for which f 0(c)¼ 0. Now

That is, the extrema maynot occur at interior pointsof (a, b).

This is an existence theorem.Often it is difficult to evaluatec explicitly.

Since ƒ is continuous on[a, b], ƒ must have both amaximum and a minimum on[a, b], by the Extreme ValuesTheorem.

f 0 xð Þ ¼ 12x3 � 6x2 � 4xþ 2

¼ 12 x2 � 1

3

� �

x� 1

2

� �

;

so that f 0 vanishes at the points 1ffiffi

3p , � 1

ffiffi

3p and 1

2in (�1, 1). Any of these three

numbers will serve for c. &

Problem 3 Verify that the conditions of Rolle’s Theorem are satisfied

by the function

f xð Þ ¼ x4 � 4x3 þ 3x2 þ 2; x 2 1; 3½ �;and determine a value of c in (1,3) for which f 0(c)¼ 0.

Problem 4 For each of the following functions, state whether Rolle’s

Theorem applies for the given interval:

(a) f xð Þ ¼ tan x; x 2 0; p½ �;(b) f xð Þ ¼ xþ 3 x� 1j j; x 2 0; 2½ �;(c) f xð Þ ¼ x� 9x17 þ 8x18; x 2 0; 1½ �;(d) f xð Þ ¼ sin xþ tan�1 x; x 2 0; 1

2p

�

:

6.4 The Mean Value Theorem

Here we continue to study the geometric properties of functions that are

differentiable on intervals, and describe some of their applications.

6.4.1 The Mean Value Theorem

First, recall the geometric interpretation of Rolle’s Theorem: Under suitable

conditions, if the chord joining the points (a, f (a)) and (b, f (b)) of the graph of f

is horizontal, then so is the tangent at some point c of (a, b).

If you imagine pushing the chord (as shown in the margin), always parallel

to itself, until it is just about to lose contact with the graph of f, it looks as

though at this point the chord becomes a tangent to the graph. Similarly, the

‘chord-pushing’ approach suggests that, even if the original chord is not

horizontal (that is, if f(a) 6¼ f(b)), there must still be some point c of (a, b) at

which the tangent is parallel to the chord.

Example 1 Consider the function

f xð Þ ¼ 3� 3xþ x3; x 2 1; 2½ �:Find a point c of (1, 2) such that the tangent to the graph of f is parallel to the

chord joining (1, f (1)) to (2, f (2)).

Solution Since f(1)¼ 3� 3þ 1¼ 1 and f(2)¼ 3� 6þ 8¼ 5, the slope of the

chord joining the endpoints of the graph is

f 2ð Þ � f 1ð Þ2� 1

¼ 5� 1

2� 1¼ 4:


Now, since f is a polynomial, it is differentiable on (1, 2) and its derivative is

f 0(x)¼�3þ 3x2; hence f 0(c)¼ 4 when 3c2¼ 7, or c ¼ffiffi

73

q

’ 1:53. Thus, at

the point (c, f(c)) the tangent to the curve is parallel to the chord joining the

end-points. &

We now generalise Rolle’s Theorem and assert that there is always a point

where the tangent to the graph is parallel to the chord joining the end-points.

This result is known as the Mean Value Theorem, so-called since

f bð Þ � f að Þb� a

can be thought of as the mean value of the derivative between a and b.

Theorem 1 Mean Value Theorem

Let f be continuous on the closed interval [a, b] and differentiable on the

open interval (a, b). Then there exists a point c in (a, b) such that

f 0 cð Þ ¼ f bð Þ � f að Þb� a

:

The idea of the proof is as follows. We define h(x) to be the vertical distance

from the chord to the curve; then h(a) and h(b) are both 0; in fact, h satisfies all

the conditions of Rolle’s Theorem. Applying Rolle’s Theorem to h, we obtain

the desired result.

Proof The slope of the chord joining the points (a, f (a)) and (b, f (b)) is

m ¼ f bð Þ � f að Þb� a

;

and so the equation of the chord is

y ¼ m x� að Þ þ f að Þ:It follows that the vertical height, h(x), between points with ordinate x on the

graph and those on the chord is given by

h xð Þ ¼ f xð Þ � m x� að Þ þ f að Þ½ �:Now h(a)¼ h(b)¼ 0, and h is continuous on [a, b] and differentiable on (a, b).

Thus h satisfies all the conditions of Rolle’s Theorem.

It follows from Rolle’s Theorem that there exists some point c in (a, b) for

which h0(c)¼ 0. But, since h0(c)¼ f 0(c)�m, it follows that

f 0 cð Þ ¼ m ¼ f bð Þ � f að Þb� a

: &

Example 2 Verify that the conditions of the Mean Value Theorem are

satisfied by the function f xð Þ ¼ x�1xþ1

; x 2

1; 72

�

; and find a value for c that

satisfies the conclusion of the theorem.

Solution The function f is a rational function whose denominator is non-zero

on

1; 72

�

; so f is continuous on

1; 72

�

and differentiable on�

1; 72

�

: Thus

f satisfies the conditions of the Mean Value Theorem.

Nowf 7

2

� �

� f 1ð Þ72� 1

¼59� 052

¼ 2

9;

Again, this is an existencetheorem.

Note that when f(a)¼ f(b), theMean Value Theorem simplyreduces to Rolle’s Theorem.

y = f (x)

1

y

x72

59


and

f 0 xð Þ ¼ 2

xþ 1ð Þ2;

so the Mean Value Theorem asserts that there exists some point c in 1; 72

� �

for

which f 0(c)¼ 29. Thus

2

cþ 1ð Þ2¼ 2

9;

and so cþ 1ð Þ2¼ 9. It follows that c¼ 2. &

Problem 1 For each of the following functions, verify that the condi-

tions of the Mean Value Theorem are satisfied, and find a value for c that

satisfies the conclusion of the theorem:

(a) f xð Þ ¼ x3 þ 2x; x 2 �2; 2½ �; (b) f xð Þ ¼ ex; x 2 0; 3½ �:

6.4.2 Positive, negative and zero derivatives

We now study some consequences of the Mean Value Theorem for functions

whose derivatives are always positive, always negative, or always zero.

First, we prove a crucial result about monotonic functions which you use

regularly to sketch the graph of a function f. It concerns the behaviour of f on a

general interval I, so here we denote the interior of I (the set of all interiorpoints of I ) by Int I.

Theorem 2 Increasing-Decreasing Theorem

Let f be continuous on an interval I and differentiable on Int I.

(a) If f 0(x)� 0 on Int I, then f is increasing on I.

(b) If f 0(x)� 0 on Int I, then f is decreasing on I.

Proof Choose any two points x1 and x2 in I, with x1< x2. The function f

satisfies the conditions of the Mean Value Theorem on the interval [x1, x2],

so there exists a point c in (x1, x2) such that

f x2ð Þ � f x1ð Þx2 � x1

¼ f 0 cð Þ:

It follows that f (x2)� f(x1) must have the same sign as f 0(c).

(a) If f 0(x)� 0 on Int I, then f(x2)� f(x1)� 0, so that f(x2)� f(x1). Thus f is

increasing on I.

(b) If f 0(x)� 0 on Int I, then f(x2)� f(x1) � 0, so that f(x2)� f (x1). Thus f is

decreasing on I. &

Remark

If the inequalities in the statement of Theorem 2 are replaced by strict inequal-

ities, the conclusions of the Theorem become the following:

(a) If f 0(x)> 0 on Int I, then f is strictly increasing on I;

(b) If f 0(x)< 0 on Int I, then f is strictly decreasing on I.

This is a quadratic equationwith roots c¼ 2 and �4; weignore the solution c¼�4,since it lies outside

�

1; 72

�

.

For example, if I¼ [0, 1), thenInt I¼ (0, 1).

The proofs of these assertionsare similar to the proofs inTheorem 2.

Problem 2 For each of the following functions f, determine whether

f is increasing, strictly increasing, decreasing or strictly decreasing:

(a) f xð Þ ¼ 3x43 � 4x; x 2 1;1½ Þ; (b) f xð Þ ¼ x� loge x; x 2 0; 1ð �:

Two useful consequences of Theorem 2 are the following corollaries.

Corollary 1 Zero Derivative Theorem

Let f be continuous on an interval I and differentiable on Int I. If f 0(x)¼ 0

for all x in Int I, then f is constant on I.

Proof Cases (a) and (b) of Theorem 2 both apply, so that f is both increasing

and decreasing on I.

Hence f is constant on I. &

As an illustration of the use of this important result, we can now prove the

claim made earlier that the function f (x)¼ lex, l an arbitrary constant, is the

only function f that satisfies the differential equation f 0(x)¼ f (x) on R . For, if

f 0(x)¼ f (x), thend

dxe�xf xð Þð Þ ¼ e�xf 0 xð Þ � e�xf xð Þ

¼ e�x f 0 xð Þ � f xð Þð Þ ¼ 0;

it then follows from Corollary 1 that e�x f(x) is just some constant l, say, so

that f (x)¼ lex.

Corollary 2 Let f and g be continuous on an interval I and differentiable

on Int I. If f 0(x)¼ g0(x) for all x in Int I, then

f(x)¼ g(x)þ c for all x in I, for some constant c.

Proof This follows immediately by applying Corollary 1 to the function

h¼ f� g, since h0(x)¼ 0 for all x in the interior of I. &

Example 3 Prove that sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

� �

, for all x2R .

Solution Let

f xð Þ ¼ sinh�1 x; x 2 R ;

g xð Þ ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

� �

; x 2 R :

Then f and g are continuous on R and differentiable on R .

Now

f 0 xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p ; x 2 R ;

also, by the Composition Rule for Derivatives

g0 xð Þ ¼ 1

xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p � 1þ 1

2� 2x x2 þ 1

� ��12

� �

¼1þ x


x2 þ 1p


x2 þ 1p

¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p ; x 2 R :

Sub-section 6.1.3.

For h is continuous on I anddifferentiable on Int I.


Hence f 0(x)¼ g0(x) for all x in R .

It follows from Corollary 2 that

sinh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1p

� �

þ c;

for some constant c. Putting x¼ 0 in the above identity, we obtain 0¼ loge(1)þ c,

so that c¼ 0. &

Problem 3 Use Corollary 1 to prove the following identities:

(a) sin�1 xþ cos�1 x ¼ 12p, for x 2 �1; 1½ �;

(b) tan�1 xþ tan�1 1x

� �

¼ 12p, for x> 0.

Next, note that second derivatives can often be used to identify whether a

point c where f 0(c)¼ 0 is a local maximum or a local minimum of a function f.

Suppose that f is defined on a neighbourhood of c, and that

f 0 cð Þ ¼ 0 and f 00 cð Þ > 0:

It can be shown that there is some punctured neighbourhood N ¼ c� r; cð Þ [c; cþ sð Þ of c such that

f 0 xð Þ � f 0 cð Þx� c

> 0; for x 2 N;

that is

f 0 xð Þx� c

> 0; for x 2 N:

We can rewrite this inequality in the form

f 0 xð Þ < 0; for x 2 c� r; cð Þ;f 0 xð Þ > 0; for x 2 c; cþ sð Þ:

It follows from the Increasing–Decreasing Theorem that f has a local

minimum at c.

A similar argument shows that, if

f 0 cð Þ ¼ 0 and f 00 cð Þ < 0;

then f has a local maximum at c.

Thus we have proved the following result.

Theorem 3 Second Derivative Test

Let f be defined on a neighbourhood of c, and f 0(c)¼ 0.

(a) If f 00(c)> 0, then f (c) is a local minimum of f.

(b) If f 00(c)< 0, then f (c) is a local maximum of f.

Remark

The theorem gives us no information in the case that f 00(c)¼ 0.

Problem 4 For the function f(x)¼ x3� 3x2þ 1, x2R , determine those

points c where f 0(c)¼ 0. Using the Second Derivative Test, determine

whether these correspond to local maxima, local minima or neither.

In Exercise 6 on this sub-section, in Section 6.7, we askyou to verify a result thatimplies this assertion.

The following diagrams arehelpful in remembering thisresult.

Inequalities

We now demonstrate how the Increasing–Decreasing Theorem can be used to

prove certain inequalities involving differentiable functions.

Example 4 Prove that, for �> 1,

1þ xð Þ�� 1þ �x; for x � �1:

Solution Let

f xð Þ ¼ 1þ xð Þ�� 1þ �xð Þ; for x 2 �1;1½ Þ:The function f is continuous on [�1,1) and differentiable on (�1,1), and

f 0 xð Þ ¼ � 1þ xð Þ��1��

¼ � 1þ xð Þ��1�1h i

:

Firstly, if �1< x< 0, then

0 < 1þ x < 1:

Since �> 1, we can then take the (�� 1)th power of each side of the inequality

1þ x< 1 to obtain

1þ xð Þ��1< 1; for �1 < x < 0;

so that

f 0 xð Þ < 0; for �1 < x < 0:

Next, if x> 0, then

1þ x > 1;

so that

1þ xð Þ��1> 1; for x > 0:

Hence

f 0 xð Þ > 0; for x > 0:

Bringing together these two arguments, we have

f 0 xð Þ < 0; for �1 < x < 0;

f 0 xð Þ > 0; for x > 0:

Also, f (0)¼ 0.

Hence, by the Increasing–Decreasing Theorem

f is decreasing on �1; 0½ �;f is increasing on ½0;1Þ;

so we have

f xð Þ � f 0ð Þ ¼ 0; for x 2 �1; 0½ �;f xð Þ � f 0ð Þ ¼ 0; for x 2 0;1½ Þ:

It follows that

1þ xð Þ�� 1þ �xð Þ � 0; for x 2 �1;1½ Þ;as required. &

This is a generalisation ofBernoulli’s Inequality

1þ xð Þn � 1þ nx;

for x�� 1, n2N , that youmet in Sub-section 1.3.3.

Since �� 1> 0, we can useRule 5 for inequalities, whichyou met in Sub-section 1.2.1.

Since �� 1> 0, we can againuse Rule 5 for inequalities.

Example 4 illustrates the following general strategy.

Strategy To prove that g(x)� h(x) on [a, b]:

1. Let

f xð Þ ¼ g xð Þ � h xð Þ;and show that f is continuous on [a, b] and differentiable on (a, b);

2. Prove that:

EITHER f að Þ � 0; and f 0 xð Þ � 0 on a; bð Þ;OR f bð Þ � 0; and f 0 xð Þ � 0 on a; bð Þ;

Problem 5 Prove the following inequalities:

(a) x� sin x, for x 2 0; 12p

�

;

(b) 23

xþ 13� x

23, for x2 [0; 1].

6.5 L’Hopital’s Rule

In order to differentiate the functions sin and exp, we needed to use the

following results

limx!0

sin x

x¼ 1 and lim

x!0

ex � 1

x¼ 1:

Each of these limits is the limit as x! 0 of a quotient in which the numerator

and denominator take the value 0 when x is 0. The evaluation of these limits

was not trivial and required considerable care.

In Analysis and in Mathematical Physics we often need to evaluate limits of

the form

limx!c

f xð Þg xð Þ ; where f cð Þ ¼ g cð Þ ¼ 0:

Such limits cannot be evaluated by the Quotient Rule for limits of functions,

because it does not apply in this situation.

For example, do the limits

limx!p

2

cos 3x

sin x� ecos xand lim

x!0

x2

cosh x� 1

exist? If they do, what are their values?

Do we have to evaluate all such limits by a direct argument, or is there a handy

rule which we can apply? We shall find that there is, called l’Hopital’s Rule.

6.5.1 Cauchy’s Mean Value Theorem

Recall that the Mean Value Theorem asserts that, under certain conditions, the

graph of a function f(x), for x2 [a, b], has the property that at some intermedi-

ate point c the tangent to the graph is parallel to the chord joining the end-

points of the graph. In other words, there exists a point c in (a, b) such that

There is a correspondingversion of the Strategy inwhich the weak inequalitiesare replaced by strictinequalities.

This also works if b is1.This also works if a is �1.

Sub-section 6.1.3.

See Sub-section 5.1.1 andProblem 8 in Sub-section 5.1.4,respectively.

The Quotient Rule for limitsof functions states that

limx!c

f xð Þg xð Þ ¼

limx!c

f xð Þlimx!c

g xð Þ ;

provided that these last twolimits exist.



f 0 cð Þ ¼ f bð Þ � f að Þb� a

: (1)

The key tool that we shall need in Sub-section 6.5.2 in the proof of

l’Hopital’s Rule is the following result.

Theorem 1 Cauchy’s Mean Value Theorem

Let f and g be continuous on [a, b] and differentiable on (a, b). Then there is

some point c in (a, b) for which

f 0 cð Þ � g bð Þ � g að Þf g ¼ g0 cð Þ � f bð Þ � f að Þf g: (2)

In particular, if g(b) 6¼ g(a) and g0(c) 6¼ 0, this equation can be written in the

form

f 0 cð Þg0 cð Þ ¼

f bð Þ � f að Þg bð Þ � g að Þ : (3)

Notice that the Mean Value Theorem is simply the special case when

g(x)¼ x. For then g0(c)¼ 1 and g(b)� g(a)¼ b� a, and equation (3) reduces

to equation (1).

However, Theorem 1 is NOT a simple consequence of the Mean Value

Theorem. For, if we apply the Mean Value Theorem separately to the functions

f and g, we establish the existence of two points c1 and c2 in (a, b) for which

f 0 c1ð Þ ¼f bð Þ � f að Þ

b� aand g0 c2ð Þ ¼

g bð Þ � g að Þb� a

:

However, since c1 and c2 are usually unequal, we cannot deduce the exist-

ence of a single point c satisfying the statement (2).

Our proof of Theorem 1 is similar to that of the Mean Value Theorem: we

choose a suitable ‘auxiliary’ function h on [a, b] for which the conditions of

Rolle’s Theorem are satisfied.

Proof Consider the function

h xð Þ ¼ f xð Þ � g bð Þ�g að Þf g�g xð Þ � f bð Þ� f að Þf g; for x2 a;b½ �:

By the Combination Rules for continuity and differentiability, h is continuous

on [a, b] and differentiable on (a, b). Also

h að Þ ¼ f að Þ � g bð Þ � g að Þf g � g að Þ � f bð Þ � f að Þf g¼ f að Þg bð Þ � g að Þ f bð Þ

This form is easier toremember.

Here two terms cancel.


and

h bð Þ ¼ f bð Þ � g bð Þ � g að Þf g � g bð Þ � f bð Þ � f að Þf g¼ f að Þg bð Þ � g að Þf bð Þ;

so that h(a)¼ h(b).

Thus h satisfies the conditions of Rolle’s Theorem on [a, b]. Therefore there

exists a point c in (a, b) for which

h0ðcÞ ¼ 0 ;

that is, these exists a point c in (a, b) for which

f 0 cð Þ � g bð Þ � g að Þf g ¼ g0 cð Þ � f bð Þ � f að Þf g:This is precisely equation (2).

Equation (3) follows immediately from equation (2). &

Example 1 By applying Cauchy’s Mean Value Theorem to the functions

f xð Þ ¼ x6 � 1 and g xð Þ ¼ 1

2x4 þ 3x3 þ 3x� 3 on ½1; 2�;

prove that the equation 3x5� 2x3� 9x2¼ 3 has at least one root in (1, 2).

Solution Since f and g are both polynomials, they are continuous on [1, 2]

and differentiable on (1, 2). It follows that Cauchy’s Mean Value Theorem

applies to the functions f and g on [1, 2].

Now, g 2ð Þ ¼ 12� 16þ 3� 8þ 3� 2� 3 ¼ 8þ 24þ 6� 3 ¼ 35 and

g 1ð Þ ¼ 12þ 3þ 3� 3 ¼ 3 1

2, so that g(2) 6¼ g(1). Also, f 0(x)¼ 6x5 and

g0(x)¼ 2x3þ 9x2þ 3 6¼ 0 on (1, 2). It therefore follows from Cauchy’s Mean

Value Theorem that there exists at least one point c in (1, 2) for which

6c5

2c3 þ 9c2 þ 3¼ 63� 0

35� 3 12

¼ 63

31 12

¼ 2:

By cross-multiplying, we see that c satisfies the equation 6c5¼ 4c3þ 18c2þ 6,

or

3c5 � 2c3 � 9c2 ¼ 3:

In other words, the equation 3x5� 2x3� 9x2¼ 3 has at least one root in

(1, 2). &

Problem 1 By applying Cauchy’s Mean Value Theorem to the functions

f xð Þ¼ x3þ x2 sinx and g xð Þ¼ xcosx� sinx on 0;p½ �;prove that the equation 3x¼ (p2� 2) sin x� xcos x has at least one root

in (0, p).

6.5.2 l’Hopital’s Rule

We are now in a position to state the key result that we need to evaluate certain

types of limits in a fairly routine way.

Again two terms cancel.

The common value of h(a)and h(b) does not matter; theimportant thing is that the twoare equal.

Here we use equation (3).


Theorem 2 l’Hopital’s Rule

Let f and g be differentiable on a neighbourhood of the point c, at which

f(c)¼ g(c)¼ 0. Then

limx!c

f xð Þg xð Þ exists and equals lim

x!c

f 0 xð Þg0 xð Þ ;

provided that this last limit exists.

Proof We assume that

limx!c

f 0 xð Þg0 xð Þ exists and equals ‘: (4)

Hence there is some punctured neighbourhood N¼ (c� r, c)[ (c, cþ s) of c

on which f 0

g0 is defined and g0(x) 6¼ 0.

We now prove that our assumption (4) implies that

limx!c

f xð ÞgðxÞ exists and equals ‘: (5)

Let y be any specific point in N for which y> c. The functions f and g are

continuous on [c, y] and differentiable on (c, y), so they satisfy the conditions of

Cauchy’s Mean Value Theorem on [c, y].

Now g(y)� g(c) 6¼ 0; for otherwise we would have g(c)¼ 0 (from our

assumptions) and so g(y)¼ 0; this would imply, by Rolle’s Theorem, that g0

would vanish somewhere in (c, y), which it does not.

It follows from the conclusion (3) of Cauchy’s Mean Value Theorem that

there exists some point z in (c, y) for which

f 0 zð Þg0 zð Þ ¼

f yð Þ � f cð Þg yð Þ � g cð Þ

¼ f yð Þg yð Þ ; since f cð Þ ¼ g cð Þ ¼ 0:

Now let y! cþ. Since c< z< y, it follows that z! cþ too.

But we know that

limz!cþ

f 0 zð Þg0 zð Þ exists and has the value ‘:

It follows that

limy!cþ

f yð Þg yð Þ exists and has the value ‘:

This is exactly the same as the statement that

limx!cþ

f xð Þg xð Þ exists and has the value ‘:

A similar argument shows that

limx!c�

f xð Þg xð Þ exists and has the value ‘:

Combining the last two statements, we obtain the desired result (5). &

We now show how we can use l’Hopital’s Rule to evaluate the two limits that

we mentioned at the start of this section.



This is just a special case of(4), with z in place of x andz! cþ in place of x! c.

Here x is simply a ‘dummyvariable’: it does not matterwhat letter we assign to thevariable in the limit.

We simply repeat the wholeargument, starting with aspecific point y in N forwhich y< c.

Example 2 Prove that limx!p

2

cos 3x

sin x � e cos x ( 6)

exists and determine its value.

Solution Let

f xð Þ ¼ cos 3x and g xð Þ ¼ sin x� ecos x; for x 2 R :

Then f and g are differentiable on R , and

fp2

� �

¼ gp2

� �

¼ 0;

so that f and g satisfy the conditions of l’Hopital’s Rule at p2.

Now

f 0 xð Þg0 xð Þ ¼

�3 sin 3x

cos xþ sin x� ecos x:

Then, by l’Hopital’s Rule, the limit (6) exists and equals

limx!p

2

f 0 xð Þg0 xð Þ ¼ lim

x!p2

�3 sin 3x

cos xþ sin x� ecos x;


By the Quotient Rule for continuous functions, we know that f 0

g0 is continuous

at p2, so that

limx!p

2


f 0 p2

� �

g0 p2

� �

¼ 3

1¼ 3:

It follows, from l’Hopital’s Rule, that the original limit (6) must also exist,

and that its value is 3. &

Example 3 Prove that limx!0

x2

cosh x � 1 ( 7)

exists and determine its value.

Solution Let

f xð Þ ¼ x2 and g xð Þ ¼ cosh x� 1; for x 2 R :


f 0ð Þ ¼ g 0ð Þ ¼ 0;

so that f and g satisfy the conditions of l’Hopital’s Rule at 0.

Now, the derivatives of f and g are

f 0 xð Þ ¼ 2x and g0 xð Þ ¼ sinh x; for x 2 R :

It follows from l’Hopital’s Rule that the limit (7) exists and equals

limx!0

2x

sinh x(8)


Now, both f 0 and g0 are differentiable on R , and f 0(0)¼ g0(0)¼ 0; thus f 0 and

g0 satisfy the conditions of l’Hopital’s Rule at 0.

Since

f 00 xð Þ ¼ 2 and g00 xð Þ ¼ cosh x; for x 2 R ;

Here we are using the fact (seeRemark 7, Sub-section 5.4.1)that, if f is continuous at c, then

limx!c


(In future, we shall notmention this fact explicitly inthis particular connection.)

We cannot assert that

limx!0


f 0 0ð Þg0 0ð Þ ;

since f 0(0)¼ g0(0)¼ 0.


it follows from l’Hopital’s Rule that the limit (8) exists and equals

limx!0

2

cosh x; (9)


The function cosh is continuous on R , and cosh 0¼ 1, so that limx!0

cosh x ¼ 1.

It follows, from the Quotient Rule for limits, that the limit (9) does exist and

that its value is

2

cosh 0¼ 2

1¼ 2:

Working backwards, we conclude that the limit (8) exists, and equals 2.

Working further backwards, we conclude that the limit (7) also exists and

equals 2. &

Before applying a theorem, it is important to check that its conditions are

satisfied; for if they are not satisfied you cannot use the theorem! For instance,

a thoughtless application of l’Hopital’s Rule can give an incorrect answer!

For example, consider the problem of evaluating

limx!1

2x2 � x� 1

x2 � x: (10)

If we put f(x)¼ 2x2� x� 1 and g(x)¼ x2� x, we might be tempted to evaluate

(10) as follows

limx!1

f xð Þg xð Þ ¼ lim

x!1


x!1

4x� 1

2x� 1

¼ limx!1


x!1

4

2

¼ 2:

(11)

In fact, the value of the limit (10) is 3; so let us review the above argument

carefully!

The line (11) in the calculation is valid, since f and g are differentiable on R

and f(1)¼ g(1)¼ 0; so the conditions of l’Hopital’s Rule are satisfied for the

first application of the Rule. However, f 0(1)¼ 3 and g0(1)¼ 1, so the condi-

tions are not satisfied for the second application of l’Hopital’s Rule!

In fact, we should have concluded directly from line (11) that

limx!1


x!1

4x� 1

2x� 1¼ 3:

The moral is that you must apply l’Hopital’s Rule with care, particularly to

make the proviso ‘provided that this last limit exists’ at the appropriate points. At

the end, you then work backwards through the chain of applications of the Rule

to reach your conclusion about the limit that you set out originally to examine.

Remark

Notice that we cannot apply l’Hopital’s Rule to evaluate the limits

limx!0

sin x

x¼ 1 and lim

x!0

ex � 1

x¼ 1;

because we used these limits to find the derivatives of sin x and ex, respectively!

Note that the carefully set outlogic of the argument here isessential, since it is aconsequence of what weknow from Theorem 2.

Note that f(1)¼ g(1)¼ 0.

Here the careful argumentsthat we gave in Examples 2and 3 have been abandoned infavour of thoughtless‘formula-pushing’!


Problem 2 Prove that the following limits exist, and evaluate them.

(a) limx!0

sinh 2xsin 3x

; (b) limx!0

1þxð Þ15� 1�xð Þ

15

1þ2xð Þ25� 1�2xð Þ

25

;

(c) limx!0

sin x2þsin x2ð Þ1�cos 4x

; (d) limx!0

sin x�x cos xx3 .

6.6 The Blancmange function

We now meet a function, called the Blancmange function, which is continuous

at each point of R but is differentiable nowhere on R . The construction of the

first function with these properties by Karl Weierstrass caused a huge excite-

ment among mathematicians!

You saw earlier that, if a function is differentiable at a point c, then it is also

continuous at c. The Blancmange function shows, in a very striking way, that

the converse result is false!

6.6.1 What is the Blancmange function?

We give the formal definition first, and then look at the underlying geometry

of the graph of the function.

Definition Let f be the function

f ðxÞ ¼ x� ½x�; 0 � x � [x] � 12,

1� ðx� ½x�Þ; 12< x � [x] < 1,

where [x] denotes the integer part of x. Then the Blancmange function is

defined on R by the formula

BðxÞ ¼ f ðxÞ þ 1

2f ð2xÞ þ 1

4f ð4xÞ þ 1

8f ð8xÞ þ � � �

¼X

1

n¼0

1

2nf ð2nxÞ:

So, what does the graph of B look like? And can we obtain any idea why the

function has its strange properties?

The function f is continuous on R , but is not differentiable at the points 12

n,

for any integer n.

You may omit this section,apart from the next twoparagraphs, at a first reading.

Sub-section 6.1.2.

This series converges by theComparison Test, sincef 2nxð Þj j � 1

2; for all x 2 R :

Next, we construct the graphs

y ¼ 12

f 2xð Þ; y ¼ 14

f 4xð Þ; y ¼ 18

f 8xð Þ; . . . ;

we obtain each graph by scaling the previous graph by a factor of 12

in both the

x-direction and the y-direction.

Each graph has twice as many peaks in any interval as its predecessor, and

each of these peaks is half the height of the peaks in its predecessor. Thus the

number of points in the interval where the function is not differentiable doubles

(approximately) at each stage.

We obtain the Blancmange function B by adding together all these functions

to form an infinite series B xð Þ ¼ f xð Þ þ 12

f 2xð Þ þ 14

f 4xð Þ þ 18

f 8xð Þ þ � � �.Thus, for example

B 12

� �

¼ f 12

� �

þ 12

f 1ð Þ þ 14

f 2ð Þ þ 18

f 4ð Þ þ � � �¼ 1

2þ 1

2� 0

� �

þ 14� 0

� �

þ 18� 0

� �

þ � � �¼ 1

2;

and

B 14

� �

¼ f 14

� �

þ 12

f 12

� �

þ 14

f 1ð Þ þ 18

f 2ð Þ þ � � �¼ 1

4þ 1

2� 1

2

� �

þ 14� 0

� �

þ 18� 0

� �

þ � � �¼ 1

4þ 1

4¼ 1

2:

To get an idea of the shape of the graph of B, we look at the graphs of

successive partial sum functions of B. To help you follow the construction, we

also draw in the graph at the previous stage (in light dashes) and the function

being added to it (in heavy dashes).

The first few of these graphsare included in the diagramsbelow.


Eventually, we obtain the following graph of B:

It looks as though B is continuous. However, it is NOT true in general that the

sum of infinitely many continuous function is itself continuous, so a proof of

the continuity of B is necessary.

Similarly, it looks as though B might not be differentiable.

For example, we have marked the points on the above graphs corresponding to

x ¼ 13. At the first stage, 1

3lies in the interval 0; 1

2

�

, and so the point 13; f 1

3

� ��

lies on a line segment of slope 1. At the next stage, 13

lies in the interval 14; 1

2

�

,

and so the point 13; f 1

3

� �

þ 12

f 23

� ��

lies on a line segment of slope 0. And so on.

At successive stages in the construction of the graph of B, the point corres-

ponding to 13

lies alternately on line segments of slope 0 and 1. Thus it seems

plausible that the slopes of chords joining the points 13; B 1

3

� ��

and (x, B(x))

do not tend to any fixed value as x tends to 13, so that B would not be

differentiable at 13.

Notice in the last diagram above that however closely we look at the graph

of the Blancmange function B, it seems to have small blancmanges growing on

it everywhere. At the nth stage, each horizontal line segment has one or two

mini-blancmanges growing on it, and each sloping line segment has one or two

‘sheared’ mini-blancmanges growing on it.

Finally, notice that the Blancmange function B is periodic, with period 1.

This occurs because f is periodic, with period 1.

Here the dashes indicate thegraphs at the early stages andalso the graphs of functionsoccurring in the summation.

We give this inSub-section 6.6.2.

We prove this inSub-section 6.6.3.


6.6.2 Continuity of the Blancmange function

To verify that B is continuous on R , we must use the power of the "� �definition of continuity.

Theorem 1 The Blancmange function B is continuous at each point c2R .

Proof We must show that


BðxÞ � BðcÞj j < "; for all x satisfying x� cj j < �: (1)

Now, it follows, from the definition of B, that B xð Þ � B cð Þ ¼P

1

n¼0

12n f 2nxð Þ � f 2ncð Þð Þ; hence, by the infinite form of the Triangle Inequality

B xð Þ � B cð Þj j �X

1

n¼0

1

2nf 2nxð Þ � f 2ncð Þj j: (2)

For all x and c, both of the numbers f(2nx) and f(2nc) lie in 0; 12

�

, so that the

modulus of their difference is at most 12; that is

f 2nxð Þ � f 2ncð Þj j � 1

2:

Now we choose an integer N such that 12N <

12". (This choice is possible, since

the sequence 12n

� �

is null.) It follows that

X

1

n¼N

1

2nf 2nxð Þ � f 2ncð Þj j �

X

1

n¼N

1

2n� 1

2

¼ 1

2N

<1

2": (3)

Next, since each of the functions x 7! f(2nx), n¼ 0, 1, . . ., N� 1, is continuous,

it follows that

for each n¼ 0, 1, . . ., N� 1, there is a positive number �n such that

f 2nxð Þ � f 2ncð Þj j < 14"; for all x satisfying x� cj j < �n:

Now we choose �¼min{�0, �1, . . ., �N�1}. Thus, for each n¼ 0, 1, . . .,N� 1, we must have

f 2nxð Þ � f 2ncð Þj j < 14"; for all x satisfying x� cj j < �:

It follows that

X

N�1

n¼0

1

2nf 2nxð Þ � f 2ncð Þj j <

X

N�1

n¼0

1

2n� 1

4"

<X

1

n¼0

1

2nþ2� "

¼ 1

2": (4)

Sub-section 3.3.1.

We apply the definition ofcontinuity to each functionin turn.

We use 14" here, in order

to obtain an " in our finalresult (1).

Substituting the upper bounds from (3) and (4) into inequality (2), we deduce

that, for all x satisfying jx� cj<�, we have

B xð Þ � B cð Þj j �X

N�1

n¼0

1

2nf 2nxð Þ � f 2ncð Þj j þ

X

1

n¼N

1

2nf 2nxð Þ � f 2ncð Þj j

<1

2"þ 1

2"

¼ ":This is the desired result (1). Hence B is continuous at c. &

6.6.3 The Blancmange function is differentiablenowhere

The proof that B is nowhere differentiable on R is more tricky, and the

following lemma about difference quotients in general (which is of interest

in its own right) plays a crucial role.

Lemma Let f be defined on an open interval I and differentiable at the

point c2 I. Let {xn} and {yn} be sequences in I converging to c such that

xn � c � yn and xn < yn; for n ¼ 0; 1; 2; . . .:

Then

limn!1

f ynð Þ � f xnð Þyn � xn

exists and equals f 0 cð Þ:

Proof We must prove that

for each positive number ", there exists some number X such that


� f 0 cð Þ�

�

�

�

�

�

�

�

< "; for all n>X: (5)

Since f is differentiable at c, we know that, for each positive number ", there

exists some positive number � such that


� f 0 cð Þ�

�

�

�

�

�

�

�

<1

4"; for all x satisfying 0 < x� cj j < �;

so that

f xð Þ� f cð Þ� f 0 cð Þ x� cð Þj j � 1

4" x� cj j; for all x satisfying x� cj j<�: (6)

Now, since xn! c and yn! c as n!1, it follows that there are numbers X1

and X2 such that

xn � cj j < � for all n>X1; and yn � cj j < � for all n>X2;

so, if we set X¼max{X1, X2}, we certainly have

xn � cj j < � and yn � cj j < � for all n>X:

It now follows from (6) that, for all n>X, we have

f xnð Þ � f cð Þ � f 0 cð Þ xn � cð Þj j � 1

4" xn � cj j and

f ynð Þ � f cð Þ � f 0 cð Þ yn � cð Þj j � 1

4" yn � cj j:

xn ync

yy = f (x)

x

This is the definition ofdifferentiability at c, butwith 1

4" in place of ", in order

to obtain an " in our finalresult (5).

In fact, this is a strictinequality if x 6¼ c. Howeverlater in the argument we willneed to allow the possibilitythat x¼ c, so we must write(6) with a weak inequalitysign.

We put x¼ xn and then x¼ yn

into (6).

Hence, we can apply the Triangle Inequality to obtain

f ynð Þ� f xnð Þ� f 0 cð Þ yn� xnð Þj j¼ f ynð Þ� f cð Þ� f 0 cð Þ yn� cð Þf g� f xnð Þ� f cð Þ� f 0 cð Þ xn� cð Þf gj j� f ynð Þ� f cð Þ� f 0 cð Þ yn� cð Þj jþ f xnð Þ� f cð Þ� f 0 cð Þ xn� cð Þj j

� 1

4" yn� cj jþ 1

4" xn� cj j

� 1

4" yn� xnj jþ 1

4" yn� xnj j

¼ 1

2" yn� xnj j; for all n> X:

Since we know that xn 6¼ yn, we can divide this inequality by the non-zero

term yn� xn to obtain, for n>X


� f 0 cð Þ�

�

�

�

�

�

�

�

� 1

2" < ":

This is the result (5) that we set out to prove. &

Remark

The hypothesis that xn and yn must lie on opposite sides of c cannot generally

be omitted. For example, consider the function

f ðxÞ ¼ x2 sin 1x

� �

; x 6¼ 0,

0; x ¼ 0,

that you saw earlier is differentiable at 0, with f 0(0)¼ 0. If we set

xn ¼1

2nþ 1ð Þp and yn ¼1

2nþ 12

� �

p; for n ¼ 0; 1; 2; . . .;

then


¼1

2nþ12ð Þ2p2

� �

� 0ð Þ

1

2nþ12ð Þp

� �

� 12nþ1ð Þp

� �

¼ 2 2nþ 1ð Þ2nþ 1

2

� �

p

! 2

pas n!1:

However, the value of this limit is not 0, the value of f 0(0).

We are now ready to prove the principal result in this sub-section.

Theorem 2 The Blancmange function B is not differentiable at any point

c2R .

Proof In order to apply the lemma, we construct two sequences {xn} and

{yn} converging to c such that the corresponding sequence of difference

quotients is not convergent. We use the method of repeated bisection to

construct {xn} and {yn}.

We first insert some terms andtake them away again.

We use the two inequalities thatwe have just verified for n>X.

Since xn� c� yn, bothjyn� cj and jxn� cj are� |yn� xn|.


y = x2

y =f (x)

y = – x2

y

xn xyn

Since B is periodic with period 1, we assume for simplicity that c2 [0, 1].

We start by defining [x0, y0]¼ [0, 1]. Then, since c lies in one of the intervals

0; 12

�

and 12; 1

� �

, we can define

x1; y1½ � ¼ 0; 12

�

;12; 1

�

;

if c 2 0; 12

�

;

if c 2 12; 1

� �

:

With this definition of [x1, y1], we have that:

1. x1; y1½ � x0; y0½ �;2. y1 � x1 ¼ 1

2;

3. x1� c� y1;

4. x1 ¼ 12

p1 and y1 ¼ 12

p1 þ 1ð Þ, for some integer p1.

We can then repeat this process, bisecting the interval [x1, y1] to obtain [x2, y2],

and so on. In this way, we obtain a sequence of closed intervals [xn, yn], for

n¼ 1, 2, . . ., such that:

1. xnþ1; ynþ1½ � xn; yn½ �;2. yn � xn ¼ 1

2

� �n;

3. xn� c� yn;

4. xn ¼ 12n pn and yn ¼ 1

2n pn þ 1ð Þ, for some integer pn.

Properties 2 and 3 imply that both the sequences {xn} and {yn} converge to c,

but we shall show that the sequence of difference quotients {Qn}, where

Qn ¼B ynð Þ � B xnð Þ

yn � xn

¼ 2n B ynð Þ � B xnð Þð Þ;

is not convergent. It will then follow, from the lemma, that B is not differenti-

able at c.

To prove that the sequence {Qn} is divergent, it is sufficient to prove that

Qnþ1 ¼ Qn � 1; for n ¼ 0; 1; 2; . . .; (7)

since it then follows that {Qn} cannot converge.

To prove (7), we put

zn ¼1

2xn þ ynð Þ ¼ 1

2nþ12pn þ 1ð Þ:

Note that

xnþ1; ynþ1½ � ¼ xn; zn½ �;zn; yn½ �;

if c 2 xn; zn½ �;if c 2 zn; ynð �:

We now claim that, for n¼ 0, 1, 2, . . .

B znð Þ ¼1

2B xnð Þ þ B ynð Þð Þ þ 1

2nþ1: (8)

It would then follow from (8) that, if [xnþ1, ynþ1]¼ [xn, zn], then

Qnþ1 ¼ 2nþ1 B znð Þ � B xnð Þð Þ¼ 2n B ynð Þ � B xnð Þð Þ þ 1 ¼ Qn þ 1;

We use Property 2 for thevalue of yn� xn.

The structure of the proof isas follows(8)) (7)) Theorem 2.


whereas, if [xnþ1, ynþ1]¼ [zn, yn], then

Qnþ1 ¼ 2nþ1 B ynð Þ � B znð Þð Þ¼ 2n B ynð Þ � B xnð Þð Þ � 1 ¼ Qn � 1:

In either case, the required result (7) would then hold. Thus, in order to

complete the proof that (7) holds, it is sufficient to prove that (8) holds.

To verify (8), note that, for k¼ 0, 1, 2, . . ., we have

2kxn ¼1

2n�kpn;

2kyn ¼1

2n�kpn þ 1ð Þ;

2kzn ¼1

2n�kþ12pn þ 1ð Þ;

9

>

>

>

>

>

=

>

>

>

>

>

;

(9)

and f ( p)¼ 0 for any integer p. It then follows from the definition of B as an

infinite series that

B xnð Þ þ B ynð Þ ¼X

1

k¼0

1

2kf 2kxn

� �

þ f 2kyn

� ��

¼X

n�1

k¼0

1

2kf 2kxn

� �

þ f 2kyn

� ��

; and

B znð Þ ¼X

1

k¼0

1

2kf 2kzn

� �

¼X

n

k¼0

1

2kf 2kzn

� �

:

We deduce from (9) that, for k¼ 0, 1, 2, . . ., n� 1, the terms 2kxn, 2kyn and

2kzn all lie in the interval

2kxn; 2kyn

�

¼ 1

2n�kpn;

1

2n�kpn þ 1ð Þ

� �

;

and so in some interval of the form 12

qk;12

qk þ 1ð Þ �

, where qk is an integer.

Now, the restriction of f to such an interval is linear, so that we have

f 2kzn

� �

¼ 12

f 2kxn

� �

þ f 2kyn

� ��

; k ¼ 0; 1; 2; . . .; n� 1;

hence

X

n�1

k¼0

1

2kf 2kzn

� �

¼ 1

2

X

n�1

k¼0

1

2kf 2kxn

� �

þ f 2kyn

� ��

: (10)

Finally

1

2nf 2nznð Þ ¼ 1

2nf pn þ

1

2

� �

¼ 1

2nþ1: (11)

If we then add (10) and (11), we obtain (8), as required.

This completes the proof of Theorem 2. &

We shall shortly use the

expressionP

1

k¼0

12k f 2kx� �

for

B(x), hence some ks nowappear.

For, f (2kxn)¼ f (2kyn)¼ 0 ifk> n� 1.

For, f (2kzn)¼ 0 if k> n.


6.7 Exercises

Section 6.1

1. Determine whether each of the following functions f is differentiable at the

specified point c; if it is, evaluate the derivative f 0(c).

(a) f xð Þ ¼ �x2; x � 0,

x2; x > 0,

c ¼ 0;

(b) f xð Þ ¼ 1; x < 1,

x2; x � 1,

c ¼ 1;

(c) f xð Þ ¼ tan x; � 12p < x < 1

3p;

x2; x � 13p;

c ¼ 13p;

(d) f ðxÞ ¼ 1� x; x < 1,

x� x2; x � 1,

c ¼ 1;

(e) f xð Þ ¼ x sin 1x2

� �

; x 6¼ 0;0; x ¼ 0;

c ¼ 0;

(f) f xð Þ ¼ sin x sin 1x

� �

; x 6¼ 0;0; x ¼ 0;

c ¼ 0:

2. Write down an expression for a function f with domain (�1, 2] and the


(a) fL0 exists at 1, and fL

0(1)¼ 1;

(b) fR0 exists at 1, and fR

0(1)¼ 2.

Verify that f has the properties (a) and (b).

Section 6.2

1. Use the rules for differentiation to verify that the function

f xð Þ ¼ loge 1þ xð Þ þ ex2

; x 2 �1;1ð Þ, is differentiable on its domain,

and determine its derivative.

2. Write down (without justification) the derivatives of the following

functions:

(a) f xð Þ ¼ x2þ1x�1

; x 2 1;1ð Þ;(b) f xð Þ ¼ loge sin xð Þ; x 2 0; pð Þ;(c) f xð Þ ¼ loge sec xþ tan xð Þ; x 2 �1

2p; 1

2p

� �

;

(d) f xð Þ ¼ cos xþsin xcos x�sin x

; x 2 0; 14p

� �

:

3. Prove that the function f(x)¼ tanh x, x2R , has an inverse function f�1 that

is differentiable on (�1, 1), and find an expression for (f�1)0(x).

4. Prove that the function f(x)¼ tan xþ 3x, x 2 �12p; 1

2p

� �

, has an inverse

function f�1 that is differentiable on R , and find the value of ( f�1)0(0).

5. Determine the derivatives of the following functions, assuming that they are

differentiable on their domains:

(a) f xð Þ ¼ coth x; x 2 R � 0f g;(b) f xð Þ ¼ loge xð Þloge x; x 2 1;1ð Þ:

Section 6.3

1. The function f has domain [�2, 2] and its graph consists of four line

segments, as shown.

Identify (a) the local minima of f, (b) the minima of f, (c) the local maxima

of f, and (d) the maxima of f, and state where these occur,

2. Let f be the function

f xð Þ ¼ 2x3 � 3x2; x 2 0; 2½ �:Find (a) the local minima and minima of f, (b) the local maxima and

maxima of f and state where these occur.

3. Prove that, of all rectangles with given perimeter, the square has the greatest

area.

4. Verify that the conditions of Rolle’s Theorem are satisfied by the function

f xð Þ ¼ 1þ 2x� x2; x 2 0; 2½ �;and determine a value of c in (0,2) for which f 0(c)¼ 0.

5. Use Rolle’s Theorem to prove that, if p is a polynomial and l1, l2, . . ., ln are

distinct zeros of p, then p0 has at least (n�1) zeros.

6. Use Rolle’s Theorem to prove that, for any real number l, the function

f xð Þ ¼ x3 � 3

2x2 þ l; x 2 R ;

never has two distinct zeros in [0, 1].

Hint: Assume that f has two distinct zeros in [0, 1], and obtain a

contradiction.

7. The function f is twice differentiable on an interval [a, b]; and, for some

point c in (a, b), f(a)¼ f(b)¼ f(c). Prove that there exists some point d in

(a, b) with f 00(d)¼ 0.

8. The function f is differentiable on a neighbourhood N of a point c; f 0 is

continuous at c and f 0(c)> 0. Prove that f is increasing on some neighbour-

hood of c.

Harder: Give an example of a function f that is differentiable on R with the

properties that (a) f 0(0)> 0 and (b) there is NO open neighbourhood of 0 on

which f is increasing.

9. Let the function f : 0; 1½ � 7! 0; 1½ � be continuous on [0, 1] and differentiable

on (0, 1). We know that there is at least one point c in [0, 1] for which

f (c)¼ c. Use Rolle’s Theorem to prove that, if f 0 xð Þ 6¼ 1 in (0, 1), then there

is exactly one such point c.

Hint: Consider the function h xð Þ ¼ f xð Þ � x; x 2 0; 1½ �; assume that two

such points c exist, and obtain a contradiction.

You saw this in Problem 2of Sub-section 4.2.1.

6.7 Exercises 253

Section 6.4

1. For each of the following functions, verify that the conditions of the Mean

Value Theorem are satisfied, and determine a value of c that satisfies the

conclusion of the theorem:

(a) f xð Þ ¼ 2x�1x�2

; x 2 �1; 1½ �; (b) f xð Þ ¼ x3 þ 2x2 þ x; x 2 0; 1½ �:2. Use the Mean Value Theorem to prove that

sin b� sin aj j � b� aj j; for a; b 2 R :

3. Use the Zero Derivative Theorem to prove that

cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p� �

; for x � 1:

4. For the function f(x)¼ x3� 2x2þ x, x2R , determine those points c where

f 0(c)¼ 0. Using the Second Derivative Test, determine whether these

correspond to local maxima, local minima, or neither.

5. Prove the following inequalities:

(a) loge x � 1� 1x; for x 2 1;1½ Þ;

(b) 4x14 � xþ 3; for x 2 0; 1½ �;

(c) loge 1þ xð Þ > x� 12

x2; for x 2 0;1ð Þ:6. Let f be defined on a neighbourhood of c, with f 0(c)> 0. Prove that there is

some punctured neighbourhood N of c such that


> 0; for x 2 N:

7. By applying the Mean Value Theorem to the function f xð Þ ¼ffiffiffi

xp

on the

interval [100, 102], prove that 10 111<


102p

< 10 110:

8. Let f be differentiable on a closed interval [a, b].

(a) Prove that, if the minimum of f on [a, b] occurs at a, then fR0(a)� 0; and

that, if the minimum of f on [a, b] occurs at b, then fL0(b)� 0.

(b) Prove that, if fR0(a)< 0 and fL

0(b)> 0, then f 0(x)¼ 0 for some x2 (a, b).

Hint: Check that the minimum of f on [a, b] occurs at some point in (a, b).

(c) By considering the function g (x)¼ f(x)� kx, x2 [a, b], prove that, if

fR0(a)< k< fL

0(b), then f 0(x)¼ k for some x2 (a, b).

Section 6.5

1. Verify that the following functions satisfy the conditions of Cauchy’s Mean

Value Theorem on [0, 2], and determine a value of c that satisfies the

conclusion of the theorem

f xð Þ ¼ x4 þ 2x2 and g xð Þ ¼ x3 þ 3x; for x 2 0; 2½ �:2. Use l’Hopital’s Rule to prove that the following limits exist, and evaluate

the limits:

(a) limx!0

sinh xþ sin xð Þsin x

; (b) limx!1

5xþ3ð Þ13� xþ3ð Þ

12

x�1;

(c) limx!0

1� cos xx2 ; (d) lim

x!0

sinh x�xsin 3x3ð Þ :

This result is known asDarboux’s Theorem, and isan intermediate value theoremfor f 0.

7 Integration

We have already used the idea of the area of a set in the plane; for example, to

define the number p and to prove that sin xx! 1 as x! 0. However, our earlier

discussions begged the question of what exactly we mean by ‘area’.

For simplicity, we shall restrict our attention in this book to defining only the

area between a graph and the x-axis.

In Section 7.1 we use the idea of lower and upper estimates to give a rigorous

definition of such an area. This is done by trapping the desired area between

underestimates and overestimates, each of which is the sum of the areas of

suitably chosen rectangles.

Then the area between the graph y¼ f (x), x2 [a, b], and the segment [a, b] of

the x-axis is defined to be A if:

the least upper bound of the underestimates¼A,

and:

the greatest lower bound of the overestimates¼A.

We call A the integral of f over [a, b], and denote it byZ b

a

f or

Z b

a

f xð Þdx:

In practice, it would be inconvenient if we had to revert to the definition in

order to tell whether a given function was integrable or not, and we shall verify

several criteria for integrability.

In Section 7.2, we identify some large classes of integrable functions, and

verify the standard rules for integrable functions.

In Section 7.3 we meet the Fundamental Theorem of Calculus which enables

us to avoid use of the definition of integrability in order to evaluate many

integrals. For instance, we verify the usual methods for integration by parts and

integration by substitution.

It follows from theFundamental Theorem ofCalculus that we can think ofintegration as the operationinverse to differentiation.

255

Often it is not possible to evaluate an integral explicitly, and the best that we

can do is to obtain upper and lower estimates for its value. In Section 7.4, we

meet a range of inequalities for integrals, including the Triangle Inequality for

integralsZ b

a

f

�

�

�

�

�

�

�

�

�Z b

a

fj j; where a < b:

We use these inequalities to prove Wallis’s Formula for p, namely that

limn!1

2

1:2

3:4

3:4

5:6

5:6

7: � � � : 2n

2n� 1:

2n

2nþ 1

� �

¼ p2;

and to establish the Maclaurin Integral Test, which enables us to determine the

convergence or divergence of series such asP

1

n¼1

1np, for p> 0, and

P

1

n¼2

1n loge n

.

Finally, in Section 7.5, we discuss the evaluation of n!. It is easy to evaluate

n! for values of n up to 10, say, by direct multiplication; but for n¼ 100 or

n¼ 200, the number n! cannot be evaluated by a standard scientific calculator.

Surprisingly, we can use integration techniques to obtain an excellent estimate

for n!, called Stirling’s Formula

n! �ffiffiffiffiffiffiffiffi

2pnp n

e

� �n

as n!1:

Before starting to read the chapter, we recommend that you refresh your

memory of greatest lower bounds and least upper bounds of functions.

Recall that, if ƒ is a function defined on an interval I�R , then:

� A real number m is the greatest lower bound, or infimum, of ƒ on I if:

1. m is a lower bound of ƒ(I);

2. if m0>m, then m0 is not a lower bound of ƒ(I).

� M is the least upper bound, or supremum, of ƒ on I if:

1. M is an upper bound of ƒ(I);

2. if M 0<M, then M 0 is not an upper bound of ƒ(I).

In Sections 7.1 and 7.2 we shall discuss some new properties of infimum and

supremum that are needed for our work on integrability.

Also, we shall recommend that you omit working your way through quite a

number of detailed proofs during your first reading of this chapter. This is not

necessarily because they are particularly difficult, but simply in order to guide

you through the key ideas first before you return to study some of the ‘gory

details’ later on once you have grasped the overall picture.

7.1 The Riemann integral

Earlier we defined the number p to be the area of the unit disc. We obtained

lower and upper estimates for p by trapping the area of the disc between the

area of a 3� 2n-sided inner polygon and the area of a 3� 2n-sided outer

polygon, and letting n!1.

At least not in 2005, whenthese words are being written.

We shall also explain theprecise meaning of thesymbol tilda: ‘�’.

You met sup and inf in Sub-section 1.4.2.

We often denote m byinf { f(x) : x2 I}, inf

x2If xð Þ,

infI

f or simply by inf f.

We often denote M bysup { f(x) : x2 I}, sup

x2I

f xð Þ,

supI

f or simply by sup f.

Sub-section 2.5.4.

256 7: Integration

For simplicity, in this book we do not consider general areas in the plane, but

restrict our attention to the area between a graph y¼ f(x), x2 [a, b], and the

interval [a, b] of the x-axis. For a continuous function f, we certainly require

such a definition to agree with our intuitive notion of area!

However, it is not obvious that we can always say that the region between a

graph and the x-axis has an area. For example, how can we define such an area

for the following functions?

(a) f(x) = x2, 0 ≤ x ≤ 1,2, 1 < x ≤ 2.{ {(c) f(x) =

1, 0 ≤ x ≤ 1, x rational,x irrational.0, 0 ≤ x ≤ 1.{(b) f(x) =

–2, 0 ≤ x < 1,3, x = 1.

Our approach is to find lower and upper estimates for the area, if such a

thing exists, that we can assign to each region by splitting it up into smaller

regions with areas which we can approximate by rectangles, and use the fact

that

Area of a rectangle¼ base� height.

We now illustrate the underlying idea of these estimates by considering the

situation when the function f is positive on [a, b] (see the diagrams below).

First, we divide up the interval [a, b] into a family of smaller intervals, called a

partition of [a, b]. Then we approximate the area by finding two sequences of

rectangles each with one of the subintervals as base. In one sequence, we

choose rectangles as large as possible so that the sum of their individual areas

forms an underestimate for the ‘area’ of the region; in the other, we choose

rectangles as small as possible so that the sum of their individual areas forms an

overestimate for the ‘area’.

Then, if there is a real number A with the properties:

the least upper bound of the underestimates¼A,

When defining p, we usedtriangles rather thanrectangles.

and:

the greatest lower bound of the overestimates¼A,

we define A to be the area between the graph and the x-axis.

We find that we can define such an area for the graphs (a) and (b) above, but

not for the graph (c).

7.1.1 The Riemann integral and integrability

We now start on a process leading to the formal definition of the integral.

Definitions A partition P of an interval [a, b] is a family of a finite

number of subintervals of [a, b]

P ¼ x0; x1½ ; x1; x2½ ; . . . ; xi�1; xi½ ; . . . ; xn�1; xn½ f g; (1)

where

a ¼ x0 < x1 < x2 < � � � < xi�1 < xi < � � � < xn�1 < xn ¼ b:

The points xi, 0� i� n, are called the partition points in P.

The length of the ith subinterval is denoted by �xi¼ xi� xi�1, and the

mesh of P is the quantity Pk k ¼ max1�i�n

�xif g.A standard partition is a partition with equal subintervals.

For example, consider the partition P of [0, 1], where

P ¼ 0;1

2

�

;1

2;3

5

�

;3

5;3

4

�

;3

4; 1

� �

:

Here

�x1 ¼1

2� 0 ¼ 1

2; �x2 ¼

3

5� 1

2¼ 1

10; �x3 ¼

3

4� 3

5¼ 3

20and

�x4 ¼ 1� 3

4¼ 1

4;

and the mesh of P is

Pk k ¼ max1

2;

1

10;

3

20;1

4

�

¼ 1

2:

P is not a standard partition of [0, 1], since not all its subintervals are of equal

length.

For brevity, we sometimesshorten this expression for Psimply to xi�1; xi½ gn

i¼1 or�

xi�1; xi½ : 1 � i � nf g:

a = x0 x1 xi–1 xn = bxi.... ....

Equivalently

P ¼ 0; 0:5½ ; 0:5; 0:6½ ;f0:6; 0:75½ ; 0:75; 1½ g:

258 7: Integration

Next, we introduce some notation, mi and Mi, associated with the height of

the graph of a bounded function f on its subintervals. Since we are intending to

develop a definition of integral that will apply to discontinuous functions as

well as to continuous functions, we use the concepts of greatest lower bound

and least upper bound of f on subintervals rather than minimum and maximum

of f on the subintervals.

We also introduce some quantities, L( f, P) and U( f, P), that correspond to

underestimates and overestimates of the possible area.

Definitions Let f be a bounded function on [a, b], and P a partition of [a, b]

given by P¼ {[xi�1, xi] : 1� i� n}. We denote by mi and Mi the quantities

mi¼ inf f xð Þ : x2 xi�1; xi½ f g and Mi¼ sup f xð Þ : x2 xi�1; xi½ f g:Then the corresponding lower and upper Riemann sums for f on [a, b] are

L f , Pð Þ ¼X

n

i¼1

mi�xi and U f , Pð Þ ¼X

n

i¼1

Mi�xi:

m1

x0

y

x x

y

x1 x2 x3 xn–1 xn xn–1 xnx0 x1 x2 x3 ......

m2 m3

mny = f (x)

y = f (x)M1

M2 M3

Mn

Example 1 Let f(x)¼ x, x2 [0, 1], and let P ¼ 0; 15

�

; 15; 1

2

�

; 12; 1

��

be a

partition of [0, 1]. Evaluate L( f, P) and U( f, P).

Solution In this case, the function f is increasing and continuous. Thus, on

each subinterval in [0, 1], the infimum of f is the value of f at the left end-point

of the subinterval and the supremum of f is the value of f at the right end-point

of the subinterval.

Hence, on the three subintervals in P, we have

m1 ¼ f 0ð Þ ¼ 0; M1 ¼ f1

5

� �

¼ 1

5; �x1 ¼

1

5� 0 ¼ 1

5;

m2 ¼ f1

5

� �

¼ 1

5; M2 ¼ f

1

2

� �

¼ 1

2; �x2 ¼

1

2� 1

5¼ 3

10;

m3 ¼ f1

2

� �

¼ 1

2; M3 ¼ f 1ð Þ ¼ 1; �x3 ¼ 1� 1

2¼ 1

2:

It then follows, from the definitions of L( f, P) and U( f, P), that

L f , Pð Þ ¼X

3

i¼1

mi�xi ¼ m1�x1 þ m2�x2 þ m3�x3

¼ 0� 1

5þ 1

5� 3

10þ 1

2� 1

2

¼ 0þ 3

50þ 1

4¼ 31

100;

We need to do this, since, if fis not continuous on a typicalsubinterval [xi�1, xi], it maynot possess a minimum or amaximum on the subinterval.

You met infimum andsupremum earlier, in Sub-section 1.4.2. The quantitiesmi and Mi exist, since f isbounded.

y

1

0

y = x

15

12

1 x

We use the fact that, if afunction is continuous on aclosed interval, then it attainsits infimum and supremumthere, by the Extreme ValuesTheorem in Sub-section 4.2.3.


U f , Pð Þ ¼X

3

i¼1

Mi�xi ¼ M1�x1 þM2�x2 þM3�x3

¼ 1

5� 1

5þ 1

2� 3

10þ 1� 1

2

¼ 1

25þ 3

20þ 1

2¼ 69

100: &

Problem 1 Evaluate L ( f, P) and U ( f, P) for the following function

and partition of [0, 1]:

f xð Þ ¼2x; 0 � x <

1

2;

1

2< x < 1;

0; x ¼ 1

2;

1; x ¼ 1;

8

>

>

<

>

>

:

x 2 0; 1½ ; and P ¼ 0; 13

�

; 13; 3

4

�

; 34; 1

��

:

Hint: Be careful over the values of m2 and M3; you may find it helpful

to sketch the graph of f.

Example 2 Evaluate L( f, Pn) and U( f, Pn) for the following function and

standard partition of [0, 1]

f xð Þ ¼ x; x 2 0; 1½ ; and

Pn ¼ 0;1

n

�

;1

n;2

n

�

; . . . ;i� 1

n;

i

n

�

; . . . ; 1� 1

n; 1

� �

;

and determine limn!1

L f ;Pnð Þ and limn!1

U f ;Pnð Þ, if these exist.

Solution In this case, the function f is increasing and continuous. Thus, on

each subinterval in [0, 1], the infimum of f is the value of f at the left end-point

of the subinterval and the supremum of f is the value of f at the right end-point

of the subinterval.

Hence, on the ith subinterval i�1n; i

n

�

in Pn, for 1� i� n, we have

mi ¼ fi� 1

n

� �

¼ i� 1

n; Mi ¼ f

i

n

� �

¼ i

n; and

�xi ¼i

n� i� 1

n¼ 1

n:

It then follows, from the definitions of L( f, Pn) and U( f, Pn), that

L f ;Pnð Þ ¼X

n

i¼1

mi�xi ¼X

n

i¼1

i� 1

n� 1

n

¼ 1

n2

X

n

i¼1

i�X

n

i¼1

1

( )

¼ 1

n2

n nþ 1ð Þ2

� n

�

¼ n� 1

2n;

Notice how a careful layingout of the calculation for thevarious terms makes itstraight-forward to calculatethese two sums.

It is quite important that youtackle this problem, to makesure that you understand thenotation being used. Youshould also read its solutioncarefully.

y

1

0

y = x

1n

2n

n –1n

1 x. . .

We follow the generalstructure of the solution toExample 1.

260 7: Integration

U f ;Pnð Þ ¼X

n

i¼1

Mi�xi ¼X

n

i¼1

i

n� 1

n

¼ 1

n2

X

n

i¼1

i

¼ 1

n2� n nþ 1ð Þ

2¼ nþ 1

2n:

:

It follows that

limn!1

L f ;Pnð Þ ¼ limn!1

n� 1

2n¼ 1

2and

limn!1

U f ;Pnð Þ ¼ limn!1

nþ 1

2n¼ 1

2: &

Problem 2 Evaluate L( f, Pn) and U( f, Pn) for the following function

and standard partition of [0, 1]

f xð Þ ¼ x2; x 2 0; 1½ ; and

Pn ¼ 0;1

n

�

;1

n;2

n

�

; . . .;i� 1

n;

i

n

�

; . . .; 1� 1

n; 1

� �

;

and determine limn!1

L f , Pnð Þ and limn!1

U f , Pnð Þ, if these exist.

Properties of Riemann sums

In all the examples that we have seen so far, the lower Riemann sum for f over

an interval has been less than or equal to the corresponding upper Riemann

sum. Also, in Example 2 we found that the value that ‘we hope for’ forR 1

0xdx,

namely 12, is greater than all the lower Riemann sums and less than all the upper

Riemann sums; it seems reasonable to ask whether such a property holds in

general.

So we now examine some of the key properties of Riemann sums. We shall

need to use some properties of least upper bounds and greatest lower bounds.

Theorem 1 For any function f bounded on an interval [a, b] and any

partition P of [a, b], L( f, P)�U( f, P).

Proof Let P¼ {[xi�1, xi] : 1� i� n}. Then, on each subinterval [xi�1, xi],

1� i� n, we have inf{ f(x) : x2 [xi�1, xi]}� sup { f(x):x2 [xi�1, xi]} – in other

words, mi�Mi.

It follows that

X

n

i¼1

mi�xi �X

n

i¼1

Mi�xi;

in other words, L( f, P)�U( f, P), as required. &

Next, we need some techniques for comparing the Riemann sums for

different partitions on the interval [a, b]; this will enable us shortly to verify

that, for two partitions P and P0 of [a, b], we must have L( f, P)�U( f, P0).

Here we are assuming that ourrigorous treatment ofintegration will give the sameanswers as those that youobtained in your initialCalculus course!

We will set these out carefullyas we need them.

For each term in the left-handsum is less than or equal to thecorresponding term in theright-hand sum.

Theorem 3, below.


Notice that this fact does NOT follow from Theorem 1, which deals only with

one partition at a time.

Definition Let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be a

partition of an interval [a, b]. Then a partition P0 of [a, b] is a refinement of

P if the partition points of P0 include the partition points of P. A partition Q

of [a, b] is the common refinement of two partitions P and P0 of [a, b] if the

partition points of Q comprise the partition points of P together with the

partition points of P0.

For example, the partition P0 ¼ 0, 12

�

, 12

, 35

�

, 35

, 34

�

, 34

, 1 ��

of [0, 1] is

a refinement of the partition P ¼ 0, 12

�

, 12

, 34

�

, 34

, 1 ��

, since it simply has

one additional partition point 35

as compared with P. Similarly,

P00 ¼ 0, 13

�

, 13

, 12

�

, 12

, 35

�

, 35

, 34

�

, 34

, 1 ��

is also a refinement of P. However the

partition Q ¼ 0, 15

�

, 15

, 12

�

, 12

, 1 ��

of [0, 1] is not a refinement of P, since its

partition points do not include all the partition points of P.

Also, the common refinement of the partitions P ¼ 0, 12

�

, 12

, 34

�

, 34

, 1 ��

and P0 ¼ 0, 13

�

, 13

, 23

�

, 23

, 1 ��

is

0;1

3

�

;1

3;1

2

�

;1

2;2

3

�

;2

3;3

4

�

;3

4; 1

� �

:

We shall need the following crucial result in our work on refinements.

Lemma 1 For any bounded function f defined on intervals I and J, with

I� J, we have

infx2J

f � infx2I

f and supx2I

f � supx2J

f :

Proof By definition of greatest lower bound, we know that infx2J

f � f xð Þ, for

x2 J. It follows, from the fact that I� J, that

infx2J

f � f xð Þ; for x 2 I:

Hence infx2J

f is a lower bound for f on I, so that infx2J

f � infx2I

f .

The proof that supx2I

f � supx2J

f is similar, so we omit it. &

Problem 3 Prove that, for any bounded function f defined on intervals

I and J where I � J; supx2I

f � supx2J

f .

Lemma 2 Let f be a bounded function on an interval [a, b]. Let P and P0 be

partitions of [a, b], where P0 is a refinement of P that contains just one

additional partition point. Then

L f , Pð Þ � L f , P0ð Þ and U f , P0ð Þ � U f , Pð Þ:

Proof Let P be the partition

P ¼ x0; x1½ ; x1; x2½ ; . . . ; xi�1; xi½ ; . . . ; xn�1; xn½ f gof [a, b], and suppose that P0 contains an additional partition point c in the

particular subinterval [�, �] of P.

The point 34

is missing from Q.

I

J

Loosely speaking, the largerinterval gives the functionmore space to get smaller andmore space to get larger.

For infx2I

f is the greatest lower

bound of f on I.

Loosely speaking, theaddition of one partition pointincreases L and decreases U.

a = x0 x1 xi–1 xn = bxi... ...

262 7: Integration

Note that c must be an interior point of [�, �]. For we are assuming that all

the partition points of partitions are distinct; and, if c were an end-point � or �,

then P0 would contain that point twice in its set of partition points.

Since [�, c] [�, �], it follows from Lemma 1 that

infx2 �;�½

f � infx2 �;c½

f and infx2 �;�½

f � infx2 c;�½

f : (2)

Now, the terms in the lower sums L ( f, P) and L ( f, P0) are the same, except that

the contribution to L ( f, P) associated with the interval [�, �] is

infx2 �;�½

f � � � �ð Þ;

whereas the contribution to L( f, P0) associated with the intervals [�, c] and

[c, �] is

infx2 �; c½

f � c� �ð Þ þ infx2 c; �½

f � � � cð Þ:

It then follows, from the inequalities (2), that

infx2 �; c½

f � c� �ð Þ þ infx2 c; �½

f � � � cð Þ � infx2 �; �½

f � c� �ð Þ þ infx2 �; �½

f � � � cð Þ

¼ infx2 �; �½

f � c� �ð Þ þ � � cð Þf g

¼ infx2 �; �½

f � � � �ð Þ:

Since the lower sums L ( f, P) and L ( f, P0) are the same, apart from these

contributions to each, it follows that L ( f, P0)� L ( f, P), as required.

The proof that U ( f, P0)�U ( f, P) is similar, so we omit it. &

Lemma 2 shows that the addition of just one point to a partition increases the

lower Riemann sum and decreases the upper Riemann sum. By applying this

fact a finite number of times, we deduce the following general result.

Theorem 2 Let f be a bounded function on an interval [a,b], and let P and

P0 be partitions of [a,b], where P0 is a refinement of P. then

L f ;Pð Þ � L f ;P0ð Þ and U f ;P0ð Þ � U f ;Pð Þ:

We now compare the Riemann sums for two different partitions, and dis-

cover that all the lower Riemann sums of a bounded function on an interval are

less than or equal to all the upper Riemann sums. This is a significant

improvement on the result of Theorem 1 that, for a given partition P of [a, b],

L( f, P)�U( f, P).

Theorem 3 Let f be a bounded function on an interval [a, b], and let P and

P0 be partitions of [a, b]. Then

L f , Pð Þ � U f , P0ð Þ:

Proof Let Q be the common partition of P and P0. Then, since Q is a

refinement of both P and P0, we have:

L( f, P)� L( f, Q), by Theorem 2,

L( f, Q)�U( f, Q), by Theorem 1,

U( f, Q)�U( f, P0), by Theorem 2.

We use � and � here ratherthan xi�1 and xi simply inorder to avoid sub-subscripts.

αc

β

Loosely speaking, refining apartition increases L anddecreases U.


It follows from this chain of inequalities that L( f, P)�U( f, P0), as

required. &

This result is exactly what makes our definitions of lower and upper

Riemann sums useful – whatever the integral of f over an interval [a, b]

might be, if it even exists, we are certainly using lower Riemann sums, all of

which provide underestimates, and upper Riemann sums, all of which provide

overestimates.

Definitions Let f be a bounded function on an interval [a, b]. Then we

define:

� the lower integral of f on [a, b] to beR

�b

af ¼ sup

P

L f , Pð Þ,

� the upper integral of f on [a, b] to be �R b

af ¼ inf

PU f , Pð Þ,

where P denotes partitions of [a, b].

Further, if the lower and upper integrals are equal, we define the integral

of f on [a, b],R b

af , to be their common value; that is

Z b

a

f ¼Z

�

b

a

f ¼�Z b

a

f :

It is all too easy to be lulled into a sense of false security by a firmly stated

definition! So we now prove the following result that assures us that the above

definitions make sense.

Theorem 4 Let f be a bounded function on an interval [a, b]. Then:

(a) The lower integralR

�b

af and the upper integral

R

�b

a f both exist;

(b)R

�b

af �

R

�ba f .

Proof

(a) Since f is bounded on [a, b], there is some number M such that j f (x)j �M

on [a, b]; in particular,

f xð Þ � M; for x 2 a; b½ :Thus for any partition P¼ {[xi�1, xi]: 1� i� n} of [a, b], we have

f xð Þ � M; for x 2 xi�1; xi½ and 1 � i � n;

in particular, we have

mi ¼ infxi�1; xi½

f xð Þ � M:

It follows that

L f ;Pð Þ ¼X

n

i¼1

mi�xi �X

n

i¼1

M�xi

¼ MX

n

i¼1

�xi ¼ M b� að Þ:

Since all the lower sums L ( f, P) are bounded above by M (b� a), it follows

that the greatest lower bound of the lower sums, supP

L f ;Pð Þ, must exist.

This greatest lower bound is precisely the lower integralR

�b

af .

Sometimes written asR

�

b

af xð Þdx.

Sometimes written asR

�b

af xð Þdx.

Often written asR b

af xð Þdx.

For, infxi�1 ; xi½

f xð Þ � f xið Þ � M:

Also,R

�b

af � M b� að Þ.

264 7: Integration

The proof of the existence of the upper integralR

�b

af is very similar, so

we omit it here.

(b) Let P and P0 be any two partitions of [a, b]. We know, from Theorem 3,

that

L f , Pð Þ � U f , P0ð Þ:

If we fix P0 for the moment, then we know that U( f, P0) serves as an upper

bound for all the lower Riemann sums L( f, P), whichever partition P may

be. It follows, then, that the least upper bound of the L( f, P), the lower

integralR

�

b

af , must satisfy the inequalityZ

�

b

a

f � U f , P0ð Þ: (3)

It follows from the inequality (3) thatR

�

b

af serves as a lower bound for all

the upper Riemann sums U( f, P0), whichever partition P0 may be. It

follows, then, that the greatest lower bound of the U( f, P0), the upper

integralR

�b

af , must satisfy the inequality

Z

�

b

a

f �Z

�b

a

f : &

Remark

It follows, from the definition of the integral (when it exists) asZ b

a

f ¼ supP

L f , Pð Þ ¼ infP

U f , Pð Þ;

that, for any partition P of [a, b]

L f ;Pð Þ �Z b

a

f � U f ;Pð Þ:

Now, we have already seen that, if f(x)¼ x, x2 [0, 1], and

Pn ¼ 0; 1n

�

; 1n; 2

n

�

; . . . ; 1� 1n; 1

��

is a standard partition of [0, 1], then

limn!1

L f , Pnð Þ ¼ 1

2and lim

n!1U f , Pnð Þ ¼ 1

2: (4)

Since f is bounded on [0, 1], we know that the lower integralR

�1

0f exists. But

L f , Pnð Þ �R

�1

0f , so that, by the Limit Inequality Rule for sequences, it follows

from (4) that

1

2�Z

�

1

0

f : (5)

Similarly, since f is bounded on [0, 1], we know that the upper integralR

�1

0f

exists. ButR

�1

0f � U f , Pnð Þ, so that, by the Limit Inequality Rule for

sequences, it follows from (4) that

Z

�1

0

f � 1

2: (6)

Example 2, above.


We then observe thatZ

�1

0

f � 1

2�Z

�

1

0

f , in light of (5) and (6), and

Z

�

1

0

f �Z

�1

0

f , by part (b) of Theorem 4.

It follows that we must haveR

�1

0f ¼

R

�1

0f ¼ 1

2, so that f is integrable on [0, 1]

andR 1

0f ¼ 1

2.

Problem 4 Use the result of Problem 2 to prove that the function

f(x)¼ x2 is integrable on [0, 1], and evaluateR 1

0f .

Problem 5 Let f be the constant function x 7! k on [0, 1], and

Pn ¼ 0; 1n

�

; 1n; 2

n

�

; . . . ; 1� 1n; 1

��

a standard partition of [0, 1]. Cal-

culate the Riemann sums L( f, Pn) and U( f, Pn). Hence prove that f is

integrable on [0, 1], and evaluateR 1

0f .

However, not all bounded functions defined on closed intervals are integrable!

Example 3 Prove that the function

f xð Þ ¼ 1; 0 � x � 1; x rational,

0; 0 � x � 1; x irrational,

is not integrable on [0, 1].

Solution Let P¼ {[x0, x1], [x1, x2], . . . , [xi�1, xi], . . . , [xn�1, xn]} be any

partition of [0, 1].

Then

L f , Pð Þ ¼X

n

i¼1

mi�xi ¼X

n

i¼1

0� �xi

¼ 0;and

U f , Pð Þ ¼X

n

i¼1

Mi�xi ¼X

n

i¼1

1� �xi

¼X

n

i¼1

�xi ¼ 1:

Since all the lower Riemann sums are 0, their least upper boundR

�1

0f is also

zero. Similarly, since all the upper Riemann sums are 1, their greatest lower

boundR

�1

0f is also 1.

SinceR

�1

0f 6¼

R

�1

0f , it follows that f is not integrable on [0, 1]. &

7.1.2 Criteria for integrability

It would be tedious to have to go through the ab initio discussion of integr-

ability on each occasion that we wished to determine whether a given function

was integrable on a given closed interval. Therefore we now meet three criteria

that we often use to avoid that process.

In other words,R 1

0xdx ¼ 1

2.

xi–1 xi

1

y

10 x

y = 1, x ∈

y = 0, x ∉

For, each subinterval [xi�1, xi]contains both rational andirrational points.

We suggest that at a firstreading you omit ALL theproofs in this sub-section butread the rest of the text.

266 7: Integration

The first criterion is of particular interest in that its statement says nothing

about the value of the integral itself: it mentions only the difference between

the upper and the lower Riemann sums.

Theorem 5 Riemann’s Criterion for integrability

Let f be a bounded function on an interval [a, b]. Then

f is integrable on a; b½

if and only if:

for each positive number ", there is a partition P of [a, b] for which

U f ;Pð Þ � L f ;Pð Þ < ": (7)

Proof Suppose, first, that f is integrable on [a, b]. Then

supP

L f ;Pð Þ ¼ infP

U f ;Pð Þ; over all partitions P of a; b½ ;

denote by I the common value of these two quantities.

It follows that, for any given positive number ", there are partitions Q and Q0

of [a, b] for which

L f ;Qð Þ> I � 1

2" and U f ;Q0ð Þ I � 1

2";

U f ;Pð Þ � U f ;Q0ð Þ < I þ 1

2":

Hence, we may deduce from subtracting these inequalities that

U f ;Pð Þ � L f ;Pð Þ < I þ 1

2"

�

� I � 1

2"

�

¼ ":This is precisely the assertion (7).

Suppose next that the statement (7) holds; that is, that for each positive

number ", there is some partition P of [a, b] for which

U f ;Pð Þ � L f ;Pð Þ < ": (9)

SinceR

�b

af � U f ;Pð Þ and

R

�b

af � L f ;Pð Þ, whatever partition P is, it follows

from (9) that

Z

�b

a

f �Z

�

b

a

f � U f ;Pð Þ � L f ;Pð Þ

< ": (10)

Since the left-hand side of (10) is some non-negative number independent of ",

it follows thatR

�b

af �

R

�b

af ¼ 0: In other words, f is integrable on [a, b], as

required. &

y

area = U( f, P) – L( f, P)

y = f (x)

xba

Recall that L( f, P)�U( f, P),for any partition P.

That is, that refining apartition increases L anddecreases U.

By (9).

Problem 6 Use Riemann’s Criterion to determine the integrability of

the following functions on [0, 1]:

(a) f xð Þ ¼ �2; 0 � x < 1,

3; x ¼ 1;

(b) f xð Þ ¼ 1; 0 � x � 1, x rational,

0; 0 � x � 1, x irrational.

Our next criterion for integrability is phrased in terms of a sequence of

partitions and its statement involves a value for the integral.

Theorem 6 Common Limit Criterion

Let f be a bounded function on an interval [a, b].

(a) If f is integrable on [a, b], then there is a sequence {Pn} of partitions of

[a, b] such that

limn!1

L f ;Pnð Þ ¼Z b

a

f and limn!1

U f ;Pnð Þ ¼Z b

a

f : (11)

(b) If there is a sequence {Pn} of partitions of [a, b] such that

limn!1


U f ;Pnð Þ both exist and are equal; (12)

then f is integrable on [a, b] and the common value of these two limits

isR b

af .

Proof

(a) Suppose first that f is integrable on [a, b], and let I denoteR b

af .

Then, corresponding to each integer n� 1, there exist some partitions

of [a, b], Qn and Qn0 say, for which

L f ;Qnð Þ > I � 1

nand U f ;Q0n

� �

< I þ 1

n: (13)

Now let Pn be the common refinement of Qn and Qn0. It follows that

I � 1

n< L f ;Qnð Þ ðby ð13ÞÞ

� L f ;Pnð Þ ðsince Pn is a refinement of QnÞ� U f ;Pnð Þ� U f ;Q0n

� �

ðsince Pn is a refinement of Q0nÞ

< I þ 1

n: ðby ð13ÞÞ

Since the sequences I � 1n

� �

and I þ 1n

� �

both converge to I as n!1,

it follows from the Squeeze Rule for sequences and the above inequal-

ities that

L f ;Pnð Þ ! I and U f ;Pnð Þ ! I as n!1:This is precisely the statement (11).

(b) Suppose next that the statement (12) holds. Let I denote the common value

of the two limits.

268 7: Integration

It follows from (12) that, for any given positive number ", there are

partitions Q and Q0 of [a, b] for which

L f ;Qð Þ> I � 1

2" and U f ;Q0ð Þ I � 1

2";

U f ;Pð Þ � U f ;Q0ð Þ < I þ 1

2":

Hence, we may deduce from subtracting these inequalities that

U f ;Pð Þ � L f ;Pð Þ < I þ 1

2"

�

� I � 1

2"

�

¼ ":It follows from Theorem 5 that the function f is therefore integrable on [a, b].

Finally, we use the sequence {Pn} of partitions mentioned in the state-

ment of part (b). For them, if we let n!1 in the inequality

L f ;Pnð Þ �Z b

a

f � U f ;Pnð Þ;

and use the assumption (12), it follows that limn!1

L f ;Pnð Þ ¼R b

af and

limn!1

U f ;Pnð Þ ¼R b

af . This concludes the proof. &

Now the statement of Theorem 6 can be loosely paraphrased as saying that a

function f is integrable on [a, b] if and only if there is a sequence of partitions

{Pn} such that the corresponding lower and upper Riemann sums tend to a

common value. This gives us no information on what sort of sequences we

should examine to verify that f is integrable. Our next result says that we only

need to examine any sequence of partitions that we please whose mesh tends to

zero. For instance, a function f is integrable on [a, b] if and only if the lower and

upper Riemann sums for the sequence of standard partitions {Pn} of [a, b] tend

to a common value.

Theorem 7 Null Partitions Criterion

Let f be a bounded function on an interval [a, b], and {Pn} any sequence of

partitions of [a, b] with jjPnjj! 0.

(a) If f is integrable on [a, b], then

limn!1

L f ;Pnð Þ ¼Z b

a

f and limn!1

U f ;Pnð Þ ¼Z b

a

f :

(b) If there is a number I such that limn!1


U f ;Pnð Þ both

exist and equal I, then f is integrable on [a, b] and I ¼R b

af :

Part (b) is simply a special case of part (b) of Theorem 6, for it is true even

without the additional assumption that jjPnjj! 0.

Part (a) must be less straight-forward, for, if we drop the assumption

that jjPnjj! 0, then it is may not be the case that limn!1

L f ;Pnð Þ ¼R b

af or

limn!1

U f ;Pnð Þ ¼R b

af . To see this, consider the function f(x)¼ x, x2 [0, 1].

We have just proved that f isintegrable on [a, b]. Here wedetermine the value of itsintegral over [a, b].

This is a wonderfulsimplication!

Often the partitions Pn will bestandard partitions of [a, b].

For inf0;1½

f ¼ 0 and sup0;1½

f ¼ 1.

There is an analogous resultfor lower Riemann sums, butwe do not need it.


xi–1 xi

a b

xi–1 xi

a b

c

P

Q

Lemma 1 (in Sub-section 7.1.1)said that, for intervals I and Jwhere I� J

infx2J

f � infx2I

f and

supx2I

f � supx2J

f :

For 0< xi� xi�1� �.

We have already seen thatR 1

0f ¼ 1

2. However, if we choose for each partition

Pn simply the trivial partition Pn¼ {[0,1]}, then L( f, Pn)¼ 0 and U( f, Pn)¼ 1

for each n.

We have not used a condition such as jjPnjj! 0 before, so we need to

establish the following preliminary result before we can tackle the proof of

part (a) of Theorem 7.

Lemma 3 Let f be a bounded function on an interval [a, b], and let � be

any positive number. Let P be a partition of [a, b] with jjPjj<�, and P0 a

partition of [a, b] with the same partition points as P together with at most N

additional partition points. Then

U f ;P0ð Þ � U f ;Pð Þ � N M � mð Þ�;where m ¼ inf

x2 a;b½ f xð Þ and M ¼ sup

x2 a;b½ f xð Þ.

Remark

We already know, since P0 is a refinement of P, that U( f, P0)�U( f, P).

Lemma 3 gives us some lower bound to how much smaller U( f, P0) can be

than U( f, P).

Proof Let c be the first partition point of P0 that is not a partition point of P.

Then, if P¼ {[x0, x1], [x1, x2], . . . , [xn�1, xn]}, there is some integer i for which

xi�1< c< xi. Denote by Q the partition of [a, b] whose partition points are

those of P together with the additional point c.

Now the upper Riemann sums U( f, P) and U( f, Q) are the same, apart from

the contributions

supx2 xi�1;xi½

f � xi � xi�1ð Þ to U f ;Pð Þ

and

supx2 xi�1; c½

f � c� xi�1ð Þ þ supx2 c; xi½

f � xi � cð Þ to U f ;Qð Þ:

Now

supx2 xi�1;xi½

f � M; by Lemma 1;

and

supx2 xi�1; c½

f � m and supx2 c; xi½

f � m;

since m is a lower bound for f on all of a; b½ :It follows that

U f ;Qð Þ � U f ;Pð Þ ¼ supx2 xi�1; c½

f � c� xi�1ð Þ þ supx2 c; xi½

f � xi � cð Þ

� supx2 xi�1; xi½

f � xi � xi�1ð Þ

� m c� xi�1ð Þ þ m xi � cð Þ �M xi � xi�1ð Þ¼ m xi � xi�1ð Þ �M xi � xi�1ð Þ¼ � M � mð Þ xi � xi�1ð Þ� � M � mð Þ�;

270 7: Integration

so that

U f ;Qð Þ � U f ;Pð Þ � M � mð Þ�:We can interpret this inequality as follows: the insertion of one additional point

into the partition P (to obtain a new partition, Q) does not decrease the upper

Riemann sum by more than (M�m)�.We can repeat this insertion process up to a further N� 1 times to end up

with the final partition P0; the effect of this is that the upper Riemann sum of P

does not decrease by more than N(M�m)�. In other words

U f ;P0ð Þ � U f ;Pð Þ � N M � mð Þ�: &

We now use this result to prove part (a) of Theorem 7.

Theorem 7, part (a) Let f be a bounded function on an interval [a, b], and

{Pn} any sequence of partitions of [a, b] with jjPnjj! 0. If f is integrable on

[a, b], then

limn!1

L f ;Pnð Þ ¼Z b

a

f and limn!1

U f ;Pnð Þ ¼Z b

a

f :

Proof We are considering a bounded function f integrable on an interval

[a,b], and {Pn} any sequence of partitions of [a, b] with jjPnjj! 0. We will use

Lemma 3 to prove that

limn!1

U f ;Pnð Þ ¼Z b

a

f :

(The proof that limn!1

L f ;Pnð Þ ¼R b

af is similar, so we omit it.)

First, note that we can assume that f is non-constant on [a, b], since the result

is clearly immediately true in that situation.

Denote by I the value of the integralR b

af , and let m¼ inf f(x) and M¼

sup f(x).

For each n� 1, we can choose a partition Qn of [a, b] for which

U f ;Qnð Þ < I þ 1

n:

Let Nn denote the number of partition points in Qn, and let

� ¼ 1

Nnn M � mð Þ :

Next, choose any partition Pn of [a, b] for which jjPnjj<�.We then apply Lemma 3, with Pn in place of P, and with Pn

0 as the common

refinement of Pn and Qn in place of P0. It follows that

U f ;P0n� �

� U f ;Pnð Þ � Nn M � mð Þ�

¼ U f ;Pnð Þ � 1

n:

We can now obtain our crucial estimate for U( f, Pn) from this inequality, as

follows

Since f is integrable on [a, b].

Such a number � exist sinceM 6¼m, for we are assumingthat f is non-constant.

Notice that Pn0 has at most

Nn extra points as comparedwith Pn.

By Lemma 3.

From the definition of �.

U f ;Pnð Þ � U f ;P0n� �

þ 1

n

� U f ;Qnð Þ þ 1

n

< I þ 2

n:

We can therefore deduce that, for each n� 1

I � U f ;Pnð Þ< I þ 2

n:

It follows, by the Squeeze Rule for sequences, that U( f, Pn)! I as n!1.&

We shall use these various criteria in the next section.

7.2 Properties of integrals

In Section 7.1 you met the definition of integrability, and saw that the constant

function and the identity function were integrable on any interval [a, b]. But we

want to know that many more functions than that are integrable! In this section

we extend our family of integrable functions to a very large family indeed –

much wider than, for example, just the continuous functions.

7.2.1 Infimum and Supremum of functions (revisited)

In order to capitalise on the definition of integrability via lower and upper

Riemann sums, we need some further properties of the infimum and supremum

of a function. So, first, we remind you of their definitions.

Definitions Let ƒ be a function defined on an interval I�R . Then:

� A real number m is the greatest lower bound, or infimum, of ƒ on I if:

1. m is a lower bound of ƒ(I );

2. if m0>m, then m0 is not a lower bound of ƒ(I ).

� A real number M is the least upper bound, or supremum, of ƒ on I if:

1. M is an upper bound of ƒ(I );

2. if M0<M, then M0 is not an upper bound of ƒ(I ).

Remark Notice that, if m is a lower bound of ƒ on I, then infI

f � m;

and, if M is an upper bound of ƒ on I, then supI

f � M.

We will also sometimes need a strategy for verifying that a particular

number is the infimum or supremum of a function on a particular interval.

This is the last inequality,rewritten.

For Pn0 is a refinement of Qn.

By our choice of Qn.

You met these earlier, in Sub-section 1.4.2 and again inSection 7.1.

These are the definitions forthe least upper bound and thegreatest lower bound of afunction on an interval I; thereis a similar definition of thesebounds on a general set S in R .

272 7: Integration

When using these strategies,we shall often use thenumbers mþ " and M� " inplace of m0 and M0, to fit inwith our increased use of " asa positive number that can beas small as we please.

Lemma 1, Sub-section 7.1.1.

Loosely speaking, the largerinterval gives the functionmore space to get smaller andmore space to get larger.

Note that K must beindependent of the choiceof x and y in I.

For sup f xð Þ : x 2 If g issimply the least upper bound.(See also the Remark above.)

For inf f yð Þ:y 2 If g is simplythe greatest lower bound. (Seealso the Remark above.)

Strategies Let ƒ be a function defined on an interval I�R . Then:

� To show that m is the greatest lower bound, or infimum, of f on I, check that:

1. f (x)�m, for all x2 I;

2. if m0>m, then there is some x2 I such that f (x)<m0.

� To show that M is the least upper bound, or supremum, of f on I, check that:

1. f (x)�M, for all x2 I;

2. if M0<M, then there is some x2 I such that f (x)>M0.

Problem 1 Let f be a bounded function on an interval I. Prove that, for

any constant k

infx2I

k þ f xð Þf g ¼ k þ infx2I

f xð Þf g and

supx2I

k þ f xð Þf g ¼ k þ supx2I

f xð Þf g:

In fact you have already met one property of inf f and sup f in your study of

integration.

Lemma 1 For any bounded function f defined on intervals I and J where

I� J, we have

infx2J

f � infx2I

f and supx2I

f � supx2J

f :

A key tool in our work will be the following innocuous-looking result.

Lemma 2 Let f be a bounded function on an interval I. If, for some

number K, f (x)� f (y)�K for all x and y in I, then supI

f � infI

f � K.

Proof Since f (x)� f (y)�K for all x and y in I, we have

f xð Þ � K þ f yð Þ; for all x and y in I: (1)

It follows from (1) that, for any choice whatsoever of y in I, then Kþ f (y) serves

as an upper bound for the set of all possible values of f (x) for x in I. It follows that

sup f xð Þ : x 2 If g � K þ f yð Þ;which we may rewrite in the form

sup f xð Þ : x 2 If g � K � f yð Þ (2)

In a similar way, it follows from (2) that sup f xð Þ : x 2 If g � K serves as a

lower bound for the set of all possible values of f(y) for y in I. It follows that

sup f xð Þ : x 2 If g � K � inf f yð Þ :y 2 If g: (3)

We may rearrange the inequality (3) in the form sup f xð Þ : x 2 If g�inf f yð Þ : y 2 If g � K. This is precisely the required result, since the letters

x and y in this last inequality are simply ‘dummy variables’. &

Remarks

1. If f is bounded on I and f (x)� f (y)<K for all x and y in I, then it may not be

true that supI

f � infI

f <K. For example, if f is the identity function on

I¼ (0, 1), then

f xð Þ � f yð Þ < 1, for all x and y in ð0,1Þ, but sup0,1½

f � inf0,1½

f ¼ 1� 0 ¼ 1:

2. The conclusion of Lemma 2 also holds if we know that, for some number K,

j f (x)� f (y)j �K, for all x and y in I, since

f xð Þ � f yð Þ � f xð Þ � f yð Þj j:

In some situations, the following result related to Lemma 2 is also useful.

Lemma 3 Let f be a bounded function on an interval I. Then

supx; y2I

f xð Þ � f yð Þf g ¼ supI

f � infI

f :

Proof We use the strategy for supremum mentioned earlier.

Let x and y be any numbers in I. Then, in particular

f xð Þ � supI

f ; for all x 2 I ;

and, since f yð Þ � infI

f , for all y2 I

� f yð Þ ��infI

f ; for all y 2 I:

If we add these inequalities, we obtain

f xð Þ � f yð Þ � supI

f � infI

f ; for all x; y 2 I;

so that

supx;y2I

f xð Þ � f yð Þf g� supI

f � infI

f :

To prove the desired result, we now need to prove that

for each positive number ", there are X and Y in I for which

f Xð Þ � f Yð Þ >�

supI

f � infI

f�

� ": (4)

Now, by the definition of infimum and supremum on I, we know that, since12" > 0, there exist X and Y in I such that

f Xð Þ> supI

f � 1

2" and f Yð Þ < inf

If þ 1

2":

It follows from these two inequalities that

f Xð Þ � f Yð Þ >�

supI

f � 1

2"�

��

infI

f þ 1

2"�

¼ supI

f � infI

f � ":


Problem 2 Let f (x)¼ x2 on the interval I¼ (�2, 3]. Determine

infI

f; supI

f; infx;y2I

f xð Þ � f yð Þf g and supx; y2I

f xð Þ � f yð Þf g:

By the earlier Remark.

By the strategy for supremum.

274 7: Integration

Let f and g be bounded functions on an interval I. Then:

Sum Rule infI

f þ gð Þ� infI

f þ infI

g and supI

f þ gð Þ � supI

f þ supI

g;

Multiple Rule infI

lfð Þ ¼ l�

infI

f�

and supI

lfð Þ ¼ l supI

f

� �

; for l> 0;

Negative Rule infI�fð Þ ¼ �

�

supI

f�

and supI

�fð Þ ¼ ��

infI

f�

.

Proof We will prove only the first part of each Rule, for the proofs of the

second parts are similar.

Sum Rule

For any x and y in I

f xð Þ � infI

f and g xð Þ � infI

g;

so that, by adding these two inequalities, we obtain

f xð Þ þ g xð Þ� infI

f þ infI

g:

Since infI

f þ infI

g is a lower bound for f(x)þ g(x) on I, it follows that

infI

f þ gð Þ � infI

f þ infI

g:

Multiple Rule

For any x in I, f xð Þ � infI

f so that

lf xð Þ � l infI

f ; for all x 2 I:

Thus

infx2I

lf xð Þð Þ � l infI

f :

To prove that infx2I

lf xð Þð Þ ¼ l�

infI

f�

, we now need to prove that:

for each positive number ", there is some X in I for which

lf Xð Þ< l�

infI

f�

þ ": (5)

Now, since "> 0 and l> 0, we have "l > 0. It follows, from the definition of

infimum of f on I, that there exists an X in I for which

f Xð Þ< infI

f þ "l:

Multiplying both sides by the positive number l, we obtain the desired

result (5).

Negative Rule

For any x in I, f xð Þ � supI

f so that

� f xð Þ� � supI

f; for all x 2 I :

Thus

infx2I�f xð Þð Þ� �sup

I

f :

Since l> 0.

To prove that infx2I�f xð Þð Þ ¼ �sup

I

f , we now need to prove that:


� f Xð Þ<� supI

f þ ": (6)

Now, since "> 0 it follows, from the definition of supremum of f on I, that

there exists an X in I for which

f Xð Þ > supI

f � "so that

� f Xð Þ<�supI

f þ ":

This is precisely the inequality (6). &

Problem 3 Prove that if f is a bounded function on an interval I, then:

(a) supI

lfð Þ ¼ l supI

f ; for l > 0; (b) supI

�fð Þ ¼ �infI

f .

We will now use these results on greatest lower bounds and least upper bounds

to address the main topic of this section: integration!

7.2.2 Monotonic and continuous functions

We now prove that bounded functions on an interval [a, b] are integrable if they

are either monotonic on [a, b] or continuous on [a, b]. This provides us with

very many integrable functions!

Neither of these two classes of functions includes the other. For example, the

function

f xð Þ ¼ x� x2; x 2 0; 1½ ;is continuous on [0,1] but is not monotonic on [0, 1]; while the function

f xð Þ ¼ x; 0 � x < 1;2; x ¼ 1;

; x 2 0; 1½ ;

is monotonic on [0,1] but is not continuous on [0, 1].

Theorem 2 Let f be a bounded function on an interval [a, b]. If f is mon-

otonic on [a, b], then it is integrable on [a, b].

Proof We shall assume that f is increasing on [a, b]; if it is decreasing, the

proof is similar. If f is constant on [a, b], it is certainly integrable on [a, b]. So,

we shall assume, in addition, that f is also non-constant on [a, b]; it follows, in

particular, that f (a) 6¼ f (b).

We will prove that:


U f ;Pð Þ � L f ;Pð Þ < ":

It then follows from the Riemann Criterion for integrability that f is integrable

on [a, b].

For any given "> 0, let P be any partition of [a, b] with mesh jjPjj such that

Pk k < "f bð Þ�f að Þ. Then, with the usual notation for Riemann sums

You should sketch the graphsof these two functions toappreciate these statements.


276 7: Integration

U f ;Pð Þ � L f ;Pð Þ ¼X

n

i¼1

Mi � mið Þ�xi ¼X

n

i¼1

f xið Þ � f xi�1ð Þð Þ�xi

� Pk k �X

n

i¼1

f xið Þ � f xi�1ð Þð Þ

¼ Pk k � f bð Þ � f að Þð Þ< ":


Notice that the brevity of the above proof hides the fact that within the proof

we are using quite a lot of work needed to prove the Riemann Criterion!

Theorem 3 Let f be a bounded function on an interval [a, b]. If f is cont-

inuous on [a, b], then it is integrable on [a, b].

Proof We will prove that:


U f ;Pð Þ � L f ;Pð Þ < ":

It then follows from the Riemann Criterion for integrability that f is integrable

on [a, b].

Our key new tool here is the fact that, since f is continuous on an interval

[a, b], which is a closed interval, it is therefore uniformly continuous on [a, b].

It follows that, for any given " > 0; "2 b�að Þ > 0, so that there is a positive

number � for which

f xð Þ � f yð Þj j < "

2 b� að Þ ; for all x and y in a, b½ satisfying x� yj j < � : (7)

Now, let P ¼ xi�1, xi½ f gni¼1 be any partition of [a, b] with mesh jjPjj such

that jjPjj<�. Then, for each i we have, for all x and y in [xi�1, xi]

x� yj j � xi � xi�1

� Pk k < �:

It follows, from (7), that

f xð Þ � f yð Þj j < "

2 b� að Þ ; for all x and y in xi�1; xi½ ;

so that, by Remark 2 following Lemma 2

supx2 xi�1;xi½

f xð Þ � infx2 xi�1;xi½

f xð Þ � "

2 b� að Þ :

Then, with the usual notation for Riemann sums


n

i¼1

Mi � mið Þ�xi

� "

2 b� að Þ �X

n

i¼1

�xi

¼ "

2 b� að Þ � b� að Þ

¼ 1

2" < ":


Since f is increasing on [a, b].

You met uniform continuityin Section 5.5; this assertion isTheorem 2 there.

Here the choice of � dependsONLY on ", the same choicewhatever x and y may be. Wechoose "

2 b�að Þ rather than ", forconvenience later on.

In Sub-section 7.2.1. Recallthat the conclusion ofLemma 2 is a weak inequality,not a strong inequality.

7.2.3 Rules for integration

Our first task is to prove the Combination Rules for integrable functions.


Let f and g be integrable on [a, b]. Then so are:

Sum Rule fþ g, andR b

af þ gð Þ ¼

R b

af þ

R b

ag;

Multiple Rule lf, andR b

alfð Þ ¼ l

R b

af ;

Product Rule fg;

Modulus Rule j f j.

Proof We use the usual notation for Riemann sums and integrals.

Sum Rule

Let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as n!1.

Since f and g are integrable on [a, b], it follows, from the Null Partitions

Criterion for integrability, that

L f ;Pnð Þ and U f ;Pnð Þ !Z b

a

f as n!1;and

L g;Pnð Þ and U g;Pnð Þ !Z b

a

g as n!1:

Then

L f ;Pnð Þ þ L g;Pnð Þ ¼X

n

i¼1

infxi�1; xi½

f � �xi þX

n

i¼1

infxi�1; xi½

g� �xi

¼X

n

i¼1

infxi�1; xi½

f þ infxi�1; xi½

g

�

� �xi

�X

n

i¼1

infxi�1; xi½

f þ gð Þ � �xi ¼ L f þ g;Pnð Þð Þ

�X

n

i¼1

supxi�1; xi½

f þ gð Þ � �xi ¼ U f þ g;Pnð Þð Þ

�X

n

i¼1

supxi�1; xi½

f þ supxi�1; xi½

g

( )

� �xi

¼ U f ;Pnð Þ þ U g;Pnð Þ:

We now let n!1 in this last set of inequalities. Then, since

L f ;Pnð Þ þ L g;Pnð Þ !Z b

a

f þZ b

a

g and U f ;Pnð Þ þ U g;Pnð Þ !Z b

a

f þZ b

a

g ;

it follows, from the Limit Inequality Rule for sequences, that

L f þ g;Pnð Þ !Z b

a

f þZ b

a

g and U f þ g;Pnð Þ !Z b

a

f þZ b

a

g; as n!1:

Hence, by part (b) of the Null Partitions Criterion for integrability, we deduce

that fþ g is integrable on [a, b] andR b

af þ gð Þ ¼

R b

af þ

R b

ag:

You may omit the detailsof all the proofs in thissub-section at a first reading.

Pn ¼ f½xi�1; xigni¼1:


By the definition of lowerRiemann sum.

By the Combination Rules forinf and sup, Theorem 1,Sub-section 7.2.1.


By the Combination Rules forinf and sup, again.

278 7: Integration

Multiple Rule


Since f is integrable on [a, b], it follows from the Null Partitions Criterion for

integrability that

L f ;Pnð Þ and U f ;Pnð Þ !Z b

a

f as n!1:

Then, if l> 0

L l f ;Pnð Þ ¼X

n

i¼1

infxi�1; xi½

lfð Þ � �xi

¼ lX

n

i¼1

infxi�1; xi½

f � �xi

¼ lL f ;Pnð Þ ! lZ b

a

f as n!1;

while, if l< 0

L lf ;Pnð Þ ¼X

n

i¼1

infxi�1; xi½

lfð Þ � �xi

¼ lX

n

i¼1

supxi�1; xi½

f � �xi

¼ lU f ;Pnð Þ ! lZ b

a

f as n!1:

It follows that, for all real l, L lf ;Pnð Þ ! lR b

af as n!1.

A similar argument shows that, for all real l, U lf ;Pnð Þ ! lR b

af as n!1.

Since the limits of the two sequences of Riemann sums are equal, it follows

from part (b) of the Null Partitions Criterion for integrability, that lf is

integrable on [a, b] and

Z b

a

lfð Þ ¼ lZ b

a

f :

Product Rule

Since f and g are bounded on [a, b], there is some number M, say, such that

f xð Þj j � M and g xð Þj j � M; for all x 2 a; b½ :

Now let {Pn} be any sequence of partitions of [a, b], where jjPnjj! 0 as

n!1. It follows that, if x, y2 [xi�1, xi] for some i, then

f xð Þg xð Þ � f yð Þg yð Þj j ¼ f xð Þg xð Þ � f xð Þg yð Þf g þ f xð Þg yð Þ � f yð Þg yð Þf gj j¼ f xð Þ g xð Þ � g yð Þf g þ g yð Þ f xð Þ � f yð Þf gj j� M g xð Þ � g yð Þj j þM f xð Þ � f yð Þj j

� M supxi�1;xi½

g� infxi�1;xi½

g

( )

þM supxi�1;xi½

f � infxi�1;xi½

f

( )

;



By the Combination Rules forinf and sup, Theorem 1,Sub-section 7.2.1.

By the Combination Rules forinf and sup.

For, if l¼ 0, the result istrivial.

so that, by Lemma 2 in Sub-section 7.2.1 and Remark 2 following that Lemma

supxi�1;xi½

fgð Þ � infxi�1;xi½

fgð Þ � M supxi�1;xi½

g� infxi�1;xi½

g

( )

þM supxi�1;xi½


f

( )

: (8)

In terms of the Riemann sums for fg, it follows that

U fg;Pnð Þ�L fg;Pnð Þ ¼X

n

i¼1

supxi�1;xi½

fgð Þ� infxi�1;xi½

fgð Þ !

�xi

�MX

n

i¼1

supxi�1;xi½

g� infxi�1;xi½

g

( )

�xiþMX

n

i¼1

supxi�1;xi½


f

( )

�xi

¼M U g;Pnð Þ�L g;Pnð Þf gþM U f ;Pnð Þ�L f ;Pnð Þf g! 0 as n!1:

It then follows, from part (b) of the Null Partitions Criterion for integrability,

that fg is integrable on [a, b].

Modulus Rule


Then, if x, y2 [xi� 1, xi] for some i, we have�

� f xð Þj j � f yð Þj j�

� � f xð Þ � f yð Þj j;so that

�

� f xð Þj j � f yð Þj j�

� � supxi�1;xi½


f

and therefore

supxi�1;xi½

fj j � infxi�1;xi½

fj j � supxi�1;xi½


f : (9)

In terms of the Riemann sums for j f j, it follows that

U fj j;Pnð Þ � L fj j;Pnð Þ ¼X

n

i¼1

supxi�1;xi½

fj j � infxi�1;xi½

fj j !

�xi

�X

n

i¼1

supxi�1;xi½


f

!

�xi

¼ U f ;Pnð Þ � L f ;Pnð Þ

! 0 as n!1:

It then follows, from part (b) of the Null Partitions Criterion for integrability,

that jf j is integrable on [a, b]. &

Our next result is often overlooked as obvious – which it is not!

Theorem 5 The Sub-interval Theorem

(a) Let f be integrable on [a, b], and let a< c< b. Then f is integrable on

both [a, c] and [c, b], andR b

af ¼

R c

af þ

R b

cf :

(b) Let f be integrable on [a, c] and [c, b], where a< c< b. Then f is

integrable on [a, b].

By definition of U and L.

By (8).

By part (a) of the NullPartitions Criterion forintegrability (Theorem 7,Sub-section 7.1.2), since f andg are integrable on [a, b].

Pn ¼ xi�1; xi½ f gni¼1:

This follows from the ‘reverseform’ of the TriangleInequality, that you met inSub-section 1.3.1.

By Remark 2 followingLemma 2, Sub-section 7.2.1.

By definition of U and L.

By (9).

By part (a) of the NullPartitions Criterion forintegrability (Theorem 7,Sub-section 7.1.2), since f isintegrable on [a, b].

a c b

280 7: Integration

Proof

(a) Let {Pn}, where Pn ¼ xi�1, xi½ f gni¼1, be any sequence of partitions of [a, b]

such that c is a partition point of each Pn, for n� 2, and such that jjPnjj! 0

as n!1.

For n� 2, let Qn consist of those subintervals in Pn that lie in [a, c], and

Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a

partition of [a, c] with jjQnjj! 0 as n!1, and {Qn0} is a partition of [c, b]

with jjQn0jj! 0 as n!1.

Now, with the usual notation for Riemann sums, we have

U f ;Qnð Þ � L f ;Qnð Þ ¼ the sum of those terms inP

n

i¼1

Mi � mið Þ�xi

that correspond to subintervals xi�1; xi½ that lie in a; c½

�X

n

i¼1

Mi � mið Þ�xi

¼ U f ;Pnð Þ � L f ;Pnð Þ ! 0 as n!1:It follows from the Null Partitions Criterion that f is integrable on [a, c].

A similar argument proves that f is integrable on [c, b].

In particular, it follows that

L f ;Qnð Þ !Z c

a

f and L f ;Q0n� �

!Z b

c

f as n!1:

So, if we let n!1 in the identity

L f ;Pnð Þ ¼ L f ;Qnð Þ þ L f ;Q0n� �

;

we deduce thatZ b

a

f ¼Z c

a

f þZ b

c

f :

(b) In the opposite direction, suppose that f is integrable on [a, c] and [c, b].

Let {Pn}, where Pn ¼ xi�1; xi½ f gni¼1, be any sequence of partitions of

[a, b] such that c is a partition point of each Pn, for n� 2, and jjPnjj! 0 as

n!1.

For n� 2, let Qn consist of those subintervals in Pn that lie in [a, c], and

Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a

partition of [a, c] with jjQnjj! 0 as n!1, and {Qn0} is a partition of [c, b]

with jjQn0jj! 0 as n!1.

Then, with the usual notation for Riemann sums, we have

U f ;Pnð Þ � L f ;Pnð Þ ¼ the sum of those terms inP

n

i¼1

Mi � mið Þ�xi that

correspond to subintervals xi�1; xi½ that lie in a; c½

þ the sum of those terms inX

n

i¼1

Mi � mið Þ�xi that

correspond to subintervals xi�1; xi½ that lie in c; b½ ¼ U f ;Qnð Þ � L f ;Qnð Þf g þ U f ;Q0n

� �

� L f ;Q0n� ��

! 0 as n!1:

It follows, from the Null Partitions Criterion, that f is integrable on

[a, b]. &

This identity follows from thedefinitions of Pn, Qn and Qn

0.

By the definition of integral.


In light of Theorem 5 we now introduce some notational conventions.

Conventions Let f be integrable on [a, b], where a< b. Then we make the

following definitionsZ a

a

f ¼ 0 and

Z a

b

f ¼ �Z b

a

f :

With these conventions, we can assert that if f is integrable on any interval that

contains the points a, b and c, thenR b

af ¼

R c

af þ

R b

cf .

7.3 Fundamental Theorem of Calculus

7.3.1 Fundamental Theorem of Calculus

It would be very tedious if, each time that we wished to evaluate an integral, we

had to calculate upper and lower sums and find their greatest lower bound and

least upper bound, respectively. Fortunately, this is usually unnecessary, as

there is a short-cut using the idea of a primitive.

Definition Let f be a function defined on an interval I. Then the function F

is a primitive of f on I if F is differentiable on I and

F0 ¼ f :

For example, if

f ðxÞ ¼ 1

4� x2; x 2 ½�1; 1;

then

FðxÞ ¼ 1

4loge

2þ x

2� x

� �

; x 2 ½�1; 1;

is a primitive of f on [�1, 1], since

F0ðxÞ ¼ 1

4� 2� x

2þ x� ð2� xÞ þ ð2þ xÞ

ð2� xÞ2

¼ 1

4� 4

ð2þ xÞð2� xÞ ¼1

4� x2:

Problem 1

(a) Prove that the function f ðxÞ ¼ 1ffiffiffiffiffiffiffiffi

x2�4p , x2 (2, 1), has a primitive

F xð Þ ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 4p� �

.

(b) Prove that the function f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

4� x2p

, x2 (�2, 2), has a primitive

F xð Þ ¼ 12

xffiffiffiffiffiffiffiffiffiffiffiffiffi

4� x2p

þ 2 sin�1 x2

� �

.

The connection between primitives of a function f and the integral of f on an

interval I is given by the following result, which is one of the most important

theorems in Analysis.

These conventions applywhere the upper and lowerlimits of the integral are thesame, or are in reverse orderon the real line.

To prove this assertion, weneed to look at all the possibleorderings of the three pointson R , which is rather tedious;so we omit it!

Sometimes we denote aprimitive of f by

R

f , and wedenote F(x) by

R

f ðxÞdx:

282 7: Integration

Theorem 1 Fundamental Theorem of Calculus

Let f be integrable on [a, b], and F a primitive of f on [a, b]. ThenZ b

a

f ¼ F bð Þ � F að Þ:

Proof Since f is integrable on [a, b], there is a sequence

Pn ¼ x0; x1½ ; x1; x2½ ; . . .; xn�1; xn½ f gof partitions of [a, b], with jjPnjj! 0 as n!1, such that

limn!1

L f ;Pnð Þ ¼Z b

a

f and limn!1

U f ;Pnð Þ ¼Z b

a

f : (1)

Now, the function F satisfies the conditions of the Mean Value Theorem on

each subinterval [xi�1, xi], for i¼ 1, 2, . . ., n. It follows that there exists some

point ci2 (xi�1, xi) such that

F xið Þ � F xi�1ð Þ ¼ F0 cið Þ xi � xi�1ð Þ¼ f cið Þ�xi: (2)

Next, since mi� f(ci)�Mi, for i¼ 1, 2 . . ., n, it follows that

X

n

i¼1

mi�xi �X

n

i¼1

f cið Þ�xi �X

n

i¼1

Mi�xi: (3)

We may now apply (2) to the general term in the sumP

n

i¼1

f cið Þ�xi; thus it

follows from (3) that

L f ;Pnð Þ �X

n

i¼1

F xið Þ � F xi�1ð Þf g � U f ;Pnð Þ:

Since the middle term is a ‘telescoping’ sum, we deduce that

L f ;Pnð Þ � F bð Þ � F að Þ � U f ;Pnð Þ: (4)

Now let n!1 in (4), using the facts in (1); this thus givesZ b

a

f � F bð Þ � F að Þ �Z b

a

f :

In other words, FðbÞ � FðaÞ ¼R b

af ; as required. &

Example 1 EvaluateR 1

02xdx.

Solution The function f(x)¼ 2x is continuous on [0, 1] and so is integrable

on [0, 1]; from the Table of Standard Primitives, it has a primitive

F xð Þ ¼ 2x

loge 2on ½0; 1:

It follows from the Fundamental Theorem of Calculus that

Z 1

0

2xdx ¼ 2x

loge 2

� 1

0

¼ 1

loge 2: &

Problem 2 Using the Table of Standard Primitives, evaluate the

following integrals:

(a)R 4

0x2 þ 9ð Þ

12dx; (b)

R e

1loge x dx:

Often F(b)�F(a) is writtenas

F xð Þ½ ba or F xð Þjba:

We use the standard notationfor lower and upper sums andpartitions.

For F 0 ¼ f and �xi¼ xi� xi�1.

Recall that

mi ¼ inf f xð Þ : x 2 xi�1; xi½ f g;Mi ¼ sup f xð Þ : x 2 xi�1; xi½ f g:

By the Limit Inequality Rulefor sequences.

The Table of StandardPrimitives appears inAppendix 2.


7.3.2 Finding primitives

The Fundamental Theorem of Calculus asserts that, if f is integrable on [a, b],

and F a primitive of f on [a, b], thenR b

af ¼ F bð Þ � F að Þ. This does not make it

clear whether a primitive of f is unique.

In fact, for any given function f, a primitive of f is NOT unique. For example,

the functionsx 7! sin�1 x and x 7! �cos�1 x; x 2 �1; 1ð Þ;

are both primitives of the function x 7! 1ffiffiffiffiffiffiffiffi

1�x2p ; so that, for instance

Z 1ffiffi

2p

0


1� x2p dx ¼ sin�1 x

�

1ffiffi

2p

0 ¼ � cos�1 x �

1ffiffi

2p

0

¼ 1

4p:

However, two primitives of a given function f are related; they can only

differ by a constant function.

Theorem 2 Uniqueness Theorem for Primitives

Let F1 and F2 be primitives of a function f on an interval I. Then there exists

some constant c such that

F2 xð Þ ¼ F1 xð Þ þ c; for x 2 I:

Proof Since F1 and F2 are primitives of f on I

F01 xð Þ ¼ f xð Þ and F02 xð Þ ¼ f xð Þ; for x 2 I;

so that

F02 xð Þ ¼ F01 xð Þ; for x 2 I:

It follows that there exists some constant c such that

F2 xð Þ ¼ F1 xð Þ þ c; for x 2 I: &

Our stock of primitives can be considerably extended by use of the following

Combination Rules. For convenience, we include here an extra rule which

applies to a composite function f �g for which the function g is just an x-scaling.


Let F and G be primitives of f and g, respectively, on an interval I, and l2R .

Then, on I:

Sum Rule fþ g has primitive FþG;

Multiple Rule lf has primitive lF;

Scaling Rule x j! f(lx) has primitive x 7! 1l F lxð Þ.

For example, it follows from the Table of Standard Primitives and the

Combination Rules that the function x 7!3 sinh 2xð Þ þ 1x, x2 (0, 1), has a

primitive x 7! 32

cosh 2xð Þ þ loge x.

Problem 3 Using the Table of Standard Primitives and the Combina-

tion Rules, find a primitive of each of the following functions:

(a) f xð Þ ¼ 4 loge x� 24þx2 ; x 2 0;1ð Þ;

(b) f xð Þ ¼ 2 tan 3xð Þ þ e2x sin x; x 2 � 16p; 1

6p

� �

:

For example

sin�1 x� �cos�1 x� �

¼ 1

2p:

By Corollary 2, Sub-section 6.4.2.

These rules are easily provedusing the corresponding rulesfor derivatives.

In applications of Theorem 3we do not usually bother tomention the theoremexplicitly.

284 7: Integration

7.3.3 Techniques of integration

We are now in a position to use the Fundamental Theorem of Calculus to give

rigorous proofs of some standard techniques for integration.

Theorem 4 Integration by Parts

If f and g are differentiable on an interval [a, b], and f 0 and g0 are continuous

on [a, b], thenZ b

a

fg0 ¼ fg½ ba �Z b

a

f 0g:

Proof We may reformulate the Product Rule for derivatives in the form

fgð Þ0 ¼ f 0gþ fg0;

in other words, that fg is a primitive on [a,b] of f 0gþ fg0.It follows from the Fundamental Theorem of Calculus that

Z b

a

f 0gþ fg0ð Þ ¼ fg½ ba;

so thatR b

afg0 ¼ fg½ ba�

R b

af 0g, as required. &

Strategy to evaluate an integral using integration by parts

1. Write the original function in the form fg0, where f is a function that you

can differentiate and g0 a function that you can integrate.

2. Use the formulaR b

afg0 ¼ fg½ ba�

R b

af 0g:

Example 2 Evaluate the integralR 1

0tan�1 xdx:

Solution Here we use a common trick: we consider g0(x) to be the factor 1,

and use integration by parts. ThusZ 1

0

tan�1 xdx ¼Z 1

0

1� tan�1 xdx

¼ x tan�1 x �1

0�Z 1

0

xdx

1þ x2

¼ tan�1 1� 1

2loge 1þ x2

� �

� 1

0

¼ 1

4p� 1

2loge 2þ 1

2loge 1 ¼ 1

4p� 1

2loge 2: &

Problem 4

(a) Find a primitive of the function f xð Þ ¼ x13 loge x; x 2 0;1ð Þ:

(b) Evaluate the integralR p

2

0x2 cos xdx:

Hint: Use integration by parts twice.

You will meet further instances of integration by parts in the next section.


Recall the notation that

fg½ ba¼ f bð Þg bð Þ � f að Þg að Þ:

There is a similar strategy forfinding a primitive byintegration by parts.

Here

f xð Þ ¼ tan�1 x; g0 xð Þ ¼ 1;

so that

f 0 xð Þ ¼ 1

1þ x2; g xð Þ ¼ x:


Note that, in this situation, forg to possess an inversefunction, g must be strictlyincreasing.

There is an analogous result inthe situation that g is strictlydecreasing.


Recall that, by hypothesis

g �; �½ ð Þ � a; b½ :

Next we look at the method of integration by substitution, and discover that

one needs to be just a little careful in making substitutions.

Theorem 5 Integration by Substitution

If f is continuous on [a, b], g differentiable on [�, �], g0 continuous on [�, �],

and g([�, �]) � [a, b], thenZ g �ð Þ

g �ð Þf xð Þdx ¼

Z �

�

f g tð Þð Þg0 tð Þdt:

If, in addition, g(�)¼ a, g(�)¼ b and g possesses an inverse function g�1 on

[a, b], thenZ b

a

f xð Þdx ¼Z g�1 bð Þ

g�1 að Þf g tð Þð Þg0 tð Þdt:

Proof Let F be a primitive of f on [a, b], and define the function h by

h tð Þ ¼ F g tð Þð Þ; t 2 �; �½ : (5)

By the Composition Rule for derivatives, applied to the composite

h(t)¼F(g(t)), we have that

h0 tð Þ ¼ F0 g tð Þð Þg0 tð Þ: (6)

By the Fundamental Theorem of Calculus applied to h on [�, �], we haveZ �

�

h0 tð Þdt ¼ h �ð Þ � h �ð Þ: (7)

If we now substitute for h and h0 from (5) and (6) into the two sides of equation

(7), we obtainZ �

�

F0 g tð Þð Þg0 tð Þdt ¼ F g �ð Þð Þ � F g �ð Þð Þ: (8)

Since F is a primitive of f on [a, b], and so a primitive of f on g([�, �]), we can

rewrite the left-hand side of equation (8) asR �

� f g tð Þð Þg0 tð Þdt:

Further, since F is a primitive of f on [a, b], and so a primitive of f on

g([�, �]), we can apply the Fundamental Theorem of Calculus to the function F

on g([�, �]) to express the right-hand side of equation (8) as

F g �ð Þð Þ � F g �ð Þð Þ ¼Z g �ð Þ

g �ð ÞF0 xð Þdx

¼Z g �ð Þ

g �ð Þf xð Þdx:

So, if we make these substitutions on both sides of equation (8), we deduce thatZ �

�

f g tð Þð Þg0 tð Þdt ¼Z g �ð Þ

g �ð Þf xð Þdx: (9)

This is the first part of the theorem.

Suppose next that g possesses an inverse function g�1 on [�, �], and

a ¼ g �ð Þ and b ¼ g �ð Þ;then

� ¼ g�1 að Þ and � ¼ g�1 bð Þ:Making these substitutions into equation (9), we obtain the second part of the

theorem. &

286 7: Integration

There is a similar strategy forfinding a primitive byintegration by substitution.

Here

when t ¼ 0; x ¼ 1;

when t ¼ 32; x ¼ 4:

Also, t ¼ 12

x� 1ð Þ.

For, if we make thesubstitution x¼ t2, then

t ¼ �1) x ¼ 1;

t ¼ 2) x ¼ 4:

Strategy to evaluate an integral �b

a f ððgððtÞÞÞÞg0ððtÞÞdt using integration by

substitution:

1. Choose a function x¼ g(t) such that dxdt¼ g0 tð Þ, and express dt in terms of

x and dx.

2. Make the necessary substitutions to give an integral in terms of x and dx.

3. Calculate this integral.

Example 3 EvaluateR 3

2

0tþ1

2tþ1ð Þ12

dt,

Solution Let x¼ g(t)¼ 2tþ 1, t2R . The function g is one–one on R .

Then dxdt¼ 2, so that dx¼ 2dt. Making the various substitutions into the

integral, we get

Z 32

0

t þ 1

2t þ 1ð Þ12

dt ¼Z 4

1

12

x� 1ð Þ þ 1

x12

1

2dx

¼Z 4

1

12

xþ 1ð Þx

12

1

2dx

¼ 1

4

Z 4

1

x12 þ x�

12

� �

dx

¼ 1

4

2

3x

32 þ 2x

12

� �4

1

¼ 1

4

16

3þ 4

� �

� 1

4

2

3þ 2

� �

¼ 5

3: &

Problem 5

(a) Find a primitive of the function f (x)¼ sin(2 sin 3x) cos 3x, x2R .

(b) Evaluate the integralR 1

0ex

1þexð Þ2 dx.

Remark

If you use the second assertion of Theorem 5, you must verify that the function

g does have an inverse; you may end up with a contradiction if you simply

calculate thoughtlessly.

For example, take�¼�1, �¼ 2, g(t)¼ t2, and f xð Þ ¼ x12; then a¼ g(�1)¼ 1

and b¼ g(2)¼ 4. ThenZ b

a

f xð Þdx ¼Z 4

1

x12dx ¼ 2

3x

32

� �4

1

¼ 16

3

� �

� 2

3

� �

¼ 14

3;

butZ 2

�1

f g tð Þð Þg0 tð Þdt ¼Z 2

�1

t2�

12 2tdt ¼

Z 2

�1

2t2dt

¼ 2

3t3

� �2

�1

¼ 16

3

� �

� � 2

3

� �

¼ 6:

This contradiction arises since the function g does NOT have an inverse on [1, 4]

that maps [1, 4] to [�1, 2].

Our second substitution technique applies when we let t¼ g(x), where the

function g has an inverse, so that x¼ g�1(t), and we express x in terms of t, and

dx in terms of t and dt.


There is a similar strategy forfinding a primitive byintegration by substitution.

Here

when x ¼ 0; t ¼ 0;when x ¼ loge 5; t ¼ 2:

One of the goals of NumericalAnalysis is to obtain goodestimates for integrals.

By the Null PartitionsCriterion, Theorem 7,Sub-section 7.1.2.

Strategy to evaluate an integral �b

a fððxÞÞdx using integration bysubstitution

1. Choose a function t¼ g(x), where g has an inverse, so that x¼ g�1(t);

express dx in terms of t and dt.

2. Make the necessary substitutions to give an integral in terms of t and dt.

3. Calculate this integral.

Example 4 EvaluateR loge 5

0e2x

ex�1ð Þ12

dx.

Solution Let t ¼ g xð Þ ¼ ex � 1ð Þ12, x2 [0, 1). The function g is one–one

on [0,1).

Then t2¼ ex� 1, so that ex¼ t2þ 1 and x¼ loge(t2þ 1). It follows that

dx

dt¼ 2t

t2 þ 1; so that dx ¼ 2t

t2 þ 1dt:

HenceZ loge 5

0

e2x

ex � 1ð Þ12

dx ¼Z 2

0

t2 þ 1ð Þ2

t� 2t

t2 þ 1dt

¼Z 2

0

2 t2 þ 1� �

dt

¼ 2

3t3 þ 2t

� 2

0

¼ 16

3þ 4 ¼ 28

3: &

Problem 6

(a) Find a primitive of the function f xð Þ ¼ 1

3 x�1ð Þ32 þ x x�1ð Þ

12

; x 2 1;1ð Þ:

(b) Evaluate the integralR loge 3

0ex


1þ exp

dx:

7.4 Inequalities for integrals and their applications

Often it is not possible to evaluate an integral explicitly, and a numerical estimate

for its value is sufficient for our purposes. This can occur both in applications

of mathematics and in the proofs of theorems that involve integration.

7.4.1 The key inequalities

Our principal tool connecting inequalities and integrals is the following.

Theorem 1 Fundamental Inequality for Integrals

If f is integrable on [a, b], and f(x)� 0 on [a, b], thenR b

af xð Þdx � 0.

Proof Since f is integrable on [a, b], the sequence {Pn} of standard partitions

of [a, b] has the property that

limn!1

L f ;Pnð Þ ¼Z b

a

f :

Now, for each value of n, we have

288 7: Integration

We use the usual notation forpartitions and lower and upperRiemann sums.


Section 7.1.1.

L f ;Pnð Þ ¼X

n

i¼1

mi�xi; where mi ¼ inf f xð Þ : x 2 xi�1; xi½ f g:

But, since f (x)� 0 on [a, b], it follows that f (x)� 0 on each subinterval [xi�1, xi],

so that 0 is a lower bound for f on [xi�1, xi], for each i. Since mi is the greatest

lower bound for f on [xi�1, xi], we must therefore have mi� 0, for each i.

Since each term mi�xi in the sum for L(f,Pn) is non-negative, it follows that

L f ;Pnð Þ ¼X

n

i¼1

mi�xi � 0: (1)

Letting n!1 in (1) and using the Limit Inequality Rule for sequences, we

deduce that

Z b

a

f ¼ limn!1

L f ;Pnð Þ � 0: &

We can now use the Fundamental Inequality for Integrals to prove the most

commonly used inequalities for integrals.

Theorem 2 Inequality Rule for Integrals

Let f and g be integrable on [a, b]. Then:

(a) If f(x) � g(x) on [a, b], thenR b

af xð Þdx �

R b

ag xð Þdx:

(b) If m � f(x) � M on [a, b], then m b� að Þ �R b

af xð Þdx � M b� að Þ:

Recall that earlier we motivated our discussion of integrals in terms of areas.

The following diagrams illustrate the results of Theorem 1 in the special case

that the functions are positive so that we can interpret the integrals as areas

between the curves and the x-axis.

7.4 Inequalities for integrals 289

Proof of Theorem 2

(a) Since f and g are integrable on [a, b], so is the function g – f.

Since f (x) � g(x) on [a, b], it follows that

g xð Þ � f xð Þ � 0; x 2 a; b½ ;so, by Theorem 1, we deduce that

Z b

a

g� fð Þ � 0;

from which it follows at once thatR b

af �

R b

ag.

(b) First, the constant function x j!M, x2 [a, b], is continuous on [a, b] and so

is integrable on [a, b].

We can therefore apply part (a) of the Theorem to the inequality f (x)�M

on [a, b] to obtainZ b

a

f �Z b

a

Mdx ¼ M b� að Þ:

Next, the constant function x j!m, x2 [a, b], is continuous on [a, b] and so

is integrable on [a, b].

We can therefore apply part (a) of the Theorem to the inequality

f (x)�m on [a, b] to obtainZ b

a

f �Z b

a

mdx ¼ m b� að Þ: &

Example 1 Prove thatR 1

0x3

2�sin4 xdx � 1

4loge 2.

Solution Since, by the Sine Inequality, jsin xj � jxj, for x2R , it follows that

sin4 x � x4, for x2R . Hence

x3

2� sin4 x� x3

2� x4; for x 2 0; 1½ :

If we apply part (a) of Theorem 2 to this inequality, we obtainZ 1

0

x3

2� sin4 xdx �

Z 1

0

x3

2� x4dx

¼ �1

4loge 2� x4

� �

� 1

0

¼ �1

4loge 1� loge 2ð Þ ¼ 1

4loge 2: &

Example 2 Prove that 3ffiffiffiffi

34p �

R 2

�1dxffiffiffiffiffiffiffiffiffi

2þ x5p � 3:

Solution Since the function x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffi

2þ x5p

is increasing on [�1, 2], we have

1ffiffiffiffiffi

34p � 1


2þ x5p � 1; for x 2 �1; 2½ :

If we apply part (b) of Theorem 2 to these inequalities, we obtain

3ffiffiffiffiffi

34p �

Z 2

�1

dxffiffiffiffiffiffiffiffiffiffiffiffiffi

2þ x5p � 3:

&

By the Combination Rules forintegrals, Theorem 4 ofSection 7.2.3.


For the length of [�1, 2] is 3.

290 7: Integration

Problem 1 Use the Inequality Rule for integrals to prove thatR 3

1x sin 1

x10

� �

dx � 4:

Problem 2 Use the Inequality Rule for integrals to prove that12�R 1

2

0ex2

dx � 23:

Hint: Use the fact that 1þ x � ex � 11�x

, for x2 [0, 1).

Our final result is of great value in many applications. We saw earlier that, if the

function f is integrable on [a, b], so is the function j f j. We can then use

Theorem 2 to obtain a connection between the values of the integrals of f and j f j.

Theorem 3 Triangle Inequality

If f is integrable on [a, b], thenZ b

a

f

�

�

�

�

�

�

�

�

�Z b

a

fj j:

Furthermore, if jf(x)j � M on [a, b], thenZ b

a

f

�

�

�

�

�

�

�

�

� M b� að Þ:

Proof From the definition of the modulus function, we have

� f xð Þj j � f xð Þ � f xð Þj j:Thus, since jfj is integrable on [a, b], it follows from part (a) of Theorem 2 that

�Z b

a

fj j �Z b

a

f �Z b

a

fj j:

We can then rewrite this pair of inequalities in the desired formZ b

a

f

�

�

�

�

�

�

�

�

�Z b

a

fj j:

Next, we assume that jf(x)j � M on [a,b]. Then, from the definition of the

modulus function, it follows that

�M � f xð Þ � M; for x 2 a; b½ :It follows from part (b) of Theorem 2 that

�M b� að Þ �Z b

a

f � M b� að Þ:

We can then rewrite this pair of inequalities in the desired formZ b

a

f

�

�

�

�

�

�

�

�

� M b� að Þ: &

Example 3 Prove thatR p

2

0

x� p2

2þ cos xdx

�

�

�

�

�

�� p2

16.

Solution By the Triangle Inequality for IntegralsZ p

2

0

x� p2

2þ cos xdx

�

�

�

�

�

�

�

�

�Z p

2

0

x� p2

�

�

�

�

2þ cos xj j dx

¼Z p

2

0

p2� x

2þ cos xdx:

By the Modulus Rule,Theorem 4, Sub-section 7.2.3.

The name arises because ofthe similarity between thisinequality and the TriangleInequality for numbers

X

n

i¼1

ai

�

�

�

�

�

�

�

�

�

�

�X

n

i¼1

aij j:

So we must now examine thislast integral.


But 2þ cos x� 2, for x 2 0; p2

�

, so that

1

2þ cos x� 1

2; for x 2 0;

p2

h i

:

By applying this inequality to the integrand in the last integral, we obtain from

the Inequality Rule for Integrals thatZ p

2

0

x� p2

2þ cos xdx

�

�

�

�

�

�

�

�

�Z p

2

0

1

2

p2� x

� �

dx

¼ 1

2

p2

x� 1

2x2

� p2

0

¼ 1

2

p2

4� p2

8

� �

¼ p2

16: &

Problem 3 Prove the following inequalities:

(a)R 4

1

sin 1xð Þ

2þ cos 1xð Þ

dx

�

�

�

�

�

�

�

�

� 3; (b)R p

4

0tan x

3� sin x2ð Þ dx�

�

�

�

�

�� 1

4loge 2:

We end this sub-section with a nice application of the Triangle Inequality,

which is closely related to the Fundamental Theorem of Calculus.

Theorem 4 Let f be continuous on [a, b], and F be defined on [a, b] by the

formula

F xð Þ ¼Z x

a

f tð Þdt:

Then F is differentiable on [a, b], and its derivative is

F0 xð Þ ¼ f xð Þ:

Proof We will prove that F is differentiable at c and F0(c)¼ f (c) in the case

that a< c< b. (The cases c¼ a and c¼ b are similar, with the appropriate one-

sided derivatives being used.)

We must prove that:



� f cð Þ�

�

�

�

�

�

�

�

< "; for all x satisfying 0 < x� cj j < �:

Now


¼ 1

x� c

Z x

a

f tð Þdt �Z c

a

f tð Þdt

�

¼ 1

x� c

Z x

c

f tð Þdt

and

f cð Þ ¼ 1

x� c

Z x

c

f cð Þdt:



We will assume that � ischosen small enough that(c� �, cþ �) [a, b].

292 7: Integration

It follows that


� f cð Þ ¼ 1

x� c

Z x

c

f tð Þ � f cð Þf gdt

so that


� f cð Þ�

�

�

�

�

�

�

�

¼ 1

x� cj j

Z x

c


�

�

�

�

�

�

�

�

:

We now suppose that x> c; it then follows from the above equation, using the

Triangle Inequality for integrals, that


� f cð Þ�

�

�

�

�

�

�

�

¼ 1

x� c

Z x

c


�

�

�

�

�

�

�

�

� 1

x� c

Z x

c

f tð Þ � f cð Þj jdt ðsince x > cÞ:

But, since f is continuous at c, we know that there must exist some positive


f tð Þ � f cð Þj j < 1

2"; for all x satisfying x� cj j < �:

It follows that, for x satisfying c< x< cþ �, we have


� f cð Þ�

�

�

�

�

�

�

�

� 1

x� c

Z x

c

1

2"dt

¼ 1

2"

< ":

A similar argument applies in the case that x< c; so that, for all x satisfying

0< jx� cj<�, we have


� f cð Þ�

�

�

�

�

�

�

�

< ":


7.4.2 Wallis’s Formula

In this sub-section we use the method of Reduction of Order together with

various inequalities between integrals to establish a remarkable formula for the

number p.

Reduction of Order method

Quite often we need to evaluate an integral In that involves a non-negative

integer n. A common approach to such integrals is to relate the value of In to the

value of In�1 or In�2 by a reduction formula, using integration by parts.

Example 4 Let In ¼R p

2

0sinn xdx, n¼ 0, 1, 2, . . ..

(a) Evaluate I0 and I1.

(b) Prove that In ¼ n�1n

In�2, for n� 2.

We omit the details.

This is just another name for arecurrence formula.

(c) Deduce from part (b) that, for n� 1

I2n ¼1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ

2:4: . . . : 2n� 2ð Þ 2nð Þ �p2

and

I2nþ1 ¼2:4: . . . : 2n� 2ð Þ 2nð Þ

3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ :

Solution

(a) We can evaluate the first two integrals easily

I0 ¼Z p

2

0

1dx ¼ p2; and

I1 ¼Z p

2

0

sin xdx ¼ � cos x½ p2

0¼ 1:

(b) We first write In in the form In ¼R p

2

0sin x sinn�1 xdx. Using integration by

parts, we find that, for n� 2

In ¼ �cos xð Þ sinn�1 x �

p2

0�Z p

2

0

�cos xð Þ n� 1ð Þ sinn�2 x cos xdx

¼ 0þ n� 1ð ÞZ p

2

0

cos2 x sinn�2 xdx

¼ n� 1ð ÞZ p

2

0

1� sin2 x� �

sinn�2 xdx

¼ n� 1ð Þ In�2 � Inf g:

We can then rewrite this result in the form nIn¼ (n� 1)In�2, so that

In ¼n� 1

nIn�2:

(c) If we replace n in the formula for In in part (b) by 2n, 2n� 2, 2n� 4, . . ., in

turn, we obtain

I2n ¼2n� 1

2nI2n�2; I2n�2 ¼

2n� 3

2n� 2I2n�4; I2n�4 ¼

2n� 5

2n� 4I2n�6; . . .:

Hence

I2n ¼2n� 1

2nI2n�2

¼ 2n� 3ð Þ 2n� 1ð Þ2n� 2ð Þ 2nð Þ I2n�4

¼ 2n� 5ð Þ 2n� 3ð Þ 2n� 1ð Þ2n� 4ð Þ 2n� 2ð Þ 2nð Þ I2n�6;

We will integrate sin x anddifferentiate sinn�1 x.

294 7: Integration

and so on. Continuing this process, we obtain

I2n ¼1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ

2:4: . . . : 2n� 2ð Þ 2nð Þ I0

¼ 1:3: . . . : 2n� 3ð Þ 2n� 1ð Þ2:4: . . . : 2n� 2ð Þ 2nð Þ �

p2:

Similarly

I2nþ1 ¼2n

2nþ 1I2n�1

¼ 2n� 2ð Þ 2nð Þ2n� 1ð Þ 2nþ 1ð Þ I2n�3

..

.

¼ 2:4: . . . : 2n� 2ð Þ 2nð Þ3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ I1

¼ 2:4: . . . : 2n� 2ð Þ 2nð Þ3:5: . . . : 2n� 1ð Þ 2nþ 1ð Þ : &

Problem 4 Let In ¼R 1

0exxndx, n¼ 0, 1, 2, . . ..

(a) Evaluate I0.

(b) Prove that In¼ e � nIn�1, for n� 1.

(c) Deduce the values of I1, I2, I3 and I4.

Wallis’s Formula

We now use the various results that we have just proved for the integral

In ¼R p

2

0sinn xdx to verify some surprising results.

Theorem 5 Wallis’s Formula

(a) limn!1

21: 2

3: 4

3: 4

5: 6

5: 6

7: . . . : 2n

2n�1: 2n

2nþ1

� �

¼ p2:

(b) limn!1

n!ð Þ222n

2nð Þ!ffiffi

np ¼

ffiffiffi

pp

:

In the next problem, we ask you to establish a number of relationships

between the terms of the two sequences in the statement of Theorem 5.

Problem 5 Let an ¼ 21: 2

3: 4

3: 4

5: 6

5: 6

7: . . . : 2n

2n� 1: 2n

2nþ 1and bn ¼ n!ð Þ222n

2nð Þ!ffiffi

np ,

n� 1.

(a) Evaluate an and bn, for n¼ 1, 2 and 3.

(b) Verify that b2n ¼ 2nþ1

nan, for n¼ 1, 2 and 3.

(c) Prove that b2n ¼ 2nþ1

nan, for all n� 1.

For I0 ¼ p2:

For I1¼ 1.

Each of the two limits iscalled Wallis’s Formula.

You may omit tackling thisproblem and the followingproof at a first reading.


Proof of Theorem 5 Let In ¼R p

2

0sinn xdx, for all n� 0, and let the sequences

{an} and {bn} be as given in Problem 5.

(a) Using the formulas for I2n and I2nþ1 in part (c) of Example 4 above, we

obtain

I2n

I2nþ1

¼ 1:3:3:5:5: . . .: 2n� 1ð Þ 2n� 1ð Þ 2nþ 1ð Þ2:2:4:4:6: . . .: 2n� 2ð Þ 2nð Þ 2nð Þ � p

2;

which we arrange in the form

an ¼I2nþ1

I2n

� p2:

Notice that, in order to complete the proof of part (a), it is sufficient to

show that

I2nþ1

I2n

! 1 as n!1: (2)

We do this as follows.

Since 0 � sin x � 1 for x 2 0; p2

�

, we have

sin2n x � sin2nþ1 x � sin2nþ2 x; for x 2 0;p2

h i

:

It follows, from the Inequality Rule for integrals, that

I2n � I2nþ1 � I2nþ2:

Thus

1 � I2nþ1

I2n

� I2nþ2

I2n

¼ 2nþ 1

2nþ 2: (3)

By letting n!1 in (3) and using the Squeeze Rule for sequences, we

deduce that the limit (2) holds, as desired.

(b) We know, from part (c) of Problem 5, that

b2n ¼

2nþ 1

nan; for all n � 1:

We also know, from the proof of part (a) of this theorem, that

an !p2

as n!1:

It follows, by applying the Product Rule for sequences to the above

formula for bn2, that

b2n ! 2� p

2¼ p as n!1:

Hence, by the continuity of the square root function, we conclude that

bn !ffiffiffi

pp

as n!1. &

Unfortunately the sequences in Theorem 5 converge rather slowly as n

increases, so they are of little value in determining p orffiffiffi

pp

to any reasonable

degree of appro ximat ion. In the next chapter , we shal l mee t prac tical way s of

estimating p.

The numerator of I2nþ1 isthe same as the denominatorof I2n.

Note that

I2nþ2 ¼2nþ 1

2nþ 2I2n;

by the reduction formulafor In.

Section 8.5.

296 7: Integration

7.4.3 Maclaurin Integral Test

We now introduce a method for determining the convergence or divergence of

certain series of the formP

1

n¼1

f nð Þ, where the terms are positive and decrease to

zero. In particular, we show that the seriesP

1

n¼1

1np converges if p> 1 and

diverges if 0< p� 1.

We use the fact that each term in such a series can be regarded as a

contribution to a lower Riemann sum or upper Riemann sum for a suitable

integral.

Theorem 6 Maclaurin Integral Test

Let f be positive and decreasing on [1, 1), and let f(x)! 0 as x!1.

Then:

(a)P

1

n¼1

f nð Þ converges if the sequenceR n

1f :n 2 N

� �

is bounded above;

(b)P

1

n¼1

f nð Þ diverges if the sequenceR n

1f :n 2 N

� �

tends to1 as n!1.

Before proving the theorem, we illustrate the underlying ideas.

Example 5 Let Pn�1 be the standard partition of [1, n] with n� 1 subintervals

½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:

(a) Determine the lower and upper Riemann sums, L( f,Pn�1) and U( f,Pn�1),

for the function f xð Þ ¼ 1x2, x 2 1;1½ Þ.

(b) Deduce that the seriesP

1

n¼1

1n2 converges, and that 1�

P

1

n¼1

1n2 � 2.

Solution

(a) Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . ., n� 1,

mi ¼ f iþ 1ð Þ ¼ 1

iþ 1ð Þ2;

Mi ¼ f ið Þ ¼ 1

i2:

Also, each subinterval in the partition has length 1.

You saw in Sub-section 3.2.1that this series converges ifp� 2.

In these results, the number 1may be replaced by anyconvenient positive integer.

In fact, the sum of this series isp2

6, but to prove this goes

outside the scope of this book.

Hence the lower and upper Riemann sums for f are

L f ;Pn�1ð Þ ¼X

n�1

i¼1

mi � 1 ¼ 1

22þ 1

32þ � � � þ 1

n2;

U f ;Pn�1ð Þ ¼X

n�1

i¼1

Mi � 1 ¼ 1

12þ 1

22þ � � � þ 1

n� 1ð Þ2:

(b) Let sn denote the nth partial sum of the seriesP

1

n¼1

1n2; thus

sn ¼1

12þ 1

22þ � � � þ 1

n� 1ð Þ2þ 1

n2:

It follows that

L f ;Pn�1ð Þ ¼ sn � 1 and U f ;Pn�1ð Þ ¼ sn �1

n2:

Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have

L f ;Pn�1ð Þ �Z n

1

f � U f ;Pn�1ð Þ;

so that

sn � 1 �Z n

1

dx

x2� sn �

1

n2: (4)

Now, the sequence {sn} is increasing, since the series has positive terms.

Also, from (4), we have

sn �Z n

1

dx

x2þ 1

¼ � 1

x

� n

1

þ1

¼ 1� 1

n

� �

þ 1 ¼ 2� 1

n� 2: (5)

Thus {sn} is bounded above.

Hence, by the Monotone Convergence Theorem for sequences, {sn} is

convergent, so that

X

1

n¼1

1

n2is convergent:

Since s1¼ 1 and {sn} is increasing, the sum of the series is at least 1.

Finally, we deduce from (5), using the Limit Inequality Rule for

sequences, that limn!1

sn � 2; so the sum of the series is at most 2. Hence

1 �X

1

n¼1

1

n2� 2: &

Example 6 Let Pn�1 be the standard partition of [1, n] with n� 1 subintervals

½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:

298 7: Integration

(a) Determine the lower and upper Riemann sums, L( f, Pn�1) and U( f, Pn�1),

for the function f xð Þ ¼ 1x; x 2 1;1½ Þ.

(b) Deduce that the seriesP

1

n¼1

1n

diverges.

Solution

(a) Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . . , n� 1

mi ¼ f iþ 1ð Þ ¼ 1

iþ 1;

Mi ¼ f ið Þ ¼ 1

i:

Also, each subinterval in the partition has length 1.



n�1

i¼1

mi � 1 ¼ 1

2þ 1

3þ � � � þ 1

n;


n�1

i¼1

Mi � 1 ¼ 1

1þ 1

2þ � � � þ 1

n� 1:

(b) Let sn denote the nth partial sum of the seriesP

1

n¼1

1n; thus

sn ¼1

1þ 1

2þ � � � þ 1

n� 1ð Þ þ1

n:

It follows that

L f ;Pn�1ð Þ ¼ sn � 1 and U f ;Pn�1ð Þ ¼ sn �1

n:


L f ;Pn�1ð Þ �Z n

1

f � U f ;Pn�1ð Þ;

so that

sn � 1 �Z n

1

dx

x� sn �

1

n;

or

sn � 1 � loge n � sn �1

n: (6)

In particular, sn� loge n. Then, since loge n!1 as n!1, it follows, from

the Squeeze Rule for sequences that tend to infinity, that sn!1 as n!1.

Hence, the seriesP

1

n¼1

1n

diverges. &

Remarks

We can in fact get more information from the above argument than just the

result of the example.

1. If we define the sequence {�n} by the expression

�n ¼ 1þ 1

2þ 1

3þ � � � þ 1

n� loge n


then

�nþ1 � �n ¼1

nþ 1� loge

nþ 1

n

¼ 1

nþ 1�Z nþ1

n

dx

x

¼Z nþ1

n

1

nþ 1� 1

x

� �

dx � 0;

so that {�n} is decreasing. Also, it follows from (6), using the fact that

�n¼ sn� loge n, that

�n �1

n;

thus the sequence {�n} is bounded below by 0.

Hence the sequence {�n} tends to a limit, � say, as n!1. The value of �clearly lies between 1 (for �1¼ 1) and 0.

This number is called Euler’s constant, and occurs often in Analysis.

In fact,

� ¼ 0:57721 56649 01532 86060 65120 90082 40243 . . .:

2. Although the sequence sn ¼ 1þ 12þ � � � þ 1

n�1þ 1

ntends to infinity, we can

make more precise statements than that, arising from (6). We can rewrite

the inequalities in (6) in the following form

1

nþ loge n � sn � 1þ loge n;

if we then divide throughout by logen and let n!1, we obtain, by the

Limit Inequality Rule for sequences, that

sn

loge n! 1 as n!1:

We can write this last limit in an equivalent form as ‘sn� logen as n!1’ –

in other words, ‘the behaviour of sn and logen are essentially the same for

large n’.

Problem 6 Use the Maclaurin Integral Test to determine the behaviour

of the seriesP

1

n¼1

1np, for p> 0, p 6¼ 1.

Problem 7 Show thatR

dx

x loge xð Þ2 ¼ �1

loge x, for x> 1, and hence prove

that the seriesP

1

n¼2

1

n loge nð Þ2 converges.

Problem 8 Show thatR

dxx loge x

¼ loge loge xð Þ, for x> 1, and hence

prove that the seriesP

1

n¼2

1n loge n

diverges.

Sequences revisited

We can apply many ideas similar to those in Example 5 to obtain useful

information about certain types of convergent sequences.

By definition of the sequence{�n}.

For loge n � sn � 1n:

We give here just the first 35decimal places.

We shall address this �notation more carefully inthe next section.

300 7: Integration

Example 7 Prove that 1nþ 1þ 1

nþ 2þ � � � þ 1

2n! loge 2 as n!1.

Solution Let f xð Þ ¼ 11þx

, x2 [0, 1]; and let Pn ¼ 0; 1n

�

; 1n; 2

n

�

; . . .; n�1n; 1

��

be the standard partition of [0, 1] into n sub-intervals of equal length 1n.

Since f is decreasing on [0, 1], it follows that on the ith sub-interval i�1n; i

n

�

we have

mi ¼ fi

n

� �

¼ 1

1þ in

:

Thus the lower Riemann sum for f on Pn is

L f ;Pnð Þ ¼X

n

i¼1

1

1þ in

� 1

n

¼X

n

i¼1

1

nþ i¼ 1

nþ 1þ 1

nþ 2þ � � � þ 1

2n:

Since f is decreasing on [0, 1], it is integrable on [0, 1]. Hence, as n!1, we

have

1

nþ 1þ 1

nþ 2þ �� þ 1

2n¼ L f ;Pnð Þ!

Z 1

0

f ¼Z 1

0

dx

1þ x

¼ loge 1þ xð Þ½ 10 ¼ loge 2:&

Example 7 illustrates a general technique that is often useful.

Strategy If f is positive and decreasing on [0, 1], then: 1n

P

n

i¼1

f in

� �

!R 1

0f as

n!1.

Problem 9 Prove that n 1n2þ12 þ 1

n2þ22 þ � � � þ 1n2þn2

� �

! p4

as n!1.

You have already met the corresponding ideas for divergent sequences in

Example 6 and Remark 2 after that example, so we state without comment

the general strategy illustrated there.

Strategy If f is positive and decreasing on [1,1), f(x)! 0 as x!1, andR n

1f !1 as n!1, then

P

n

i¼1

f ið Þ �R n

1f as n!1.

Remark

In fact under the hypotheses of this strategyP

n

i¼1

f ið Þ � ðR n

1f Þ þ c, for any fixed

number c; in particular situations we may choose c in any convenient way.

Problem 10 Prove that 1þ 1ffiffi

2p þ 1

ffiffi

3p þ � � � þ 1

ffiffi

np � 2

ffiffiffi

np

as n!1.

Recall that

mi ¼ inf f xð Þ : x 2 i�1n; i

n

��

:

The trick in applying thisstrategy is to make a goodchoice of f.

Recall that this means that

P

n

i¼1

f ið ÞR n

1f! 1 as n!1:

The reason for this will beexplained in Section 7.5.1.

Here you will use both thestrategy and the subsequentremark.


Proof of the Maclaurin Integral Test

Theorem 6 Maclaurin Integral Test

Let f be positive and decreasing on [1,1), and let f(x)! 0 as x!1. Then:

(a)P

1

n¼1

f nð Þ converges if the sequenceR n

1f : n 2 N

� �

is bounded above;

(b)P

1

n¼1

f nð Þ diverges if the sequenceR n

1f : n 2 N

� �

tends to1 as n!1.

Proof Let In denote the integralR n

1f , let sn¼ f(1)þ f(2)þ � � � þ f(n) denote

the nth partial sum of the series, and let Pn�1 be the standard partition of [1, n]

with n� 1 subintervals

½1; 2; ½2; 3; . . .; ½i; iþ 1; . . .; ½n� 1; nf g:

Since f is decreasing on [1,1), it follows that, for i¼ 1, 2, . . ., n� 1

mi ¼ f iþ 1ð Þ and Mi ¼ f ið Þ:Also, each subinterval in the partition has length 1.



n�1

i¼1

mi � 1 ¼ f 2ð Þ þ f 3ð Þ þ � � � þ f nð Þ

¼ sn � f 1ð Þ;


n�1

i¼1

Mi � 1 ¼ f 1ð Þ þ f 2ð Þ þ � � � þ f n� 1ð Þ

¼ sn � f nð Þ:


L f ;Pn�1ð Þ � In � U f ;Pn�1ð Þ;

so that

sn � f 1ð Þ � In � sn � f nð Þ: (7)

Case 1: {In} is bounded

We are now assuming that, for some M, In�M, for all n. It follows from

(7) that

sn � f 1ð Þ þM; for all n:

302 7: Integration

Thus the increasing sequence {sn} is bounded above, and so, by the Monotone

Convergence Theorem, it is convergent.

Hence the seriesP

1

n¼1

f nð Þ is convergent.

Case 2: {In} is not bounded

The sequence {In} is increasing, since Inþ1 � In ¼R nþ1

nf � 0. Since we are

now assuming that {In} is not bounded, it follows that In!1 as n!1.

Now, from (7), sn� In; so, by the Squeeze Rule for sequences which tend to

infinity,

sn !1 as n!1:

Hence the seriesP

1

n¼1

f nð Þ is divergent. &

7.5 Stirling’s Formula for n!

For small values of n, we can evaluate n! directly by multiplication or by using

a scientific calculator.

Problem 1 Complete the following table of values of n!

n n! n n! n n!

1 1 6 720 20 2 2 7 5040 30 3 6 8 40 320 40 4 24 9 362 880 50 5 120 10 3 628 800 60

As n increases, n! grows very quickly; for instance:

� around 14! seconds have elapsed since the birth of Christ;

� around 18! seconds have elapsed since the formation of the Earth.

A calculator soon becomes useless for evaluating n!; thus the author’s

current scientific calculator (2005) gives that 69!’ 1.7� 1098, but an error

message when asked the value of 70!.

Stirling’s Formula gives us a way of estimating n! for large values of n. In order

to state the formula, though, we must first introduce some notation that enables us

to compare the behaviour of positive functions of n for large values of n.

7.5.1 The tilda notation

For many numbers that arise in the normal way, we often use the symbol ‘’’ to

denote ‘is approximately equal to’. For example,ffiffiffi

2p¼ 1:4142135623730950

4880 . . . (where we have given only the first 20 decimal places); for many

purposes it is sufficient to use estimates such asffiffiffi

2p’ 1:414 or even

ffiffiffi

2p’ 1:4,

14! ’ 8:7� 1010

18! ’ 6:4� 1015

Often an estimate is sufficientfor our purposes.


since the error involved is small. But what do we mean by ‘the error involved is

small’? Sometimes we mean that the error, the difference between the exact

value and the approximation, is a small number; sometimes we mean that the

percentage error, namely

percentage error ¼ actual error

exact value� 100;

is a small number.

To handle the behaviour of positive functions of n for large values of n, we

introduce a very precise mathematical notation.

Notation For positive functions f and g with domain N , we write

f nð Þ � g nð Þ as n!1;

to mean thatf nð Þg nð Þ ! 1 as n!1.

For example, n2þ 1000nþ 10� n2 as n!1, since n2þ1000nþ 10n2 !1 as

n!1; and sin 1n

� �

� 1n

as n!1, sincesin 1

nð Þ1n

! 1 as n!1.

Notice that, if f(n)� g(n) as n!1, then g(n)� f(n) as n!1.

Also, if f(n)� g(n) as n!1, g(n)!1 and c is a given number, then

f(n)þ c� g(n) as n!1, since, as n!1f nð Þ þ c

g nð Þ ¼f nð Þg nð Þ þ

c

g nð Þ ! 1þ 0 ¼ 1:

Notice, however, that the statement f(n)� g(n) as n!1 does NOT mean

that f(n)� g(n)! 0 or even that f(n)� g(n) is bounded. For example,

n2þ 1000nþ 10� n2 as n!1, yet

n2 þ 1000nþ 10� �

� n2� �

¼ 1000nþ 10!1 as n!1:

In situations like this example, ‘�’ compares the sizes of the dominant terms

in f and g.

Problem 2 Find two pairs from the following functions such that

fi(n)� fj(n) as n!1f1 nð Þ ¼ sin n2ð Þ; f2 nð Þ ¼ sin 1

n2

� �

; f3 nð Þ ¼ 1� cos 1n

� �

;f4 nð Þ ¼ 2

n2 ; f5 nð Þ ¼ 1n2 ; f6 nð Þ ¼ 1� 1

n; f7 nð Þ ¼ 1

2n2 :

Remark

If ‘ is finite and positive, then the statements

‘f nð Þ ! ‘ as n!1’ and ‘f nð Þ � ‘ as n!1’

are equivalent, since each is equivalent to the statement

‘f nð Þ‘! 1 as n!1’:

We can also, in this situation, legitimately say that f(n)� ‘ for large n.

The Combination Rules for handling tilda are direct consequences of the

corresponding Combination Rules for sequences.

‘For example, the error is lessthan 1’.

For example, ‘the percentageerror is less than 1%’.

The notation is ONLY usedfor positive functions.

We say ‘f tilda g’ or ‘ftwiddles g’.

You can think of this as sayingthat the percentage errorinvolved by replacing f(n) byg(n) is small for large n.

For, limx!0

sin xx¼ 1:

For,f nð Þg nð Þ ! 1, g nð Þ

f nð Þ ! 1.

For example, we may write

1þ 1

n

� �n

� e as n!1;

since

1þ 1

n

� �n

! e as n!1:

Note that we only use � forpositive funtions.

304 7: Integration

Combination Rules

Let f1(n)� g1(n) and f2(n)� g2(n) as n!1. Then:

Sum Rule f1(n)þ f2(n)� g1(n)þ g2(n);

Multiple Rule lf1(n)� lg1(n), for any number l> 0;

Product Rule f1(n)� f2(n)� g1(n)� g2(n);

Reciprocal Rule 1f1 nð Þ � 1

g1 nð Þ.

Example 1 Prove that, if f(n)� g(n) as n!1, then f nð Þð Þ1n � g nð Þð Þ

1n

as n!1.

Solution Since f(n)� g(n) as n!1, we have

f nð Þg nð Þ ! 1 as n!1;

so that

loge

f nð Þg nð Þ ! 0 as n!1;

and therefore

1

nloge

f nð Þg nð Þ ! 0 as n!1:

But

1

nloge

f nð Þg nð Þ ¼ loge

f nð Þg nð Þ

� �1n

¼ loge

f nð Þð Þ1n

g nð Þð Þ1n

;

so that

loge

f nð Þð Þ1n

g nð Þð Þ1n

! 0 as n!1:

We now use the fact that the exponential function is continuous at 0; it follows

from the previous limit that

f nð Þð Þ1n

g nð Þð Þ1n

! 1 as n!1;

which is precisely the statement that f nð Þð Þ1n � g nð Þð Þ

1n as n!1. &

Problem 3 Give specific examples of functions f and g to show that

f nð Þð Þ1n � g nð Þð Þ

1n as n!1 does not imply that f(n)� g(n) as n!1.

Hint: Try f(n)¼ n2 and g(n)¼ n.

7.5.2 Stirling’s Formula

Stirling’s Formula was discovered in the eighteenth century, in an analysis of

a gambling problem!

For example, if

f1 nð Þ ¼ n2 þ n; g1 nð Þ ¼ n2;

f2 nð Þ ¼ n3 þ n; g2 nð Þ ¼ n3;

then

n2 þ n� �

þ n3 þ n� �

� n2 þ n3;

5 n2 þ n� �

� 5n2 l ¼ 5ð Þ;n2 þ n� �

� n3 þ n� �

� n2 � n3 ¼ n5;

1

n2 þ n� 1

n2:

Here we use the fact thatthe function loge is continuousat 1.


Theorem 1 Stirling’s Formula


2pnp n

e

� �n

as n!1:

Problem 4 Use your calculator to evaluateffiffiffiffiffiffiffiffi

2pnp

ne

� �nfor n¼ 5. For

this value of n does the expression approximate n! to within 1%?

For small values of n, Stirling’s Formula gives reasonable approximations to

n!, and the percentage error quickly decreases as n increases:

n n! Stirling’s approximation Error

10 3 628 800 3 598 696 0.83% (1 in 120)

20 2.433� 1018 2.423� 1018 0.42% (1 in 240)

52 8.066� 1067 8.053� 1067 0.16% (1 in 620)

100 9.333� 10157 9.325� 10157 0.09% (1 in 1170)

We illustrate the use of Stirling’s Formula as follows.

If 200 coins are tossed, then the probability of there being exactly 100 heads

and 100 tails is

200

100

� �

� 1

2

� �200

¼ 200!

100!ð Þ2�2200:

It follows from Stirling’s Formula that this probability is

200!

100!ð Þ2�2200’


400pp

� 200e

� �200


200pp

� 100e

� �100� �2

� 2200

¼ 1

10ffiffiffi

pp

¼ 1

17:724 . . .:

In other words, the probability of there being exactly 100 heads and 100 tails

is about 1 in 18 – rather higher than you might expect.

Problem 5 Prove that limn!1

nn

n!

� �1n ¼ e.

Hint: Use the result of Example 1.

Problem 6 Use Stirling’s Formula to estimate the following numbers

to two significant figures:

(a)300

150

� �

� 1

2

� �300

; (b)300!

100!ð Þ3� 1

3

� �300

:

Problem 7 Use Stirling’s Formula to determine a number l such that

4n

2n

� ��

2n

n

� �

� l22n as n!1:

This fact is easily provedusing Probability Theory.

Here we substitute n¼ 200and n¼ 100 into Stirling’sFormula.

306 7: Integration

7.5.3 Proof of Stirling’s Formula

Theorem 1 Stirling’s Formula


2pnp n

e

� �n

as n!1:

Proof We divide up the proof into a number of steps, for clarity.

Step 1: Setting things up

We consider the function

f ðxÞ ¼ loge x; x 2 1; n½ ;and the standard partition Pn�1 of the interval [1, n] with n� 1 subintervals

1; 2½ ; 2; 3½ ; . . .; i; iþ 1½ ; . . .; n� 1; n½ f g:We consider also the sequence of numbers cnf g1n¼2, where cn is the total area

between the concave curve y¼ loge x, for x2 [1, n], and the polygonal line with

vertices (1, 0), (2, loge 2), (3, loge 3), . . ., (n, loge n), as illustrated below. This

consists of n� 1 small slivers.

y

x1 2 3 ... ...i

total area ofsilvers = cn

y = loge x y = loge x

(i, logei)

(i + 1,loge(i + 1))

contributionto cn

i + 1 n

Step 2: Calculating areas

The area between the curve y¼ loge x and the x-axis, for x2 [1, n], isZ n

1

loge xdx ¼ x loge x� x½ n1

¼ n loge n� n� 1ð Þ: (1)

Next, the area between the polygonal line and the x-axis is

12

L f , Pn�1ð Þ þ U f ;Pn�1ð Þf g; (2)

and, since f is increasing, we have

L f , Pn�1ð Þ ¼ loge 1þ loge 2þ � � � þ loge n� 1ð Þ¼ loge n� 1ð Þ!¼ loge n!� loge n (3)

and

U f , Pn�1ð Þ ¼ loge 2þ � � � þ loge n

¼ loge n!: (4)

Substituting from (3) and (4) into (2), we find that the area between the

polygonal line and the x-axis is


loge n!� 12

loge n: (5)

It follows from (1) and (5) that

cn ¼ n loge n� n� 1ð Þ � loge n!þ 12

loge n

¼ loge

nnþ12

en�1n!: (6)

Step 3: Behaviour of the sequence {cn}

It is obvious from the definition of cn that the sequence {cn} is positive and

increasing. In order to apply the Monotone Convergence Theorem to the

sequence, we must prove next that {cn} is bounded above.

So, let i be an integer with 1� i� n� 1, and let

A ¼ i; loge ið Þ;B ¼ iþ 1; logeðiþ 1Þð Þ; and C ¼ iþ 1; loge ið Þ:Then, as illustrated in the diagram in the margin

BC ¼ loge iþ 1ð Þ � loge i

¼ loge 1þ 1

i

� �

:

Next, let the tangent at A to the curve meet the line BC at D. Since AC¼ 1

and the slope of the line AD is 1i, it follows that CD ¼ 1

i. Hence

BD ¼ CD� BC

¼ 1

i� loge 1þ 1

i

� �

� 1

2i2:

Hence the contribution to the area cn of this sliver between the lines x¼ i and

x¼ iþ 1 is at most

area of � ABD ¼ 1

2AC � BD

� 1

2� 1� 1

2i2¼ 1

4i2:

If we now sum such areas over i¼ 1, 2, . . ., n� 1, we obtain

cn �1

4

X

n�1

i¼1

1

i2

� 1

4

X

1

i¼1

1

i2:

Since this infinite series converges, it follows that the sequence {cn} is

bounded above.

It follows from the Monotone Convergence Theorem that the sequence {cn}

is convergent.

Step 4: Properties of the sequence {an}, where an ¼ ecn

Since {cn} is convergent and the exponential function is continuous, it follows

that, if we set an ¼ ecn , for n¼ 2, 3, . . ., then the sequence {an} is also

convergent. Thus, using the expression (6) for cn, we have

an ¼nnþ1

2

en�1n!! L; as n!1; (7)

for some non-zero number L.

Here we use the inequality

loge 1þ xð Þ > x� 12x2;

for x > 0;

that you met in Section 6.7,Exercise 5(c) on Section 6.4.

The actual value of the bounddoes not matter here.

We now introduce {an} sothat we can use Wallis’sFormula in the next Step.

L 6¼ 0 since the exponentialdoes not take the value 0.

308 7: Integration

It follows from the formula for an in (7) that

a2n

a2n

¼ n2nþ1

e2n�2 n!ð Þ2� e2n�1 � 2nð Þ!

2nð Þ2nþ12

¼ 2nð Þ!n12e

n!ð Þ222nþ12

; (8)

after some cancellation.

Step 5: Proving Stirling’s Formula

We can rewrite (8) in the form

a2n

a2n

¼ 2nð Þ!n12

n!ð Þ222n� e

ffiffiffi

2p : (9)

But, by Wallis’s Formula, the first quotient on the right-hand side of (9) tends

to 1ffiffi

pp as n!1. It follows that, if we let n!1 on both sides of (9), we obtain

L2

L¼ 1

ffiffiffi

pp � e

ffiffiffi

2p ;

so that

L ¼ effiffiffiffiffiffi

2pp :

Knowing the value of L, we can then rewrite (7) in the form

nnþ12

en�1n!! e

ffiffiffiffiffiffi

2pp ;

by the Reciprocal Rule for sequences, it follows that

en�1n!

nnffiffiffi

np !

ffiffiffiffiffiffi

2pp

e:

. By the definition of tilda, this limit is exactly equivalent to the relation


2pnp n

e

� �n

as n!1;

that we set out to prove. &

Remark

It is quite remarkable how many of the techniques that we have met so far in the

book are needed in order to prove this apparently straight-forward result.

7.6 Exercises

Section 7.1

1. Sketch the graph of the function

f xð Þ ¼ 1� xj j; �1 < x < 1;1; x ¼ �1:

Determine the minimum, maximum, infimum and supremum of f on [�1, 1].

You met Wallis’s Formulain Theorem 5 of Sub-section 7.4.2.

limn!1

n!ð Þ222n

2nð Þ!ffiffiffi

np ¼

ffiffiffi

pp

:

7.6 Exercises 309

f xð Þ ¼ xj j; � 1 < x < 1;12; x ¼ �1:

Evaluate L( f, P) and U( f, P) for each of the following partitions P of [�1, 1]:

(a) P ¼ �1;� 12

�

; � 12; 0

�

; 0; 12

�

; 12; 1

��

;

(b) P ¼ �1;� 14

�

; � 14; 1

3

�

; 13; 1

��

:


f xð Þ ¼ 1� x; 0 � x < 1;2; x ¼ 1:

(a) Using the standard partition Pn of [0, 1] with n equal subintervals,

evaluate L( f, Pn) and U( f, Pn).

(b) Deduce that f is integrable on [0, 1], and evaluateR 1

0f :

4. Let f be the function f(x)¼ x3, x2 [0, 1].

(a) Using the standard partition Pn of [0, 1] with n equal subintervals,

evaluate L( f, Pn) and U( f, Pn).

(b) Deduce that f is integrable on [0, 1], and evaluateR 1

0f :

5. Let f be the function f (x)¼ sin x, x2 0; p2

�

:

(a) Using the standard partition Pn of 0; p2

�

with n equal subintervals,

evaluate L ( f, Pn) and U ( f, Pn).

(b) Deduce that f is integrable on 0; p2

�

, and evaluateR p

2

0f :

Hint: Use the formula

sin Aþ sin Aþ Bð Þ þ � � � þ sin Aþ n� 1ð ÞBð Þ

¼sin 1

2nB

� �

sin 12

B� � sin Aþ n� 1

2B

� �

; B 6¼ 0:

6. Prove that the function

f xð Þ ¼ 1þ x; 0 � x � 1; x rational;1� x; 0 � x � 1; x irrational;

is not integrable on [0, 1].

7. Prove that Dirichlet’s function

f xð Þ ¼1q; if x is a rational number p

q, in lowest terms, with q > 0,

0; if x is irrational.

is integrable on [0, 1].

Hint: Verify that there are at most 12

n nþ 1ð Þ points x in [0, 1] for which

f xð Þ > 1n:

Section 7.2

1. For bounded functions f and g on an interval I, prove that

supI

f þ gð Þ � sup fI

þ supI

g:

For the interval I¼ [�1, 2], write down functions f and g for which strict

inequality holds in this result.

This formula can be proved byMathematical Induction.

You met this function in Sub-section 5.4.2.

310 7: Integration

2. Write down functions f and g and an interval I for which it is not true that

supI

fgð Þ ¼ sup fI

� supI

g:

3. Write down functions f and g on [0, 1] that are not integrable on [0, 1] but

such that fþ g is integrable on [0, 1].

4. (a) Write down functions f and g on [0, 1] such that f is integrable, g is not

integrable, and fg is integrable.

(b) Write down a function f on [0, 1] such that jfj is integrable but f is not

integrable on [0, 1].

5. Prove that, if the functions f and g are integrable on [a, b], then so is the

function max{f, g}.

Hint: Use the formula max a; bf g ¼ 12

aþ bþ a� bj jf g, for a, b2R .

6. Prove that, if f and g are integrable on [a, b], then

Z b

a

fg

� �2

�Z b

a

f 2

� �

Z b

a

g2

� �

:

Hint: Use the Cauchy-Schwarz InequalityP

n

i¼1

aibi

� �2

�P

n

i¼1

a2i

� �

P

n

i¼1

b2i

� �

:

Section 7.3

1. Write down a primitive of each of the following functions:

(a) f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 9p

; x 2 3;1ð Þ;(b) f xð Þ ¼ sin 2xþ 3ð Þ � 4 cos 3x� 2ð Þ; x 2 R ;

(c) f xð Þ ¼ e2x sin 3xð Þ; x 2 R :

2. Using the result of part (c) of Exercise 1, write down a primitive F of the

function f xð Þ ¼ e2x sin 3xð Þ, x2R , for which F(p)¼ 0.

3. Show that the following functions are all primitives of the function

f(x)¼ sech x, x 2 � p2; p

2

� �

:

(a) F1 xð Þ ¼ tan�1 sinh xð Þ; (b) F2 xð Þ ¼ 2 tan�1 exð Þ;(c) F3 xð Þ ¼ sin�1 tanh xð Þ; (d) F4 xð Þ ¼ 2 tan�1 tanh 1

2x

� ��

:

4. Let In ¼R e

1x loge xð Þndx, for n � 0:

(a) Prove that In ¼ 12

e2 � 12

nIn�1, for n� 1.

(b) Evaluate I0, I1, I2 and I3.

5. Evaluate each of the following integrals, using the suggested substitution

where given:

(a)R p

2

0tan sin xð Þ cos xdx; (b)

R 1

0

tan�1 xð Þ21þx2 dx ðtry u ¼ tan�1 xÞ;

(c)R 1

0x5 þ 2x2� �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x6 þ 4x3 þ 4p

dx;

(d)R p

2

0dx

2þcos xdx try u¼ tan 1

2x

� �

and use the identity cos x¼ 1� tan2 12xð Þ

1þ tan2 12xð Þ

� �

;

(e)R e

18x7 loge xdx; (f)

R e2

eloge loge xð Þ

xdx;

(g)R p

2

0sin 2xð Þ

1þ3 cos2 xdx; (h)

R 4

1dx

1þxð Þffiffi

xp ðtry u¼

ffiffiffi

xpÞ:

This is known as theCauchy–Schwarz Inequalityfor integrals.

You met this in Sub-section 1.3.3.

7.6 Exercises 311

Section 7.4

1. Prove the following inequalities:

(a)R 1

0x3

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 1þ x99ð Þp

dx � 12; (b)

R 1

0x4ffiffiffiffiffiffiffiffiffiffiffi

1þ3x97p dx � 1

10:

2. Prove that 12�R 1

01þx30

2�x99 dx � 2:

3. Prove thatR 2

0

x2 x�3ð Þ sin 99xð Þ1þx20 dx

�

�

�

�

�

�� 4:

4. Show thatR

dx

x loge xð Þ32

¼ �2 loge xð Þ�12, and deduce that

P

1

n¼2

1

n loge nð Þ32

is

convergent.

5. Show thatR

dx

x loge xð Þ12

¼ 2 loge xð Þ12, and deduce that

P

1

n¼2

1

n loge nð Þ12

is divergent.

6. Prove that limn!1

n 1

nþ1ð Þ2 þ1

nþ2ð Þ2 þ � � � þ1

2nð Þ2� �

¼ 12:

7. Determine the convergence or divergence of the following series:

(a)P

1

n¼2

1

loge nð Þloge n; (b)P

1

n¼2

1

loge nð Þloge loge nð Þ:

8. Prove thatP

n

k¼2

loge kk� 1

2loge nð Þ2, as n!1.

9. Prove that, if f is integrable on [a, b], where a< b, and f(x)> 0, for all x2 [a, b],

thenR b

af > 0:

Hint: Use the following steps:

(a) Assume that the result is false, so that in factR b

af ¼ 0;

(b) Verify that, if c; d½ � a; b½ , thenR d

cf ¼ 0;

(c) Prove that there is a partition P of [a, b] with U( f, P)� b� a, and

deduce that there is some subinterval [a1, b1] of [a, b] with a1< b1 and

supa1,b1½

f � 1;

(d) Using the fact thatR b1

a1f ¼ 0, prove that there is a sequence

an, bn½ f g1n¼ 1 of intervals with anþ1, bnþ1½ an, bn½ and supan,bn½

f � 1n;

(e) Prove that there is some point c in [a, b] such that an! c as n!1, and

that c2 [an, bn], for all n� 1;

(f) Verify that f(c)¼ 0.

Section 7.5

1. Prove that limn!1

3nð Þ!n3n

� �1n¼ 27

e3 :

2. Use Stirling’s Formula to estimate each of the following numbers to two

significant figures:

(a)�

400

200

�

� 1

2

� �400

; (b)400!


800pð Þp

100!ð Þ44400:

3. Use Stirling’s Formula to determine the number l such that

8nð Þ!2nð Þ!ð Þ4

� l216n

n32

as n!1:

312 7: Integration

8 Power series

The evaluation of given functions at given points in their domains is of great

importance. If we are dealing with a polynomial function, the calculation of the

function’s values presents no problem: it is simply a matter of arithmetic. For

example, if

f xð Þ ¼ 1þ 1

2x� 1

2x2 � 1

6x3þ 1

4x4;

then

f 1ð Þ ¼ 1þ 1

2� 1

2� 1

6þ 1

4¼ 13

12:

On the other hand, the sine function is rather different: there is no way of

calculating its values precisely merely by the use of arithmetic.

It is important to be able to estimate values of functions that cannot be

evaluated exactly, and to know how close the estimates are to the actual values

of the function.

In this chapter we are primarily concerned with a procedure for calculating

approximate values of functions, like the sine function, whose values cannot be

calculated easily at all points in their domains. We see how, in principle, we

can use a certain sequence of polynomials to calculate the values of the sine

function, for example, to any desired degree of accuracy; and how we can

represent sin x, for any x, as the sum of a series. We shall see, for instance, that

the polynomial p xð Þ ¼ x� 16

x3 approximates f (x)¼ sin x to within 10�5 for all

x in the interval [0, 0.1], and that, in general

sin x ¼ x� x3

3!þ x5

5!� x7

7!þ � � �; for x 2 R :

In Section 8.1 we define the Taylor polynomial Tn(x) and discuss some

particular examples of functions for which the Taylor polynomials appear to

provide useful approximations.

In Section 8.2 we investigate how closely Taylor polynomials approximate a

given function in the neighbourhood of a point a in its domain, and we establish

a criterion for when we can say that

f xð Þ ¼ limn!1

Tn xð Þ ¼X

1

n¼0

an x� að Þn:

In this case, we say that f is the sum function of the power seriesP

1

n¼0

an x� að Þn.

In Sections 8.3 and 8.4 we look at the behaviour of power series in their own

right; that is, we consider functions which are defined by power series. In

particular, in Section 8.3 we see that a power seriesP

1

n¼0

an x� að Þn behaves in

one of three ways: it converges only for x¼ a, or it converges for all x, or it

Section 8.2, Example 1, andTheorem 3.

313

converges in an interval (a�R, aþR), where R is a positive number called the

radius of convergence.

In Section 8.4 we discuss various rules for power series, including the Sum,

Multiple and Product Rules. We also find that it is valid to differentiate or

integrate a given power series term-by-term, and that this does not affect the

radius of convergence.

Finally, in Section 8.5 we look briefly at various methods for estimating the

number p, and prove that p is irrational.

Some of the proofs in this chapter are not particularly illuminating, and you

may wish to leave them till a second reading of the chapter. You will need a

calculator handy while you work through Sections 8.1 and 8.2.

8.1 Taylor polynomials

8.1.1 What are Taylor polynomials?

Let f be a continuous function defined on an open interval I containing the

point a. It follows from the definition of continuity that

limx!a

f xð Þ ¼ f að Þ:

Thus we can write

f xð Þ ’ f að Þ; for x near a:

In geometrical terms, this means that we can approximate the graph y¼ f(x)

near a by the horizontal line y¼ f(a) through the point (a, f(a)). For most

continuous functions, this does not give a very good approximation.

However, if the function is differentiable on I, then we can obtain a better

approximation by using the tangent line through (a, f(a)) instead of the

horizontal line. The tangent to the graph at (a, f(a)) has equation

y� f að Þx� a

¼ f 0 að Þ

or

y ¼ f að Þ þ f 0 að Þ x� að Þ;so, for x near a, we can write

f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ:This approximation is called the tangent approximation to f at a.

Notice that the function f and the approximating polynomial

f að Þ þ f 0 að Þ x� að Þhave the same value at a and the same first derivative at a.

Example 1 Determine the tangent approximation to the function f(x)¼ ex at 0.

Solution Here

f xð Þ ¼ ex; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ ex; f 0 0ð Þ ¼ 1:

That is, I is a neighbourhoodof a.

We can think of the tangent at(a, f(a)) as the line of bestapproximation to the graphnear a.

314 8: Power series

Hence the tangent approximation to f at 0 is

ex’ f 0ð Þ þ f 0 0ð Þ x� 0ð Þ ¼ 1þ x: &

Problem 1 Determine the tangent approximation to each of the

following functions f at the given point a:

(a) f xð Þ ¼ ex; a ¼ 2; (b) f xð Þ ¼ cos x; a ¼ 0:

So far we have seen two approximations to f(x) for x near a:

f xð Þ ’ f að Þ ða constant functionÞ;f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ ða linear functionÞ:

If the function f is twice differentiable on a neighbourhood I of a, then we can

find an even better approximation to f(x) by considering the expression

f xð Þ � f að Þ þ f 0 að Þ x� að Þf gx� að Þ2

:

Let

F xð Þ ¼ f xð Þ � f að Þ þ f 0 að Þ x� að Þf g; for x 2 I;

G xð Þ ¼ x� að Þ2; for x 2 I:

Then F and G are differentiable on I, and

F0 xð Þ ¼ f 0 xð Þ � f 0 að Þ;G0 xð Þ ¼ 2 x� að Þ:

Also, F að Þ ¼ G að Þ ¼ 0:Hence, by l’Hopital’s Rule, it follows that

limx!a

F xð ÞG xð Þ exists and equals lim

x!a

F0 xð ÞG0 xð Þ ;

provided that this latter limit exists.

Now

limx!a

F0 xð ÞG0 xð Þ ¼ lim

x!a

f 0 xð Þ � f 0 að Þ2 x� að Þ ;

and, since the function f is twice differentiable at a, this limit exists and equals12

f 00 að Þ. Hence

limx!a

f xð Þ � f að Þ þ f 0 að Þ x� að Þf gx� að Þ2

exists and equals1

2f 00 að Þ:

We can reformulate this result as follows: for x near a

f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ þ 1

2f 00 að Þ x� að Þ2 ða quadratic functionÞ:

Notice that the function f and the approximating polynomial f að Þþf 0 að Þ x� að Þ þ 1

2f 00 að Þ x� að Þ2 have the same value at a and the same

first and second derivatives at a.



This suggests that, if the function is n-times differentiable on I, then we may

be able to find a better approximating polynomial of degree n whose value at a

and whose first n derivatives at a are equal to those of f. This leads to the

following definition.

Definition Let f be n-times differentiable on an open interval containing

the point a. Then the Taylor polynomial of degree n for f at a is the

polynomial

Tn xð Þ ¼ f að Þ þ f 0 að Þ1!

x� að Þ

þ f 00 að Þ2!

x� að Þ2þ � � � þ f ðnÞ að Þn!

x� að Þn:

Notice that

Tn að Þ ¼ f að Þ; T 0n að Þ ¼ f 0 að Þ; . . .; T nð Þn að Þ ¼ f nð Þ að Þ;

that is, f and Tn have the same value at a and have equal derivatives at a of all

orders up to and including n.

Example 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x) for the

function f(x)¼ sin x at each of the following points:

(a) a¼ 0; (b) a ¼ p2�

Solution Here

f xð Þ ¼ sin x; f 0ð Þ ¼ 0; fp2

� �

¼ 1;

f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1; f 0p2

� �

¼ 0;

f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0; f 00p2

� �

¼ �1;

f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1; f 000p2

� �

¼ 0:

Hence:

(a) T1 xð Þ ¼ x; T2 xð Þ ¼ x and T3 xð Þ ¼ x� x3

3! ;

(b) T1 xð Þ ¼ 1; T2 xð Þ ¼ 1� 12

x� p2

� �2and T3 xð Þ ¼ 1� 1

2x� p

2

� �2: &

Problem 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x)

for each of the following functions f at the given point a:

(a) f xð Þ ¼ ex; a ¼ 2; (b) f xð Þ ¼ cos x; a ¼ 0:

Problem 3 Determine the Taylor polynomial of degree 4 for each of

the following functions f at the given point a:

(a) f xð Þ ¼ 7� 6xþ 5x2 þ x3; a ¼ 1; (b) f xð Þ ¼ 11�x

; a ¼ 0;

(c) f xð Þ ¼ loge 1þ xð Þ; a ¼ 0; (d) f xð Þ ¼ sin x; a ¼ p4

;

(e) f xð Þ ¼ 1þ 12

x� 12

x2 � 16

x3 þ 14

x4; a ¼ 0:

Problem 4 Determine the percentage error involved in using the

Taylor polynomial of degree 3 for the function f (x)¼ tan x at 0 to eva-

luate tan 0.1. (Use your calculator to evaluate tan 0.1.)

Strictly speaking, we shoulduse more complicatednotation to indicate that theTaylor polynomial Tn(x)depends on n, a and f.

We do not usually multiplyout such brackets, since thatwould make the results lessclear.

316 8: Power series

8.1.2 Approximation by Taylor polynomials

We now look at some specific examples of functions to investigate the asser-

tion that Taylor polynomials provide good approximations for a large class of

functions.

The function f ðxÞ ¼ 1þ 12 x � 1

2 x2 � 16 x3 þ 1

4 x4

It follows from the result of Problem 3(e) above that the Taylor polynomials of

degrees 1, 2, 3 and 4 for f at 0 are:

T1 xð Þ ¼ 1þ 1

2x; T2 xð Þ ¼ 1þ 1

2x� 1

2x2;

T3 xð Þ ¼ 1þ 1

2x� 1

2x2� 1

6x3; and T4 xð Þ ¼ 1þ 1

2x� 1

2x2� 1

6x3þ 1

4x4:

Since f nð Þ 0ð Þ ¼ 0 for n� 5, it follows that for n� 5 the Taylor polynomial of

degree n for f at 0 is just the same as the Taylor polynomial of degree 4 for f at 0.

The graphs of these Taylor polynomials are as follows:

In this case, the polynomials T2(x) and T3(x) provide good approximations to

f(x) near 0, and Tn(x)¼ f(x) for all n� 4.

The function f(x)¼ sin x

By calculating higher derivatives of the function f(x)¼ sin x at 0, we can show

that the following are Taylor polynomials for f at 0

T1 xð Þ ¼ T2 xð Þ ¼ x; T3 xð Þ ¼ T4 xð Þ ¼ x� x3

3!;

T5 xð Þ ¼ T6 xð Þ ¼ x� x3

3!þ x5

5!; T7 xð Þ ¼ T8 xð Þ ¼ x� x3

3!þ x5

5!� x7

7!:

In general, if f is a polynomialof degree N, then

Tn xð Þ ¼ f xð Þ for all n � N:


The following graphs illustrate how the approximation to f(x) given by Tn(x)

gets better as n increases.

For example, the graph of T5 appears to be very close to the graph of sine

over the interval �p2

, p2

� �

, so that sin x and T5(x) do not differ by very much for

values of x in this interval. Thus T5(x) seems to be a good approximation to

sin x in this interval.

It appears that, as the degree of the Taylor polynomial increases, so its

graph becomes a good approximation to that of sine over more and more of

R . For instance, in the above diagrams the shaded area covers the interval of the

x-axis on which the Taylor polynomial Tn(x) agrees with sin x to three decimal

places.

The function f ðxÞ ¼ 11�x

By repeated differentiation it is easy to verify that, for k¼ 1, 2, . . .

f kð Þ xð Þ ¼ k!

1� xð Þkþ1;

thus, in particular, f kð Þ 0ð Þ ¼ k!.Hence the Taylor polynomial of degree n for f at 0 is

Tn xð Þ ¼X

n

k¼0

f kð Þ 0ð Þk!

xk

¼X

n

k¼0

xk ¼ 1þ xþ x2 þ � � � þ xn:

The following diagram shows the graphs of the Taylor polynomials for f at 0 of

degrees 1, 2, 4 and 7.

318 8: Power series

For jxj � 1, the sequence{Tn(x)} does not converge,and so cannot provide anapproximation to f(x).

The graphs show that the nature of the approximation is very different from

the previous examples. For sine, the interval over which the approximation is

good seems to expand indefinitely as the degree of the polynomials increases;

but for f ðxÞ ¼ 11�x

the interval of good approximation always seems to be

contained in the interval (�1, 1).

Notice, however, that for this function f, the Taylor polynomials Tn(x) are

just the nth partial sums of the geometric seriesP

1

k¼0

xk; this series converges

with sum 11�x

for jxj< 1, and diverges for jxj � 1. It follows that, if jxj< 1, then

Tn xð Þ ! f xð Þ as n!1;

so that, if jxj< 1, the polynomials Tn(x) do provide better and better approx-

imations to f(x) as n increases.

Which functions can be approximated by Taylorpolynomials?

In the next section we obtain a criterion for determining those functions f for

which the Taylor polynomials provide useful approximating polynomials, and

the intervals on which the approximation occurs. We also introduce certain

basic power series which correspond to the functions in the following problem.

Problem 5 Determine the Taylor polynomial of degree n for each of

the following functions at 0:

(a) f xð Þ ¼ 11�x

; (b) f xð Þ ¼ loge 1þ xð Þ; (c) f xð Þ ¼ ex;

(d) f xð Þ ¼ sin x; (e) f xð Þ ¼ cos x:

8.2 Taylor’s Theorem

8.2.1 Taylor’s Theorem and approximation

In Section 8.1 we demonstrated how to find the Taylor polynomial Tn(x) of

degree n for a function f at a point a. This polynomial and its first n derivatives

agree with f and its first n derivatives at a, and the polynomial appears to

approximate f at points near a. But how good an approximation is it? What

error is involved if we replace f by a Taylor polynomial at a? The answer to

these questions is given by the following important result.

Theorem 1 Taylor’s Theorem

Let f be (nþ 1)-times differentiable on an open interval containing the

points a and x. Then

f xð Þ ¼ f að Þ þ f 0 að Þ x� að Þ

þ f 00 að Þ2!

x� að Þ2þ � � � þ f nð Þ að Þn!

x� að ÞnþRn xð Þ;

where

Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;

and c is some point between a and x.

Remarks

1. When n¼ 0, Taylor’s Theorem reduces to the assertion

f xð Þ ¼ f að Þ þ f 0 cð Þ x� að Þ;which we can rewrite (for x 6¼ a) in the form

f xð Þ � f að Þx� a

¼ f 0 cð Þ; for some c between a and x:

But this is just the Mean Value Theorem! It follows that Taylor’s Theorem

can be considered as a generalisation of the Mean Value Theorem.

2. The result of Theorem 1 can be expressed in the form

f xð Þ ¼ Tn xð Þ þ Rn xð Þ;where Rn(x) is thought of as a ‘remainder term’ or ‘error term’ involved in

approximating f(x) by the estimate Tn(x).

Strictly speaking, we shoulduse more complicatednotation to indicate that theremainder term Rn(x) dependson n, a and f.

320 8: Power series

Proof For simplicity, we assume that x> a; the proof is similar if x< a.

We use the auxiliary function

h tð Þ ¼ f tð Þ � Tn tð Þ � A t � að Þnþ1; t 2 a; x½ �; (1)

where Tn is the Taylor polynomial of degree n for f at a, and A is a constant

chosen so that

h að Þ ¼ h xð Þ: (2)

Now

f að Þ ¼ Tn að Þ; f 0 að Þ ¼ T 0n að Þ; . . .; and f nð Þ að Þ ¼ T nð Þn að Þ;

so that

h að Þ ¼ 0; h0 að Þ ¼ 0; . . .; and h nð Þ að Þ ¼ 0:

The function h is continuous on the closed interval [a, x] and differentiable on

the open interval (a, x); also, h(a)¼ h(x). Hence, by Rolle’s Theorem, there

exists some number c1 between a and x for which

h0 c1ð Þ ¼ 0:

Next, we apply Rolle’s Theorem to the function h0 on the interval [a, c1]. The

function h0 is continuous on [a, c1] and differentiable on (a, c1); also,

h0(a)¼ h0(c1)¼ 0. Hence, by Rolle’s Theorem, there exists some number c2

between a and c1 for which

h00 c2ð Þ ¼ 0:

In turn, we apply Rolle’s Theorem to the functions

h00; h000; . . .; h nð Þ

on the intervals

a; c2½ �; a; c3½ �; . . . ; a; cn½ �;where c2 > c3 > c4 > � � � > cn > a:

At the last stage, we find that there exists some point c between a and cn for

which

h nþ1ð Þ cð Þ ¼ 0: (3)

By differentiating equation (1) (nþ 1) times, we obtain

h nþ1ð Þ tð Þ ¼ f nþ1ð Þ tð Þ � A nþ 1ð Þ!: (4)

From (3) and (4), we deduce that

0 ¼ f nþ1ð Þ cð Þ � A nþ 1ð Þ!;so that

A ¼ f nþ1ð Þ cð Þnþ 1ð Þ! : (5)

Finally, it follows from equations (1) and (2), with h(a)¼ 0, that

f xð Þ ¼ Tn xð Þ þ A x� að Þnþ1:

Hence, by equation (5)

f xð Þ ¼ Tn xð Þ þ f nþ1ð Þ cð Þnþ 1ð Þ! x� að Þnþ1;

as required. &


This choice of A is made sothat we can apply Rolle’sTheorem to h on [a, x].

Problem 1 Obtain an expression for R1(x) when Taylor’s Theorem is

applied to the function f xð Þ ¼ 11�x

at a¼ 0. Calculate the value of c

when x ¼ 34.

Problem 2 What can you say about Rn(x) when f is a polynomial of

degree at most n?

Problem 3 By applying Taylor’s Theorem to the function f (x)¼ cos x

at a¼ 0, prove that cos x ¼ 1� 12

x2 þ R3 xð Þ, x 2 R , where

R3 xð Þj j � 124

x4:

In most applications of Taylor’s Theorem, we do not know the value of c

explicitly. However, for many purposes this does not matter, since we can

show that jRn(x)j is small by finding an estimate for f nþ1ð Þ cð Þ�

�

�

� which is valid

for all c between a and x, and then applying the following result.

Corollary 1 Remainder Estimate

Let f be (n+1)-times differentiable on an open interval containing the points

a and x. If

f nþ1ð Þ cð Þ�

�

�

� � M;

for all c between a and x, then

f xð Þ ¼ Tn xð Þ þ Rn xð Þ;

where

Rn xð Þj j � M

nþ 1ð Þ! x� aj jnþ1:

Proof From Taylor’s Theorem, we have


where c is some point between a and x. It follows that

Rn xð Þj j ¼f nþ1ð Þ cð Þ�

�

�

�

nþ 1ð Þ! x� aj jnþ1

� M

nþ 1ð Þ! x� aj jnþ1:&

Example 1 By applying the Remainder Estimate to the function f (x)¼ sin x,

with a¼ 0 and n¼ 3, calculate sin 0.1 to four decimal places.

Solution Here

f xð Þ ¼ sin x; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0;

f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1:

Strictly speaking, we shoulduse more complicatednotation to indicate that theupper bound M depends on n,a and f.

322 8: Power series

Hence the Taylor polynomial of degree 3 for f at 0 is

T3 xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ f 00 0ð Þ2!

x2 þ f 000 0ð Þ3!

x3

¼ x� 1

6x3:

Also, f 4ð Þ xð Þ ¼ sin x, so that

f 4ð Þ cð Þ�

�

�

� ¼ sin cj j � 1; for c 2 R :

Taking M¼ 1 in the Remainder Estimate, we deduce that

R3 0:1ð Þj j � 1

4!� 0:1ð Þ4

¼ 1

24� 10�4

<1

2� 10�5:

Now

sin 0:1 ’ T3 0:1ð Þ

¼ 0:1� 1

6� 10�3

¼ 0:1� 0:0001666 . . .

¼ 0:0998333 . . .:

It follows from the Remainder Estimate that T3(0.1) gives an estimate for sin 0.1

to four decimal places. Hence sin 0.1¼ 0.0998 (to four decimal places). &

Problem 4 By applying the Remainder Estimate to the function

f xð Þ ¼ loge 1þ xð Þ, with a¼ 0 and n¼ 2, calculate loge 1:02 to four

decimal places.

In many practical situations we do not know how many terms of the power

series are needed in order to calculate the value of a given function to a

prescribed number of decimal places. In such cases, we can use the

Remainder Estimate to determine how many terms are needed.

Example 2 By applying the Remainder Estimate to the function f (x)¼ ex,

with a¼ 0, calculate the value of e to three decimal places.

Solution Since f kð Þ xð Þ ¼ ex; for k ¼ 0; 1; . . .; we have

f kð Þ 0ð Þ ¼ 1; for k ¼ 0; 1; . . .:

It follows that, for each n

Tn xð Þ ¼ 1þ xþ x2

2!þ � � � þ xn

n!:

Also, f nþ1ð Þ xð Þ ¼ ex, for all x, so that, for all c2 (0,1), we have

f nþ1ð Þ cð Þ�

�

�

� � e < 3:

It follows from the Remainder Estimate, with x¼ 1 and M¼ 3, that

Rn 1ð Þj j � 3

nþ 1ð Þ!� 1nþ1:

We proved that e< 3 in Sub-section 2.5.3.

To calculate e to three decimal places, we must choose n so that

3

nþ 1ð Þ! < 2� 10�4; or 15;000 < nþ 1ð Þ!:

Since 7!¼ 5,040 and 8!¼ 40,320, we may safely choose n¼ 7.

It follows that

e ’ T7 1ð Þ

¼ 1þ 1

1!þ 1

2!þ 1

3!þ 1

4!þ 1

5!þ 1

6!þ 1

7!

¼ 1þ 1þ 0:5þ 0:16þ 0:1416þ 0:0083þ 0:00138

þ 0:000198412 . . .

¼ 2:7182 . . .:

Hence, e¼ 2.718 (to three decimal places). &

Problem 5 By applying the Remainder Estimate to the function f(x)¼cos x, with a¼ 0, calculate cos 0.2 rounded to four decimal places.

Our next example illustrates how to obtain an approximation valid over an

interval.

Example 3 Calculate the Taylor polynomial T3(x) for f xð Þ ¼ 1xþ2

at 1.

Show that T3(x) approximates f (x) with an error less than 5� 10�3 on the

interval [1, 2].

Solution Here

f xð Þ ¼ 1

xþ 2; f 1ð Þ ¼ 1

3;

f 0 xð Þ ¼ �1

xþ 2ð Þ2; f 0 1ð Þ ¼ � 1

9;

f 00 xð Þ ¼ 2

xþ 2ð Þ3; f 00 1ð Þ ¼ 2

27;

f 000 xð Þ ¼ �6

xþ 2ð Þ4; f 000 1ð Þ ¼ � 2

27:

Hence the Taylor polynomial of degree 3 for f at 1 is

T3 xð Þ ¼ 1

3� 1

9x� 1ð Þ þ 1

27x� 1ð Þ2� 1

81x� 1ð Þ3:

Also, f 4ð Þ xð Þ ¼ 24

xþ2ð Þ5, so that

f 4ð Þ cð Þ�

�

�

� � 24

35; for c 2 1; 2ð Þ:

Taking M ¼ 2435 in the Remainder Estimate, we deduce that

R3 xð Þj j � 24

35� 2� 1ð Þ4

4!

¼ 1

35¼ 0:0041 . . .; for x 2 1; 2½ �:

Since the remainder term is less than 0.005, it follows that T3(x) approximates

f (x) with an error less than 5� 10�3 on [1, 2]. &

324 8: Power series

Problem 6 Calculate the Taylor polynomial T4(x) for f (x)¼ cos x at p.

Show that T4(x) approximates f (x) with an error less than 3� 10�3 on the

interval 34p; 5

4p

� �

.

8.2.2 Taylor’s Theorem and power series

From Taylor’s Theorem, we know that, if a function f can be differentiated

as often as we please on an open interval containing the points a and x, then,

for any n

f xð Þ ¼ Tn xð Þ þ Rn xð Þ

¼X

n

k¼0

f kð Þ að Þk!

x� að Þk þ Rn xð Þ;

where


for some c between a and x. It follows that, if Rn(x)! 0 as n!1, then we can

express f (x) as a power series in (x� a).

Theorem 2 Let f have derivatives of all orders on an open interval con-

taining the points a and x. If Rn(x)! 0 as n!1, then

f xð Þ ¼X

1

n¼0

f nð Þ að Þn!

x� að Þn:

We call f the sum function of the power seriesP

1

n¼0

f nð Þ að Þn! x� að Þn, and we call

this power series the Taylor series for f at a.

Warning A dramatic example of what happens when the remainder term

does not tend to zero is given by the function

f xð Þ ¼ e�1

x2 ; x 6¼ 0;0; x ¼ 0:

For this function, f (0)¼ 0, f 0(0)¼ 0, f 00(0)¼ 0, . . .. Thus the Taylor polyno-

mial Tn(x) is identically zero for each n, although the function f is not identi-

cally zero. In this case, Rn(x)¼ f (x) for all n, and so the Taylor series for f at 0

X

1

n¼0

f nð Þ 0ð Þn!

xn ¼ 0þ 0xþ 0x2 þ � � �;

converges to f (x) only at 0!

We can use Theorem 2 to obtain the following basic power series.

Theorem 3 Basic power series

(a) 11�x¼1þ xþ x2 þ x3 þ � � � ¼

P

1

n¼0

xn; for xj j < 1;

(b) loge 1þ xð Þ ¼ x� x2

2þ x3

3� � � � ¼

P

1

n¼1

�1ð Þnþ1xn

n; for xj j < 1;

–3 –2 –1 1

1

2 3 x

y

y = e –1/x2

0

We indicate a proof of thisassertion in Section 8.6,Exercise 6 on Section 8.2.

(c) ex ¼ 1þ xþ x2

2! þ x3

3!þ � � � ¼P

1

n¼0

xn

n!; for x 2 R ;

(d) sin x ¼ x� x3

3! þ x5

5! � � � � ¼P

1

n¼0

�1ð Þn x2nþ1

2nþ1ð Þ!; for x 2 R ;

(e) cos x ¼ 1� x2

2! þ x4

4! � � � � ¼P

1

n¼0

�1ð Þn x2n

2nð Þ!; for x 2 R :

Proof

(a) Here we use the fact that, for x 6¼ 1

1

1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ xnþ1

1� x:

But the sequence xnþ1

1�x

n o

is null, if jxj< 1. Thus, by letting n!1, we

obtain

1

1� x¼X

1

n¼0

xn; for xj j < 1:

(b) Similarly

1

1þ t¼1� t þ t2 � � � � þ �1ð Þntnþ �1ð Þnþ1

tnþ1

1þ t:

Integrating both sides from 0 to x, where jxj< 1, we obtain

Z x

0

dt

1þ t¼Z x

0

1� t þ t2 � � � � þ �1ð Þntn þ �1ð Þnþ1t nþ1

1þ t

!

dt;

so that

loge 1þ tð Þ½ �x0¼ t � t2

2þ t3

3� � � � þ �1ð Þn t nþ1

nþ 1

�x

0

þ �1ð Þnþ1

Z x

0

t nþ1

1þ tdt:

Hence

loge 1þ xð Þ¼ x� x2

2þ x3

3� � � � þ �1ð Þn xnþ1

nþ 1

þ �1ð Þnþ1

Z x

0

tnþ1

1þ tdt:

When 0� x< 1, we have, by the Inequality Rule for integrals

�1ð Þnþ1

Z x

0

tnþ1

1þ tdt

�

�

�

�

�

�

�

�

¼Z x

0

tnþ1

1þ tdt �

Z x

0

tnþ1dt

¼ tnþ2

nþ 2

�x

0

¼ xnþ2

nþ 2� 1

nþ 2! 0 as n!1:

When�1< x< 0, we put T¼�t and X¼�x, so that 0< T< 1. Then

This identity is easily provedby multiplying both sides by1� x:


326 8: Power series

�1ð Þnþ1

Z x

0

tnþ1

1þ tdt

�

�

�

�

�

�

�

�

¼Z X

0

Tnþ1

1� TdT � 1

1� X

Z X

0

Tnþ1dT

¼ Xnþ2

1� Xð Þ nþ 2ð Þ �1

1� Xð Þ nþ 2ð Þ ! 0 as n!1:

Combining these two results, we deduce that

loge 1þ xð Þ ¼X

1

n¼1

�1ð Þnþ1 xn

n; for�1 < x < 1:

(c) This was one of our definitions of the exponential function.

(d) Let f (x)¼ sin x. Then, by Taylor’s Theorem, we have sin x ¼ Tn xð ÞþRn xð Þ, where

Rn xð Þ ¼ f nþ1ð Þ cð Þnþ 1ð Þ! x nþ1;

for some number c between 0 and x. We saw earlier that

f nþ1ð Þ cð Þ ¼ � sin c or � cos c;

so that, in particular, we can be sure that f nþ1ð Þ cð Þ�

�

�

� � 1. It follows from

the Remainder Estimate, with M¼ 1, that

Rn xð Þj j � 1

nþ 1ð Þ! xj jnþ1! 0 as n!1;

so that, in particular, Rn(x)! 0 as n!1.

Hence, by letting n!1 in the equation sin x ¼ Tn xð Þ þ Rn xð Þ, we obtain

sin x ¼ x� x3

3!þ x5

5!� x7

7!þ � � � ¼

X

1

n¼0

�1ð Þn x2nþ1

2nþ 1ð Þ!; for x 2 R :

(e) The proof is similar to that of part (d), so we omit it. &

Remark

Probably the first definitions of sin x and cos x that you met were expressed in

terms of a right-angled triangle, but those definitions only make sense for

x 2 0; p2

� �

. We have now shown that sin x and cos x can be represented by the

power series in parts (d) and (e) of Theorem 3, and we can use these power

series to define sin x and cos x for all x2R .

Finally, we use Taylor’s Theorem to prove an interesting limit which is a

generalisation of two limits that you met earlier

limn!1

1þ 1

n

� n

¼ e and limn!1

1þ x

n

� �n

¼ ex; for any x 2 R :

Example 4

(a) Calculate T1(x) and R1(x) for the function f xð Þ ¼ loge 1þ xð Þ; x 2 �1; 1ð Þ,at 0.

(b) By replacing x in part (a) by �x, where jxj> j�j, prove that, for any real

numbers � and �

For, 0<X< 1 and

11�T� 1

1�Xfor T2 [0, X].

See Sub-section 3.4.3.

Problem 5(d),Sub-section 8.1.2.

Sub-section 2.5.3.

limx!1

1þ �x

� ��x

¼ e��:

(c) Deduce from part (a) that

n loge 1þ 1

n

�

! 1 as n!1:

Solution

(a) For the function f xð Þ ¼ loge 1þ xð Þ, x 2 �1; 1ð Þ, we have

f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ 1

1þ x; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ �1

1þ xð Þ2:

Hence

T1 xð Þ ¼ f 0ð Þ þ f 0 0ð Þx ¼ x and R1 xð Þ ¼ f 00 cð Þ2!

x2 ¼ �x2

2 1þ cð Þ2;

for some number c between 0 and x.

(b) Replacing x by �x

in the equation

loge 1þ xð Þ ¼ x� x2

2 1þ cð Þ2; for xj j < 1;

we obtain

loge 1þ �x

� �

¼ �x� �2

2 1þ cð Þ2x2; for xj j > �j j:

Hence

�x loge 1þ �x

� �

¼ �� 2�

2 1þ cð Þ2x; for xj j > �j j:

Since 1x! 0 as x!1, it follows that

limx!1

�x loge 1þ �x

� �

¼ ��;

so that

limx!1

loge 1þ �x

� ��x

¼ ��:

Since the exponential function is continuous on R , and

exp loge 1þ �x

� ��x�

¼ 1þ �x

� ��x

;

it follows, from the Composition Rule for limits, that

limx!1

1þ �x

� ��x

¼ e��:

(c) For x2 (�1, 1), we have, from part (a), that loge 1þ xð Þ ¼ x� x2

2 1þcð Þ2,for some number c between 0 and x. Putting x ¼ 1

n, we obtain

loge 1þ 1

n

�

¼ 1

n� 1

2n2 1þ cnð Þ2;

for some number cn between 0 and 1n.

328 8: Power series

Hence

n loge 1þ 1

n

�

¼ 1� 1

2n 1þ cnð Þ2

! 1 as n!1;

since

0 � 1

2n 1þ cnð Þ2� 1

2n

and 12n

� �

is a null sequence. &

8.3 Convergence of power series

8.3.1 The radius of convergence

In the previous section you saw that certain standard functions can be

expressed as the sum functions of power series; for example

1

1� x¼X

1

n¼0

xn and loge 1þ xð Þ ¼X

1

n¼1

�1ð Þnþ1xn

n; for xj j < 1:

These power series are the Taylor series for the given functions at 0.

Conversely, power series can be used to define functions. Thus, for instance,we defined the exponential function x 7! ex, x2R , to be the sum function for

the power seriesP

1

n¼0

xn

n!.

But there are many other functions which are defined as the sum functions of

power series (as distinct from a power series obtained as the Taylor series for a

given function). For example, the Bessel function

J0 xð Þ ¼X

1

n¼0

�1ð Þn x2

� �2n

n!ð Þ2; for x 2 R ;

arises in analyses of the vibrations of a circular drum and of the radiation from

certain types of radio antenna.

However, all such uses of power series depend on knowledge of those

numbers x for which a power seriesP

1

n¼0

an x� að Þn converges. In each of the

above examples, the series converges on an interval; indeed, all power series

have this property.

Theorem 1 Radius of Convergence Theorem

For a given power seriesP

1

n¼0

an x� að Þn, precisely one of the following

possibilities occurs:

(a) The series converges only when x¼ a;

(b) The series converges for all x;

Section 3.4.

See Exercise 6 on Section 8.4,in Section 8.6.

We give the proof ofTheorem 1 in Sub-section 8.3.3.

(c) There is a number R> 0, called the radius of convergence of the power

series, such that the series converges if jx� aj<R and diverges if

jx� aj>R.

For example:

(a)P

1

n¼0

n!xn converges only when x¼ 0;

(b)P

1

n¼0

xn

n! converges for all x;

(c)P

1

n¼0

xn converges if jxj< 1 and diverges if jxj> 1 – its radius of conver-

gence is 1.

Remarks

1. Sometimes we abuse our notation by writing R¼ 0, if a series converges

only for x¼ a, or R¼1, if a series converges for all x.

2. It is important to remember that Theorem 1, part (c), makes no assertion

about the convergence or divergence of the power series at the end-points

a�R, aþR of the interval (a�R, aþR).

For example, we shall see shortly that the three power series

X

1

n¼1

xn;X

1

n¼1

xn

nand

X

1

n¼1

xn

n2

all have radius of convergence 1. However, the intervals on which these

series converge are, respectively, (�1, 1), [�1, 1) and [�1, 1].

The interval of convergence of the power series is the interval (a�R, aþR),

together with any end-points of the interval at which the series converges.

The following diagram illustrates the various types of interval of conver-

gence ofP

1

n¼0

an x� að Þn:

Theorem 1 is not concerned with the problem of evaluating the radius of con-

vergence of a power series. However, a power series is simply a particular type of

series, and so all the convergence tests for series can be applied to power series.

In practice, we can tackle most commonly arising power series by using the

following version of the Ratio Test.

Theorem 2 Ratio Test for Radius of Convergence

Suppose thatP

1

n¼0

an x� að Þn is a given power series, and that

�

�

�

anþ1

an

�

�

�! L; as n!1:

(a) If L is1, the series converges only for x¼ a.

For all non-zero x, the seriesP

1

n¼0

n!xn diverges, by the Non-

null Test, as�

1n!xn

�

¼n 1

xð Þn

n!

o

is

a basic null sequence.

Example 2.

Sections 3.1–3.3.

We give the proof ofTheorem 2 in Sub-section 8.3.3.

R¼ 0.

330 8: Power series

(b) If L¼ 0, the series converges for all x.

(c) If L> 0, the series has radius of convergence 1L:

Example 1 Determine the radius of convergence of each of the following

power series:

(a)P

1

n¼1

nn xþ1ð Þnn! ; (b)

P

1

n¼1

x�2ð Þnn! :

Solution

(a) Here an ¼ nn

n! for all n, so

�

�

�

anþ1

an

�

�

�¼ nþ 1ð Þnþ1

nn� n!

nþ 1ð Þ! ¼ 1þ 1

n

� n

; for all n:

Thus�

�

�

anþ1

an

�

�

�! e as n!1:

Hence, by the Ratio Test, the radius of convergence is 1e�

(b) Here an ¼ 1n!, for all n, so

�

�

�

anþ1

an

�

�

�¼ n!

nþ 1ð Þ! ¼1

nþ 1; for all n:

Thus�

�

�

anþ1

an

�

�

�! 0 as n!1:

Hence, by the Ratio Test, the power seriesP

1

n¼1

x�2ð Þnn! converges for all x. &

Problem 1 Determine the radius of convergence of each of the follow-

ing power series:

(a)P

1

n¼0

2n þ 4nð Þxn; (b)P

1

n¼1

n!ð Þ22nð Þ! xn;

(c)P

1

n¼1

nþ 2�nð Þ x� 1ð Þn; (d)P

1

n¼1

xn

n!ð Þ1n:

Hint for part (d): Use Stirling’s Formula and the result of Example 1,

Sub-section 7.5.1.

If we wish to find the interval of convergence (as opposed to the radius of

convergence), then we may need to use some of the other tests for series in

order to determine the behaviour at the end-points of the interval.

Example 2 Determine the interval of convergence of each of the following

power series:

(a)P

1

n¼1

xn; (b)P

1

n¼1

xn

n; (c)

P

1

n¼1

xn

n2.

Solution

(a) Here an¼ 1, for all n, so�

�

�

anþ1

an

�

�

�¼ 1; for all n:

R¼1.

R¼ 1L�


Hence, by the Ratio Test, the radius of convergence is 1; in other wordsX

1

n¼1

xn converges for�1 < x < 1:

Next, we consider the behaviour of the power series at the end-points of

this interval, namely �1 and 1. Since the sequences 1nf gand �1ð Þnf g are

both non-null, it follows thatP

1

n¼1

xn diverges when x ¼ �1, by the Non-

null Test.

Hence the interval of convergence ofP

1

n¼1

xn is (�1, 1).

(b) Here an ¼ 1n, for all n, so�

�

�

anþ1

an

�

�

�¼ n

nþ 1¼ 1

1þ 1n

; for all n:

Thus�

�

�

anþ1

an

�

�

�! 1 as n!1:

Hence, by the Ratio Test, the radius of convergence is 1; in other wordsX

1

n¼1

xn

nconverges for�1 < x < 1:

But we know thatP

1

n¼1

1n

diverges andP

1

n¼1

�1ð Þnn

converges. It follows that the

interval of convergence of the power seriesP

1

n¼1

xn

nis �1; 1½ Þ.

(c) Here an ¼ 1n2, for all n, so�

�

�

anþ1

an

�

�

�¼ n2

ðnþ 1Þ2¼ 1

ð1þ 1nÞ2; for all n:

Thus�

�

�

anþ1

an

�

�

�! 1 as n!1:

Hence, by the Ratio Test, the radius of convergence is 1; in other words

X

1

n¼1

xn

n2converges for �1 < x < 1:

But we know thatP

1

n¼1

1n2 converges; so, by the Absolute Convergence Test,

the seriesP

1

n¼1

�1ð Þnn2 also converges. It follows that the interval of conver-

gence of the power seriesP

1

n¼1

xn

n2 is �1; 1½ �. &

Now, the Radius of Convergence Theorem tells us that, if a power seriesP

1

n¼0

an x� að Þn has radius of convergence R, then it converges for jx� aj<R. In

fact, we can say more than this – namely, that, for all points x with jx� aj<R,

the power series is actually absolutely convergent!

Theorem 3 Absolute Convergence Theorem

Let the power seriesP

1

n¼0

an x� að Þn have radius of convergence R. Then it is

absolutely convergent for all x with jx� aj<R.

By Example 2 ofSub-section 3.2.1, andat the start of Sub-section 3.3.2, respectively.


332 8: Power series

Such a choice is alwayspossible – for example,choose X to be the midpoint ofx and the nearest end-point ofthe interval (a�R, aþR).

ForP

1

n¼0

x� aX� a

�

�

�

�

nis a geometric

series.

A similar result holds forpower series of the formP

1

n¼0

an x� að Þn; a 6¼ 0.

Equivalently

limx!1�

P

1

n¼0

anxn

�

¼P

1

n¼0

an:


Proof Let x be a number such that jx� aj<R, and chose a number X such

that jx� aj< jX� aj<R.

Now, the seriesP

1

n¼0

an X � að Þn is convergent, so that an X � að Þn! 0 as

n!1. In particular, there exists some number N such that an X � að Þnj j < 1,

for all n>N. It follows that

an x� að Þnj j ¼ x� a

X � a

�

�

�

�

�

�

n

� an X � að Þnj j

� x� a

X � a

�

�

�

�

�

�

n

; for n > N:

But the seriesP

1

n¼0

x� aX� a

�

�

�

�

nis convergent, so that the series

P

1

n¼0

an x� að Þnj j is also

convergent, by the Comparison Test for series.

It follows that the power seriesP

1

n¼0

an x� að Þn is absolutely convergent, as

required. &

Problem 2 Determine the interval of convergence of each of the

following power series:

(a)P

1

n¼0

nxn; (b)P

1

n¼1

1n3n xn:

Problem 3 Determine the radius of convergence of the power series

1þ �xþ � �� 1ð Þ2!

x2 þ � � � ¼X

1

n¼0

� �� 1ð Þ . . . �� nþ 1ð Þn!

xn;

where� 6¼ 0; 1; 2; . . .:

8.3.2 Abel’s Limit Theorem

We can sometimes use the ideas of power series to sum interesting and

commonly arising series such as

X

1

n¼1

�1ð Þnþ1

n¼1� 1

2þ 1

3� 1

4þ � � � and

X

1

n¼0

�1ð Þn

2nþ 1¼1� 1

3þ 1

5� 1

7þ � � �:

A major tool in this connection in the following result.

Theorem 4 Abel’s Limit Theorem

Let f (x) be the sum function of the power seriesP

1

n¼0

anxn, which has radius of

convergence 1; and letP

1

n¼0

an be convergent. Then

limx!1�

f xð Þ ¼X

1

n¼0

an:

Proof This proof is a wonderful illustration of the sheer power and magic of

the "! � method for proving results in Analysis! We use the standard

approach for proving that a limit exists as x! 1�.

Let sn ¼ a0 þ a1 þ � � � þ an�1 be the nth partial sum of the seriesP

1

n¼0

an; and

let s denote the sumP

1

n¼0

an.

Now,

a0 ¼ s1 and an ¼ snþ1 � sn; for all n � 1: (1)

It follows that, for xj j51

1� xð ÞX

1

n¼0

snþ1xn ¼ 1� xð Þ s1þ s2xþ s3x2þ �� þ snþ1xnþ ��

¼ s1þ s2xþ s3x2þ s4x3 � � � þ snþ1xnþ� s1x� s2x2� s3x3� � � �� snxn� snþ1xnþ1� �� ¼ s1þ s2� s1ð Þxþ s3� s2ð Þx2þ �� þ snþ1� snð Þxnþ ��

¼X

1

n¼0

anxn ¼ f ðxÞ:

We now study closely the identity

1� xð ÞX

1

n¼0

snþ1xn ¼ f ðxÞ: (2)

Next, we want to prove that:


f xð Þ � sj j < ", for all x satisfying 1� � < x< 1.

Now, we use equation (2) in the following way to get a convenient expres-

sion for f (x)� s

f xð Þ � s ¼X

1

n¼0

anx n �X

1

n¼0

an

¼ 1� xð ÞX

1

n¼0

snþ1x n � s

¼ 1� xð ÞX

1

n¼0

snþ1x n � 1� xð ÞX

1

n¼0

sx n

¼ 1� xð ÞX

1

n¼0

snþ1 � sð Þx n: (3)

Next, choose a number N such that snþ1 � sj j < 12" for all n>N. We can

then apply the Triangle Inequality to equation (3), for x 2 0; 1ð Þ, to see that

f xð Þ� sj j ¼ 1� xð ÞX

N

n¼0

snþ1� sð ÞxnþX

1

n¼Nþ1

snþ1� sð Þxn

�

�

�

�

�

�

�

�

�

�

� 1� xð ÞX

N

n¼0

snþ1� sð Þxn

�

�

�

�

�

�

�

�

�

�

þ 1� xð ÞX

1

n¼Nþ1

snþ1� sð Þxn

�

�

�

�

�

�

�

�

�

�

� 1� xð ÞX

N

n¼0

snþ1� sj jxnþ 1� xð Þ� 1

2"�

X

1

n¼Nþ1

xn

Using (1).

This is simply the definitionof lim

x!1�f xð Þ ¼ s:

Here we use the definitionof s, equation (2), and the factthat the sum of the seriesP

1

n¼0

x n is 11�x

:

Such a choice of N is possible,since snþ1 ! s as n!1.

We split the infinite sum intotwo parts.

Using the Triangle Inequality.

For snþ1 � sj j < 12", for all

n<N; also, we use that

0< x< 1 so thatP

1

n¼0

xn ¼ 11�x

.

334 8: Power series

� 1� xð ÞX

N

n¼0

snþ1 � sj jxn þ 1

2"

� 1� xð ÞX

N

n¼0

snþ1 � sj j þ 1

2" (4)

Finally, since the linear function x 7! 1� xð ÞP

N

n¼0

snþ1 � sj j is continuous on

R , and takes the value 0 at 1, it follows that we can choose a positive number �(with � 2 (0, 1)) such that

1� xð ÞX

N

n¼0

snþ1 � sj j< 1

2"; for all x satisfying 1� � < x < 1:

If we substitute this upper bound 12" for 1� xð Þ

P

N

n¼0

snþ1 � sj j into the

inequality (4), we find that we have proved that


f xð Þ � sj j < "; for all x satisfying 1� � < x < 1:

This is what we set out to prove. &

As an application of Abel’s Limit Theorem, we use it to evaluate

X

1

n¼1

�1ð Þnþ1

n¼ 1� 1

2þ 1

3� 1

4þ � � �:

Let f xð Þ ¼P

1

n¼1

�1ð Þnþ1

nxn. We have seen already that this power series converges

in (�1, 1) to the sum loge 1þ xð Þ. Further, we saw earlier that the seriesP

1

n¼1

�1ð Þnþ1

nis convergent, by the Alternating Test. It follows, from Abel’s

Limit Theorem, that

X

1

n¼1

�1ð Þnþ1

n¼ lim

x!1�

X

1

n¼1

�1ð Þnþ1

nxn ¼ lim

x!1�loge 1þ xð Þ ¼ loge 2:

Remark

It is always necessary to check that the conditions of Abel’s Theorem apply

before using it. For example, a thoughtless application to the identity

1� xþ x2 � x3 þ � � � ¼X

1

n¼0

�1ð Þnxn ¼ 1

1þ x; where xj j < 1;

of taking the limit as x! 1�, would give the following absurd conclusion

1� 1þ 1� 1þ � � � ¼X

1

n¼0

�1ð Þn ¼ 1

2!

Problem 4 Use Abel’s Limit Theorem to evaluateP

1

n¼0

�1ð Þn2nþ1¼ 1� 1

3þ 1

5� 1

7þ � � �.

Hint: Use the facts that tan�1 x ¼ x� x3

3þ x5

5� x7

7þ � � � for jxj< 1, and

that the latter series has radius of convergence 1.

Again we use that 0< x< 1.

We add in the extrarequirement that � < 1 in orderto ensure that we are onlyconsidering values of xin (0, 1).

Part (b) of Theorem 3, Sub-section 8.2.2.

We did not check that the

seriesP

1

n¼0

�1ð Þnxn was

convergent at 1!

We prove these facts in Sub-section 8.4.1.

8.3.3 Proofs

We now supply the proofs that we omitted from Sub-section 8.3.1.

Theorem 1 Radius of Convergence Theorem

For a given power seriesP

1

n¼0

an x� að Þn, precisely one of the following

possibilities occurs:

(a) The series converges only when x¼ a;

(b) The series converges for all x;

(c) There is a number R> 0 such that the series converges if jx� aj<R and

diverges if jx� aj>R.

Proof For simplicity, we shall assume that a¼ 0.

Clearly the possibilities (a) and (b) are mutually exclusive; we shall there-

fore assume that for a given power series neither possibility occurs, and then

prove that possibility (c) must occur.

First, let

S ¼ x :X

1

n¼0

anxn converges

( )

:

Since the possibility (b) has been excluded, we deduce that S must be bounded.

For, ifP

1

n¼0

anXn diverges, then, in view of the Absolute Convergence Theorem

for power series,P

1

n¼0

anxn cannot converge for any x with jxj> jXj, since

otherwiseP

1

n¼0

anXn would then have to be convergent – which is not the case.

Also, S is non-empty, since the power seriesP

1

n¼0

anxn converges at 0.

It then follows, from the Least Upper Bound Property of R , that the set S

must have a least upper bound, R say. We now prove that the seriesP

1

n¼0

anxn is

convergent if jxj<R and divergent if jxj>R.

First, notice that R> 0. For, since possibility (a) has been excluded, there must

be at least one non-zero value of x, x1 say, such thatP

1

n¼0

anxn1 is convergent; hence

we have x12 S. It follows, from the Absolute Convergence Theorem, thatP

1

n¼0

anxn must converge for those x with jxj< jx1j, so that � x1j j; x1j jð Þ lies in S.

Now, choose any x for which jxj<R. Then by the definition of R as sup S,

there exists some number x2 with jxj< x2<R such thatP

1

n¼0

anxn2 converges. It

follows, from the Absolute Convergence Theorem, thatP

1

n¼0

anxn converges.

Finally, choose any x for which xj j > R ¼ sup S. It follows that x =2 S, so thatP

1

n¼0

anxn must be divergent. &

You may omit this sub-section at a first reading.

For the general proof, replacex throughout by (x� a).


Here we identify a number Rthat will be the desired radiusof convergence.

336 8: Power series

In view of the Absolute Convergence Theorem, we can slightly strengthen

the conclusion of the Radius of Convergence Theorem, as follows.

Corollary If the power seriesP

1

n¼0

an x� að Þn has radius of convergence

R> 0, then it is absolutely convergent if jx� aj<R. If the series converges

for all x, then it is absolutely convergent for all x.

Finally we prove the Ratio Test for Radius of Convergence of power series.

Theorem 2 Ratio Test for Radius of Convergence

Suppose thatP

1

n¼0

an x� að Þn is a given power series, and that

�

�

�

anþ1

an

�

�

�! L as n!1:

(a) If L is1, the series converges only for x = a.

(b) If L¼ 0, the series converges for all x.

(c) If L> 0, the series has radius of convergence 1L:

Proof For simplicity, we shall assume that a¼ 0.

(a) Suppose that janþ1

anj ! 1 as n!1. Then, for any non-zero value of x, the

sequence janþ1xnþ1

anxn j ! 1, so that the sequence {anxn} is unbounded. It

follows, from the Non-null Test, that the seriesP

1

n¼0

anxn must be divergent.

(b) Suppose that janþ1

anj ! 0 as n!1. Then, for any non-zero value of x

anþ1xnþ1

anxn

�

�

�

�

�

�

�

�

¼�

�

�

anþ1

an

�

�

�� xj j

! 0� xj j ¼ 0;

so thatP

1

n¼0

anxn is absolutely convergent, and so is convergent.

(c) Suppose that�

�

�

anþ1

an

�

�

�! L as n!1, where L> 0.

First, suppose that xj j > 1L. Then

anþ1xnþ1

anxn

�

�

�

�

�

�

�

�

¼�

�

�

anþ1

an

�

�

�� xj j

! L� xj j > 1;

so thatP

1

n¼0

anxn is not absolutely convergent. It follows, from the above

Corollary to the Radius of Convergence Theorem, that the radius of

convergence ofP

1

n¼0

anxn must be less than or equal to 1L.

Next, suppose that xj j < 1L. Then

anþ1xnþ1

anxn

�

�

�

�

�

�

�

�

¼�

�

�

anþ1

an

�

�

�� xj j

! L� xj j< 1;


This result is sometimes alsocalled the AbsoluteConvergence Theorem.

For the general proof, replacex throughout by (x�a).

so thatP

1

n¼0

anxn is absolutely convergent, and so is convergent.

Hence the radius of convergence ofP

1

n¼0

anxn must be at least equal to 1L.

Combining these two facts, it follows that the desired radius of conver-

gence is exactly 1L. &

8.4 Manipulating power series

It would be tedious to have to apply Taylor’s Theorem every time that we wished

to determine the Taylor series of a given function. While sometimes this really

has to be done, in many commonly arising situations we can use standard rules

for power series and the list of basic power series to avoid most of the effort.

We now set out to establish the rules for manipulating power series.

8.4.1 Rules for power series

Many of the rules for manipulating power series are similar to the correspond-

ing rules for manipulating ‘ordinary’ series.


Let

f xð Þ ¼X

1

n¼0

an x� að Þn; for x� aj j < R; and

g xð Þ ¼X

1

n¼0

bn x� að Þn; for x� aj j < R0:

Then:

Sum Rule f þ gð Þ xð Þ ¼P

1

n¼0

an þ bnð Þ x� að Þn; for x� aj j < r;

where r ¼ min R;R0f g;

Multiple Rule lf xð Þ ¼P

1

n¼0

lan x� að Þn; for x� aj j < R; where l 2 R :

We do not supply a proof, as Theorem 1 is a simple restatement of the Sum and

Multiple Rules for ‘ordinary’ series.

We can use the Combination Rules to find the Taylor series at 0 for the

function cosh x. We start with the power series for the exponential function

ex ¼ 1þ xþ x2

2!þ x3

3!þ � � � ¼

X

1

n¼0

xn

n!; for x 2 R :

It follows that e�x ¼ 1� xþ x2

2! � x3

3! þ � � � ¼P

1

n¼0

�1ð Þn xn

n!; for x 2 R . We

may then use the Sum Rule to obtain

ex þ e�x ¼ 2� 1þ x2

2!þ x4

4!þ � � �

�

¼ 2�X

1

n¼0

x2n

2nð Þ!; for x 2 R ;


Thus, if the power series forboth f and g converge at aparticular point x, then so doesthe series for fþ g.



For the odd-powered termscancel.

338 8: Power series

so that, by using the Multiple Rule with l ¼ 12, we obtain

cosh x¼ 1

2exþ e�xð Þ ¼ 1þ x2

2!þ x4

4!þ � � � ¼

X

1

n¼0

x2n

2nð Þ!; for x 2 R :

Problem 1 Find the Taylor series at 0 for each of the following func-

tions, and state its radius of convergence:

(a) f xð Þ ¼ sinh x; x 2 R ;

(b) f xð Þ ¼ loge 1� xð Þ þ 2 1� xð Þ�1; xj j < 1:

Problem 2 Find the Taylor series at 0 for each of the following func-

tions, and state its radius of convergence:

(a) f xð Þ ¼ sinh xþ sin x; x 2 R ; (b) f xð Þ ¼ loge1þx1�x

� �

; xj j < 1;(c) f xð Þ ¼ 1

1þ2x2 ; x 2 R :

Remark

In Theorem 1, the radius of convergence of the power series for fþ g may be

larger than r ¼ min R;R0f g. For example, we can use the basic power series

and the Combination Rules to verify that the Taylor series at 0 for the functions

f xð Þ ¼ 11�x

and gðxÞ ¼ �11�xþ 1

1�x2

are

f xð Þ ¼ 1þ xþ x2 þ x3 þ x4þ � � � ¼X

1

n¼0

xn and

g xð Þ ¼ � 1

2x� 3

4x2 � 7

8x3 � 15

16x4� � � � ¼

X

1

n¼0

�1þ 1

2n

�

xn;

each with radius of convergence 1. It follows, from Theorem 1, that the power

series for the function f þ gð Þ xð Þ ¼ 11�x

2is

f þ gð Þ xð Þ ¼ 1þ 1

2xþ 1

4x2 þ 1

8x3 þ 1

16x4 þ � � � ¼

X

1

n¼0

1

2nxn;

and that this power series has radius of convergence at least 1. In fact, it has

radius of convergence 2.

We can find the Taylor series at 0 for the function f xð Þ¼ 1þ x1� x

, for jxj< 1, by

writing 1þx1�x

in the form2� 1�xð Þ

1�x¼ 2

1�x� 1. Since the Taylor series at 0 for the

function 21�x

is 2þ 2xþ 2x2 þ 2x3þ � � � ¼P

1

n¼0

2xn, with radius of conver-

gence 1, it follows that the Taylor series for f at 0 is 1þ 2xþ 2x2þ2x3þ � � � ¼ 1þ

P

1

n¼1

2xn, with radius of convergence 1.

However we could obtain the same result by multiplying together the Taylor

series for the functions x 7! 1þ x and x 7! 11�x

, since

1þ xð Þ 1þ xþ x2þ x3þ� � ��

¼ 1þ xþ x2þ x3þ� � �þ x 1þ xþ x2þ x3þ� � ��

¼ 1þ 2xþ 2x2þ 2x3þ� � �:In fact, we can justify such multiplication together of power series to obtain

further power series.

We simply add the powerseries term-by-term.

We now ‘collect together’ themultiples of successivevarious powers of x.

Let

f xð Þ ¼X

1

n¼0

an x� að Þn; for x� aj j < R; and

g xð Þ ¼X

1

n¼0

bn x� að Þn; for x� aj j < R0:

Then

fgð Þ xð Þ ¼X

1

n¼0

cn x� að Þn; for x� aj j < r;where r ¼ min R;R0f g;

and

c0 ¼ a0b0; c1 ¼ a0b1 þ a1b0 and

cn ¼ a0bn þ a1bn�1 þ � � � þ an�1b1 þ anb0:

Theorem 2 is an immediate consequence of the Product Rule for series, applied

to the two seriesP

1

n¼0

an x� að Þn andP

1

n¼0

bn x� að Þn, both of which are absolutely

convergent for jx� aj< r – by the Absolute Convergence Theorem.

Example 1 Determine the Taylor series at 0 for the function f xð Þ ¼ 1þ x

1�xð Þ2.

Solution We use our knowledge of the power series at 0 for the functions

x 7! 11� x

and x 7! 1þ x1� x

, both of which have radius of convergence 1. Thus, by

the Product Rule, we have

1þ x

1� xð Þ2¼ 1

1� x� 1þ x

1� x

¼ 1þ xþ x2 þ x3 þ x4 þ � � ��

� 1þ 2xþ 2x2 þ 2x3 þ 2x4 þ � � ��

¼X

1

n¼0

cnxn;

where c0¼ 1� 1¼ 1, c1¼ 1� 2þ 1� 1¼ 3 and cn¼ 1� 2þ 1� 2þ . . .þ1� 2þ 1� 1¼ 2nþ 1.

Thus the required power series for f at 0 is 1þ x

1� xð Þ2 ¼P

1

n¼0

2nþ 1ð Þxn: &

Problem 3 Determine the Taylor series at 0 for:

(a) f xð Þ ¼ 1þ xð Þ loge 1þ xð Þ; for xj j < 1;

(b) f xð Þ ¼ 1þx

1�xð Þ3 ; for xj j < 1:

Next, we have seen already that the hyperbolic functions have the following

Taylor series at 0

sinh x ¼ xþ x3

3!þ x5

5!þ � � � and cosh x ¼ 1þ x2

2!þ x4

4!þ � � � ;

each with radius of convergence1. Notice that the derivative of the function

sinh x is cosh x, and that term-by-term differentiation of the series

xþ x3

3! þ x5

5! þ � � � gives the series 1þ x2

2! þ x4

4! þ � � �. It looks as though we can

As in Theorem 1, the radius ofconvergence of the product

seriesP

1

n¼0

cn x� að Þn may be

greater than r.



Also, its radius ofconvergence is at least 1.(In fact it is easy to check thatits radius of convergence isexactly 1.)

At the start of this sub-section.

340 8: Power series

obtain the Taylor series for the derivative of a function f simply by differentiat-

ing the Taylor series for f itself.

Our next result states that we can differentiate or integrate the Taylor series

of a function f term-by-term to obtain the Taylor series of the corresponding

function f 0 orR

f , respectively.

Theorem 3 Differentiation and Integration Rules

Let f xð Þ ¼P

1

n¼0

an x� að Þn; for x� aj j < R. Then:

Differentiation Rule f 0 xð Þ ¼P

1

n¼1

nan x� að Þn�1; for x� aj j < R;

Integration RuleR

f xð Þdx ¼P

1

n¼0

anx� að Þnþ1

nþ 1þ constant;

for x� aj j < R:

All three series have the same radius of convergence.

For example, consider the Taylor series at 0 for the function tan�1. We know that

1

1þ x2¼ 1� x2 þ x4 � x6 þ � � �

¼X

1

n¼0

�1ð Þnx2n; with radius of convergence 1;

and thatd

dxtan�1 x ¼ 1

1þ x2; for x 2 R :

It follows, from the Integration Rule, that

tan�1 x¼ x� x3

3þ x5

5� x7

7þ�� þc; for xj j< 1;and some constant c:

Substituting x¼ 0 into this equation, we find that c¼ tan�1 0¼ 0. It follows that

tan�1 x ¼ x� x3

3þ x5

5� x7

7þ � � �; for xj j < 1:

Problem 4 Find the Taylor series at 0 for the following functions:

(a) f xð Þ¼ 1� xð Þ�2; for xj j< 1; (b) f xð Þ¼ 1� xð Þ�3; for xj j< 1;(c) f xð Þ¼ tanh�1 x; for xj j< 1:

Problem 5 Find the first three non-zero terms in the Taylor series at 0

for the function f (x)¼ ex(1� x)�2, jxj< 1. State its radius of convergence.

Problem 6 Let f be the function f xð Þ ¼ xþ x3

1:3þ x5

1:3:5þ � � � þx2nþ1

1:3:��: 2nþ1ð Þ þ � � �; x 2 R .

Determine the Taylor series at 0 for each of the following functions:

(a) f 0 xð Þ; (b) f 0 xð Þ � xf xð Þ:

Problem 7 Determine the Taylor series at 0 for the function

f xð Þ ¼ e�x2

, x 2 R .

Deduce thatR 1

0e�x2

dx ¼ 1� 13þ 1

10� 1

42þ � � � þ �1ð Þn

2nþ 1ð Þ�n!þ � � �:

To find the constant, put x¼ a.

They may, however, behavedifferently at the end-points oftheir intervals of convergence.

This is a geometric series.

By the Integration Rule, thefinal Taylor series musthave the same radius ofconvergence as the originalTaylor series, namely 1.

We have now found a whole variety of techniques for identifying Taylor series:

Taylor’s Theorem;

Combination, Product, Differentiation and Integration Rules.

But how do we know that different techniques will always give us the same

expression as the Taylor series? The following result states that there is only

one Taylor series for a function f at a given point a – any valid method will give

the same coefficients.

Theorem 4 Uniqueness Theorem

IfX

1

n¼0

an x�að Þn¼X

1

n¼0

bn x�að Þn; for x�aj j<R; then an¼ bn:

Proof Let f xð Þ ¼P

1

n¼0

an x� að Þn and g xð Þ ¼P

1

n¼0

bn x� að Þn, for jx� aj<R.

If we differentiate both equations n times, using the Differentiation Rule,

and put x¼ a, we obtain

f nð Þ að Þ ¼ n!� an and g nð Þ að Þ ¼ n!� bn;

Since f(x)¼ g(x), for jx� aj<R, we must have f (n)(a)¼ g(n)(a). It follows that

an¼ bn, for all n� 0. &

8.4.2 General Binomial Theorem

You will have already met the Binomial Theorem, which states that, for each

positive integer n

1þ xð Þn¼X

n

k¼0

n

k

�

xk; wheren

k

�

¼ n n� 1ð Þ . . . n� kþ 1ð Þk!

¼ n!

k! n� kð Þ! :

In the Binomial Theorem, the power n is a positive integer and the result is true

for all x2R . In fact, a similar result holds for more general values of the power

but with a restriction on the values of x for which it is valid.

Theorem 5 General Binomial Theorem

For any �2R

1þ xð Þ�¼X

1

n¼0

�n

�

xn; where�n

�

¼� ��1ð Þ . . . ��nþ1ð Þn!

and xj j< 1:

For example

1þ 2xð Þ�6 ¼X

1

n¼0

�6

n

�

2xð Þn; where

�6

n

�

¼ �6ð Þ �7ð Þ . . . �6� nþ 1ð Þn!

and xj j < 1

2;

See, for example, Appendix 1.

By convention,�0

�

¼ 1.

342 8: Power series

so that

1þ 2xð Þ�6 ¼ 1� 12xþ 84x2 � � � �; for xj j < 1

2:

Another important type of application occurs when the power � is not an

integer. For example

1þ xð Þ�13¼X

1

n¼0

� 13

n

�

xn; where� 1

3

n

�

¼� 1

3

� �

� 43

� �

� 73

� �

. . . � 13� nþ 1

� �

n!;

so that

1þ xð Þ�13¼ 1� 1

3xþ 2

9x2 � 14

81x3þ � � �; for xj j < 1:

Problem 8 Use the General Binomial Theorem to determine the first

four non-zero terms in the Taylor series at 0 for the function f xð Þ ¼1þ 6xð Þ

14, xj j< 1

6. State the radius of convergence of the Taylor series.

Problem 9

(a) Determine the Taylor series at 0 for the function f xð Þ ¼ 1� xð Þ�12,

jxj< 1.

(b) Hence determine the Taylor series at 0 for the function

f xð Þ ¼ sin�1 x, jxj< 1.

In our proof of the General Binomial Theorem, we use the following lemma.

Lemma Forany�2R andanyn2N , n�n

�

þ nþ1ð Þ �nþ1

�

¼� �n

�

.

Proof Since�k

�

¼ � ��1ð Þ: : : : : ��kþ1ð Þk! , for any k2N , we may rewrite the

left-hand side of the desired identity in the form

n�

n

�

þ nþ1ð Þ�

nþ1

�

¼ n�

n

�

þ nþ1ð Þ� ��1ð Þ : : : ��nð Þnþ1ð Þ!

¼ n�

n

�

þ� ��1ð Þ . . . ��nð Þn!

¼ n�

n

�

þ ��nð Þ� ��1ð Þ . . . ��nþ1ð Þn!

¼ n�

n

�

þ ��nð Þ�

n

�

¼��

n

�

: &

Theorem 5 General Binomial Theorem

For any �2R

1þxð Þ�¼X

1

n¼0

�n

�

xn; where�n

�

¼� ��1ð Þ . . . ��nþ1ð Þn!

and xj j< 1:

You may omit the remainderof this sub-section at a firstreading.

Here we cancel the term(nþ 1).

We now bring the term(�� n) in front of the fraction.

The last fraction is just�n

�

.

Proof Let

f xð Þ ¼X

1

n¼0

�n

�

xn and g xð Þ ¼ f xð Þ 1þ xð Þ��; for xj j < 1:

We want to prove that g(x)¼ 1, for all x with jxj< 1.

Using the rules for differentiation, we may differentiate the expression for

g to obtain

g0 xð Þ ¼ f 0 xð Þ 1þ xð Þ��f xð Þ 1þ xð Þ��1

¼ 1þ xð Þf 0 xð Þ � �f xð Þ½ � 1þ xð Þ��1:

Now

1þ xð Þf 0 xð Þ ¼ 1þ xð ÞX

1

n¼1

n�

n

�

xn�1

¼X

1

n¼1

n�

n

�

xn�1 þX

1

n¼1

n�

n

�

xn

¼X

1

n¼0

nþ 1ð Þ�

nþ 1

�

xn þX

1

n¼0

n�

n

�

xn

¼X

1

n¼0

nþ 1ð Þ�

nþ 1

�

þ n�

n

� �

xn

¼ �X

1

n¼0

�

n

�

xn

¼ �f xð Þ:It follows from the earlier expression for g0(x) that

g0 xð Þ ¼ 1þ xð Þf 0 xð Þ � �f xð Þ½ � 1þ xð Þ��1

¼ 0� 1þ xð Þ��1¼ 0:

So, g( x) is a constant. Hence

g xð Þ ¼ g 0ð Þ¼ f 0ð Þ ¼ 1;

as required. &

8.4.3 Proofs

Differentiation Rule The series

f xð Þ ¼X

1

n¼0

an x� að Þn andX

1

n¼1

nan x� að Þn�1

have the same radius of convergence, R say. Also, f is differentiable on

(a�R, aþR), and

f 0 xð Þ ¼X

1

n¼1

nan x� að Þn�1:

Here we differentiate thepower series term-by-term.Then we split the initialbracket.

We now replace n by nþ 1 inthe first sum, and check whatthe limits of summation are inboth sums.

Now we combine the twosums into one.

We can apply the result of theLemma to this square bracket.

By the definition of f(x).

You may omit this Sub-sectionat a first reading.

This is one part of Theorem 3,Sub-section 8.4.1.

344 8: Power series

Proof For simplicity, we assume that a¼ 0.

Let the seriesP

1

n¼0

anxn andP

1

n¼1

nanxn�1 have radii of convergence R and R0,

respectively.

We start by proving that R¼R0.

We first show that, if jxj<R, then the power seriesP

1

n¼1

nanxn�1 is conver-

gent. (By the Corollary to the Radius of Convergence Theorem, this shows

that R0 �R.)

To prove this, choose a real number c with jxj< c<R. Then the seriesP

1

n¼0

ancn is convergent, and so the sequence {ancn} is a null sequence. Thus

there is a number K such that

ancnj j � K; for n ¼ 0; 1; 2; . . .: (1)

Then

nanxn�1�

�

�

�¼ ancnj j � n

c� x

c

�

�

�

�

�

�

n�1

� K

c� n� x

c

�

�

�

�

�

�

n�1

; (2)

by (1). Since xc

�

�

�

� < 1, the seriesP

1

n¼1

n xc

�

�

�

�

n�1converges. Hence, by the Comparison

Test for series, the seriesP

1

n¼1

nanxn�1 is absolutely convergent, and so is

convergent. This proves that R0 �R.

Next, suppose that R0>R. Let c be any number such that R< c<R0. ThenP

1

n¼1

nancn�1 is absolutely convergent, by the Absolute Convergence Theorem.

But

ancnj j ¼ nancn�1�

�

�

�� c

n

� c� nancn�1�

�

�

�:

Hence, by the Comparison Test for series, the seriesP

1

n¼1

ancn is absolutely

convergent, and so is convergent. This contradicts the definition of R, and thus

shows that R0 6> R.

It follows that R¼R0.Next, we show that f is differentiable on (�R, R), and that f 0 is of the stated

form.

So, choose any point c2 (�R, R). Then choose a positive number r<R such

that c 2 (�r, r). Then, for all non-zero h such that jhj< r� jcj

f cþ hð Þ � f cð Þh

�X

1

n¼1

nancn�1 ¼ 1

h

X

1

n¼1

an cþ hð Þn� cn�hncn�1� �

: (3)

Now, we apply Taylor’s Theorem to the function x 7! xn on the interval with

end-points c and cþ h. We obtain

cþ hð Þn¼ cn þ nhcn�1 þ 1

2n n� 1ð Þh2cn�2

n ;

Sub-section 8.3.3.

For example, c ¼ 12

xj j þ Rð Þ.


For example, c ¼ 12

Rþ R0ð Þ.

For example, r ¼ 12

cj j þ Rð Þ.

where cn is some number between c and cþ h (and, in particular, with

cnj j � r).

Now

cþ hð Þn�cn � nhcn�1�

�

�

� � 1

2n n� 1ð Þh2r n�2: (4)

It follows from (3) and (4) and the Triangle Inequality that


�X

1

n¼1

nancn�1

�

�

�

�

�

�

�

�

�

�

� 1

2hj jX

1

n¼2

n n� 1ð Þ anj jrn�2: (5)

Since the seriesP

1

n¼1

nanxn�1 has radius of convergence R, it follows that so

does the seriesP

1

n¼2

n n� 1ð Þanxn�2; we conclude that the seriesP

1

n¼2

n n� 1ð Þ

anrn�2 is (absolutely) convergent.

Hence, it follows from (5) and the Limit Inequality Rule that

limh!0


¼X

1

n¼1

nancn�1: &

Integration Rule The series

f xð Þ ¼X

1

n¼0

an x� að Þn and F xð Þ ¼X

1

n¼0

an

nþ 1x� að Þnþ1

have the same radius of convergence, R say. Also, f is integrable on

(a�R, aþ R), andZ

f xð Þdx ¼ F xð Þ:

Proof The two power series have the same radius of convergence, by the

Differentiation Rule for power series, applied to F.

It also follows, from the Differentiation Rule for power series, that F is

differentiable and F0 ¼ f, so that F is a primitive of f. &

8.5 Numerical estimates for p

One of the most interesting problems in the history of mathematics throughout

the last two thousand years has been the determination to any pre-assigned

degree of accuracy of naturally arising irrational numbers such asffiffiffi

2p

, e and p.

Here we look at various way of estimating p, and prove that p is irrational.

8.5.1 Calculating p

It is easy to check that p is slightly larger than 3 by wrapping a string round a

circle of radius r and measuring the length of the string corresponding to one

circumference; for this length 2pr is just slightly larger than 6r.

For we have shown that term-by-term differentiation doesnot alter the radius ofconvergence of a powerseries.

This is the other part ofTheorem 3, Sub-section 8.4.1.

That is, F is a primitive of f on(a�R, aþR)

346 8: Power series

Over hundreds of years mathematicians devised more sophisticated methods

for estimating the value of p. Some of their results are as follows:

the Babylonians c. 2000 BC p ¼ 3 18¼ 3:125

the Egyptians c. 2000 BC p ¼ 3 1381’ 3:160

the Old Testament c. 550 BC p ¼ 3

Archimedes c. 300–200 BC between 3 17

and 3 1071

, p ’ 3:141

the Chinese c. 400–500 AD p ’ 3:1415926

the Hindus c. 500–600 AD p ’ffiffiffiffiffi

10p

’ 3:16

With the development of the Calculus in the seventeenth century, new for-

mulas for estimating p were discovered, including:

Wallis’s Formula p2¼ 2

1� 2

3� 4

3� 4

5� 6

5� 6

7� � � � � 2n

2n�1� 2n

2nþ1� � � �

Leibniz’s Series tan�1 x ¼ x� x3

3þ x5

5� x7

7þ � � �

Gregory’s Series p4¼ 1� 1

3þ 1

5� 1

7þ � � �

However, neither Wallis’s Formula nor Gregory’s Series is very useful for

calculating p beyond a few decimal places, as they converge far too slowly.

However, we can use the Taylor series for tan�1 effectively for calculating pby choosing values of x closer to 0; in fact, the smaller the value of x, the faster

the series converges, and so the fewer the number of terms we need to calculate

p to a given degree of accuracy.

For example, if we choose x ¼ 1ffiffi

3p in Leibniz’s Series, we obtain

p6¼ tan�1 1

ffiffiffi

3p�

¼ 1ffiffiffi

3p � 1

3

1ffiffiffi

3p� 3

þ 1

5

1ffiffiffi

3p� 5

� 1

7

1ffiffiffi

3p� 7

þ � � �

¼ 1ffiffiffi

3p 1� 1

9þ 1

45� 1

189þ � � �

�

:

This formula can be used to calculate p to several decimal places without much

effort; we gain about one extra place for every two extra terms.

To obtain series that are more effective for calculating p, we can use the

Addition Formula for tan�1

tan�1 xþ tan�1 y ¼ tan�1 xþ y

1� xy

�

;

provided that tan�1 xþ tan�1 y lies in the interval � p2; p

2

� �

.

Problem 1 Use the Addition Formula for tan�1 to prove that

tan�1 1

3

�

þ tan�1 1

4

�

þ tan�1 2

9

�

¼ p4:

Similar applications of the Addition Formula give further expressions for p4,

such as

6 tan�1 1

8

�

þ 2 tan�1 1

57

�

þ tan�1 1

239

�

¼ p4

;

4 tan�1 1

5

�

� tan�1 1

239

�

¼ p4:

1 Kings vii, 23;2 Chronicles iv, 2.


Just before Problem 4, Sub-section 8.4.1.


This is known as Sharp’sFormula.

Problem 3(a) on Section 4.3,in Section 4.5.

This is known as Machin’sFormula.


Such formulas have been used to calculate p to great accuracy: 1 million decimal

places were achieved in 1974. Recently, improved methods of numerical ana-

lysis have been used to calculate p correct to several million decimal places.

In 1770, Lambert showed that p is irrational; that is, p is not the solution of a

linear equation a0þ a1x¼ 0, where the coefficients a0 and a1 are integers.

Then, in 1882, Lindemann showed that p is transcendental; that is, p is not a

solution of any polynomial equation a0þ a1xþ a2x2þ � � � þ anxn¼ 0, where

all the coefficients are integers.

We end with a couple of mnemonics that can be used to recall the first few

digits for p; the word lengths give the successive digits:

May I have a large container of coffee?3: 1 4 1 5 9 2 6

and

How I need a drink; alcoholic of course;3: 1 4 1 5 9 2 6

after all those formulas involving tangent functions!

5 3 5 8 9 7 9

8.5.2 Proof that p is irrational

We now prove that p is irrational. Our proof is rather intricate; and, surprisingly,

it uses many of the properties of integrals that you met in Chapter 7! It depends

on the properties of a suitably chosen integral that we examine in Lemmas 1–3.

Lemma 1 Let In ¼R 1

�11� x2ð Þncos 1

2px

� �

dx, n¼ 0, 1, 2, . . .. Then:

(a) In > 0; (b) In � 2.

Proof

(a) The function f xð Þ ¼ 1� x2ð Þncos 12px

� �

, for x2 [�1, 1], is non-negative,

for n¼ 0, 1, 2, . . .. It follows, from the Inequality Rule for integrals, that

In� 0, for each n.

To prove that In> 0 we need to examine the integral in more detail.

By the Inequality Rule for integrals, we haveZ 1

�1

f xð Þdx ¼Z �1

2

�1

þZ 1

2

�12

þZ 1

12

!

f xð Þdx

�Z 1

2

�12

f xð Þdx:

Now, if x 2 �12; 1

2

� �

, then 1� x2ð Þncos 12px

� �

� 34

� �ncos 1

4p

� �

, so that

Z 1

�1

f xð Þdx �Z 1

2

�12

3

4

� n

cos1

4p

�

dx

¼ 3

4

� n

cos1

4p

�

> 0;

as required.

For interest only, we list thefirst 1000 decimal places ofp in Appendix 3.

This solved the ancient Greekproblem of finding a ruler andcompass method to constructa square with an area equal tothat of a given circle.

You may omit this at a firstreading.

ForR�1

2

�1f xð Þdx � 0 and

R 112

f xð Þdx � 0:

By the Inequality Rule forintegrals.

For the length of the intervalof integration is 1.

348 8: Power series

(b) For x 2 �1; 1½ �, 1� x2ð Þncos 12px

� �

� 1� 1 ¼ 1, so thatZ 1

�1

f xð Þdx �Z 1

�1

1dx ¼ 2: &

Problem 2 Prove that pI0 ¼ 4 and p3I1 ¼ 32.

Next, we obtain a reduction formula for In, using integration by parts.

Lemma 2 The integral In satisfies the following reduction formula

p2In ¼ 8n 2n� 1ð ÞIn�1 � 16n n� 1ð ÞIn�2:

Proof Using integration by parts twice, we obtain

In ¼ 1� x2� �n2

psin

1

2px

� �1

�1

� 2

p

Z 1

�1

n �2xð Þ 1� x2� �n�1

sin1

2px

�

dx

¼ 4n

p

Z 1

�1

x 1� x2� �n�1

sin1

2px

�

dx

¼ 4n

px 1� x2� �n�1 � 2

p

�

cos1

2px

� �1

�1

þ 8n

p2

Z 1

�1

1� x2� �n�1� 2x2 n� 1ð Þ 1� x2

� �n�2n o

cos1

2px

�

dx

¼ 8n

p2In�1 �

16

p2n n� 1ð Þ

Z 1

�1

x2 1� x2� �n�2

cos1

2px

�

dx

¼ 8n

p2In�1 �

16

p2n n� 1ð Þ

Z 1

�1

1� x2� �n�2� 1� x2

� �n�1n o

cos1

2px

�

dx

¼ 8n

p2In�1 �

16

p2n n� 1ð Þ In�2 � In�1f g:

Multiplying both sides by p2, we obtain

p2In ¼ 8nIn�1 � 16n n� 1ð ÞIn�2 þ 16n n� 1ð ÞIn�1

¼ 8n 2n� 1ð ÞIn�1 � 16n n� 1ð ÞIn�2: &

We now use the result of Lemma 2 to prove the crucial tool in our proof that

p is irrational.

Lemma 3 For n¼ 0, 1, 2, . . ., there exist integers a0, a1, a2, . . ., an such

that

p2nþ1

n!In ¼ a0 þ a1pþ a2p2þ � � � þanpn ¼

X

n

k¼0

akpk

!

:

Proof For simplicity, let Jn ¼ p2nþ1

n! In. It follows from Problem 2 that

J0 ¼ pI0 ¼ 4 and J1 ¼ p3I1 ¼ 32: (1)

By the Inequality Rule forintegrals.

The value of the first term iszero.

The value of the first term iszero.

For

x2 1� x2� �n�2

¼ 1� 1� x2� ��

1� x2� �n�2

¼ 1� x2� �n�2� 1� x2

� �n�1:


Next, we rewrite the reduction formula in Lemma 2 in terms of Jn as

Jn ¼p2nþ1

n!In

¼ 8n 2n� 1ð Þn!

p2n�1In�1 �16n n� 1ð Þ

n!p2n�1In�2

¼ 8 2n� 1ð ÞJn�1 � 16p2Jn�2: (2)

The desired result now follows by Mathematical Induction. We know that

it holds for n¼ 0 and for n¼ 1, by the statements (1) above. Using the

reduction formula (2) we can prove that the statement of the Lemma holds

for all n� 2. &

Finally, we can use the fact that, for any integer p, the sequence p2nþ1

n!

n o

is

null, to prove that p is irrational.

Theorem 1 p is irrational.

Proof Suppose, on the contrary, that p is rational; so that p ¼ pq, for p, q2N .

Then, the conclusion of Lemma 3 can be written in the form

p2nþ1

q2nþ1n!In ¼

X

n

k¼0

ak

pk

qk;

where the coefficients ak are integers.

If we now multiply both sides by the expression q2nþ1, we find that

p2nþ1

n!In ¼

X

n

k¼0

akpkq2nþ1�k;

so that

p2nþ1

n!In is an integer: (3)

However, we know that the sequence p2nþ1

n!

n o

is a null sequence; and we know,

from Lemma 1, that jInj� 2. It follows, from the Squeeze Rule for sequences, that

p2nþ1

n!In ! 0 as n!1: (4)

It follows from (4) that eventually p2nþ1

n! In < 1; and so, from (3), that in fact In

must equal 0. This contradicts the result of Lemma 1, part (a). This contra-

diction proves that, in fact, p must be rational. &

8.6 Exercises

Section 8.1

1. Determine the tangent approximation to the function f(x)¼ 2� 3xþ x2þ ex

at the given point a:

(a) a¼ 0; (b) a¼ 1.

We omit ‘the gory details’!

The sequencep2ð Þnn!

�

is a

basic null sequence.


350 8: Power series

2. Determine the Taylor polynomial of degree 3 for each of the following

functions f at the given point a:

(a) f(x)¼ loge(1þ x), a¼ 2; (b) f xð Þ ¼ sin x, a ¼ p6;

(c) f xð Þ ¼ 1þ xð Þ�2, a ¼ 1

2; (d) f xð Þ ¼ tan x, a ¼ p

4:

3. Determine the Taylor polynomial of degree 4 for each of the following

functions f at the given point a:

(a) f(x)¼ cosh x, a¼ 0; (b) f(x)¼ x5, a¼ 1.

4. Determine the percentage error involved in using the Taylor polynomial of

degree 3 for the function f(x)¼ ex at 0 to evaluate e0.1.

Section 8.2

1. Obtain an expression for the remainder term R1(x) when Taylor’s Theorem is

applied to the function f(x)¼ ex at 0. Show that, when x¼ 1 and n¼ 1, then

the value of c in the statement of Taylor’s Theorem is approximately 0.36.

2. By applying Taylor’s Theorem to the function f(x)¼ sin x with a ¼ p4, prove

that

sin x ¼ 1ffiffiffi

2p 1þ x� p

4

� �

� 1

2x� p

4

� �2�

þ R2 xð Þ; x 2 R ;

where R2 xð Þj j � 16

x� p4

�

�

�

�

3:

3. By applying the Remainder Estimate to the function f(x)¼ sinh x with

a¼ 0, calculate sinh 0.2 to four decimal places.

4. Calculate the Taylor polynomial T3(x) for the function f xð Þ ¼ xxþ3

at 2.

Show that T3(x) approximates f(x) to within 10�4 on the interval 2; 52

� �

:

5. (a) Determine the Taylor polynomial of degree n for the function

f(x)¼ loge x at 1.

(b) Write down the remainder term Rn(x) in Taylor’s Theorem for this

function, and show that Rn(x)! 0 as n!1 if x2 (1, 2).

(c) By using Theorem 2 in Section 8.2, determine a Taylor series for f at 1

which is valid when x2 (1, 2).

6. Let f xð Þ ¼ e�1

x2 ; x 6¼ 0;0; x ¼ 0:

(a) Prove that f 0 0ð Þ ¼ 0:(b) Prove that, for x 6¼ 0, f 0(x) is of the form

(a polynomial of degree at most 3 in 1x)�e�

1

x2 .

(c) Prove that, for x 6¼ 0, f (n)(x) is of the form

(a polynomial of degree at most 3n in 1x)�e�

1

x2 .

(d) Prove that, for any positive integer n, f (n)(0)¼ 0.

Section 8.3

1. Determine the radius of convergence of each of the following power series:

(a)P

1

n¼1

3nð Þ!n!ð Þ2 xþ 1ð Þn; (b)

P

1

n¼1

nn

n! x� 1e

� �n;

For this function, the Taylorpolynomial Tn(x) isidentically zero for each n,although the function f is notidentically zero. In this case,Rn(x)¼ f(x), for all n; andso the Taylor series for f

at 0,P

1

n¼0

f nð Þ 0ð Þn! xn, converges to

f(x) only at 0.

8.6 Exercises 351

(c)P

1

n¼1

nþ 1ð Þ�nxn; (d) 1

2xþ 1:3

2:5 x2 þ 1:3:52:5:8 x3 þ � � �;

(e) 1þ a:b1:c xþ a aþ1ð Þ:b bþ1ð Þ

1:2:c cþ1ð Þ x2 þ � � �, where a, b, c > 0:

2. Determine the Taylor series for the function f(x)¼ loge(2þ x) at the given

point a; in each case, indicate the general term and state the radius of

convergence of the series:

(a) a¼ 1; (b) a¼�1.

Hint: Use the Taylor series at 0 for the function x 7! loge 1þ xð Þ, with

t¼ x� 1 and t¼ xþ 1, respectively.

3. Determine the interval of convergence of each of the following power

series:

(a)P

1

n¼1

2n

n! xn; (b)P

1

n¼1

�1ð Þn 2n

nx� 1ð Þn:

4. Give an example (if one exists) of each of the following:

(a) a power seriesP

1

n¼0

anxn which diverges at both x¼ 1 and x¼�2;

(b) a power seriesP

1

n¼0

anxn which diverges at x¼ 1 but converges at x¼�2;

(c) a power seriesP

1

n¼0

anxn which converges at x¼ 1 but diverges at x¼�2.

5. Give an example (if one exists) of each of the following:

(a) a power seriesP

1

n¼0

anxn which converges only if �3 < x < 3;

(b) a power seriesP

1

n¼0

anxn which converges only if �3 � x < 3;

(c) a power seriesP

1

n¼0

anxn which converges only if �3 < x � 3;

(d) a power seriesP

1

n¼0

anxn which converges only if �3 � x � 3.

Section 8.4

1. Determine the Taylor series for each of the following functions at 0; in each

case, indicate the general term and state the radius of convergence of the

series:

(a) f xð Þ ¼ loge 1þ xþ x2ð Þ ¼ loge1�x3

1�x

� �

; (b) f xð Þ ¼ cosh x1�x

:

2. Determine the first three non-zero terms in the Taylor series for each of the

following functions at 0; and state the radius of convergence:

(a) f(x)¼ cos(ex� 1); (b) f(x)¼ loge(1þ sin x); (c) f(x)¼ ex sin x.

3. For the function f xð Þ ¼ xþ 23

x3 þ 2:43:5 x5 þ 2:4:6

3:5:7 x7 þ � � �, for jxj< 1, prove

that

1� x2� �

f 0 xð Þ � xf xð Þ ¼ 1:

4. By using the Integration Rule for power series, prove that

sinh�1 x ¼ x� 1

2

x3

3þ 1:3

2:4

x5

5� 1:3:5

2:4:6

x7

7þ � � �; for xj j < 1:

This series is often called thehypergeometric series.

In fact, f xð Þ ¼ sin�1 xffiffiffiffiffiffiffiffi

1�x2p .

352 8: Power series

5. By using the Integration Rule for power series, find an infinite series with

sumR 1

0sin t2ð Þdt.

6. The Bessel function J0 is defined by the power series

J0 xð Þ¼ 1� x2

22: 1!ð Þ2þ x4

24: 2!ð Þ2� x6

26: 3!ð Þ2þ�� þ �1ð Þnx2n

22n: n!ð Þ2þ�� ; for x2R :

(a) Determine power series for J00 xð Þ and J000 xð Þ, indicating the general term

in each.

(b) Prove that xJ000 xð Þ þ J00 xð Þ þ xJ0 xð Þ ¼ 0.

J0 is not the Taylor series at 0of any familiar function.

8.6 Exercises 353

Appendix 1: Sets, functions and proofs

Sets

A set is a collection of objects, called elements. We use the symbol 2 to mean ‘is a

member of’ or ‘belongs to’; thus ‘x2A’ means ‘the element x is a member of the set A’.

Similarly, we use the symbol 62 to mean ‘is not a member of’ or ‘does not belong to’;

thus ‘x 62A’ means ‘the element x is not a member of the set A’.

We often use curly brackets (or braces) to list the elements of a set in some way. Thus

{a, b, c} denotes the set whose three elements are a, b and c. Similarly, {x : x2¼ 2}

denotes the set whose elements x are such that x2¼ 2; we would read this symbol in

words as ‘the set of x such that x2¼ 2’. When we define a set in this latter way, the

symbol x is a dummy variable; that is, if we replace that symbol x by any other symbol,

such as y, the set is exactly the same set; thus, for instance, the sets {x : x2¼ 2} and

{y : y2¼ 2} are identical.

Two sets are equal if they contain the same elements. We say that a set A is a subset

of a set B, if all the elements of A are elements of B, and we denote this by writing

‘A�B’. We may wish to indicate specifically the possibility that the subset A is equal to

B by writing ‘A�B’, or that A is a proper subset of B (that is, A is a subset of B but

A 6¼B), by writing ‘A�6¼

B’.

Notice that, to show that two sets A and B are equal, it is necessary to prove that A is

a subset of B and that B is a subset of A. In symbols

A ¼ B is equivalent to the two properties both holding: A � B and B � A:

The empty set is the set that contains no elements; it is denoted by the symbol ‘Ø’.

It may seem strange to define such a thing; but, in practice, it is often a convenient set to use.

The union of two sets A and B consists of the set of elements that belong to at least

one of A and B; in symbols

A [ B ¼ x : x 2 A or x 2 Bf g:

The intersection of two sets A and B consists of the set of elements that belong to

both A and B; in symbols

A \ B ¼ x : x 2 A and x 2 Bf g:

We denote the set of elements that belong to A but not to B by A�B; that is

A� B ¼ x : x 2 A; x =2Bf g:There are some special symbols used that are used to denote commonly arising sets of

real numbers:

N , the set of all natural numbers; thus N ¼ {1, 2, 3, . . .};

Z, the set of all integers; thus Z ¼ {0, �1, �2, �3, . . .};

Q , the set of all rational numbers; that is, numbers of the form pq, where p, q2Z

but q 6¼ 0;

R , the set of all real numbers;

Rþ, the set of all positive real numbers.

On the real line R , an interval I is a set of real numbers such that, if a and b both lie in I,

then all numbers between a and b also lie in I.

Thus the colon ‘:’ means‘such that’.

Here we also allow x to belongto both A and B.

354

There are nine types of intervals in R :

a; b½ � ¼ x : x 2 R ; a � x � bf g;a; bð Þ ¼ x : x 2 R ; a < x < bf g;a; b½ Þ ¼ x : x 2 R ; a � x < bf g;a; bð � ¼ x : x 2 R ; a < x � bf g;a;1½ Þ ¼ x : x 2 R ; x � af g;a;1ð Þ ¼ x : x 2 R ; x > af g;�1; bð � ¼ x : x 2 R ; x � bf g;�1; bð Þ ¼ x : x 2 R ; x < bf g;

R .

An interval is said to be closed if it contains both of its end-points, open if it contains

neither of its end-points, and half-closed or half-open if it is neither closed nor open.

Functions

A function is a mapping of some element of R onto another element of R . Thus a

function f is defined by specifying:

a set A, called the domain of f;

a set B, called the codomain of f;

a rule x 7! f xð Þ that associates with each element x of A a unique element f(x) of B.

Symbolically, we write this in the following way:

f : A! B

x 7! f xð Þ:

The element f(x) is called the image of x under f, and the set f (A)¼ {y : y¼ f (x) for

some x2A} the image of A under f.

A particularly simple function is the identity function that maps each element of a

set to itself. Thus, the identity function on a set A, iA, is the function

iA : A! B

x 7! x:

Sometimes every point of the codomain is in the image. We say that a function

f : A!B is an onto function if f (A)¼B.

Sometimes every point of the image is the image of precisely one point of the

domain. We say that a function f : A!B is a one–one function if:

whenever f(x)¼ f(y) for elements x, y of A, then necessarily x¼ y.

Notice that this does NOT mean that f is a one–one mapping of A onto B.

A given function f : A!B may be onto, one–one, both onto and one–one, or neither

onto nor one–one. A function f : A!B is called a bijection if it is both onto and

one–one.

If a function f : A!B is one–one, then it has an inverse function f�1 : f (B)!A with

the defining rule

f�1 yð Þ ¼ x; where y ¼ f xð Þ:

Notice that a function f only has an inverse function f�1 if it is bijective.

Sometimes we want to ‘change the domain’ of a function to a larger set or a

smaller set.

The first four types ofintervals are boundedintervals.

The final five types ofintervals are unboundedintervals.

Sets, functions and proofs 355

If f : A!B, and C is a subset of A, we say that the function g : C!B is the restriction

of f to C if

g xð Þ ¼ f xð Þ; for all x 2 C:

Similarly, if f : A!B, and A is a subset of C, we say that the function g : C!B is the

extension of f to C if

g xð Þ ¼ f xð Þ; for all x 2 A:

Finally, the composite function or composite g f of two functions

f : A! B and g : C ! D;

where f (A)�C, is the function

g f : A! D

x 7! g f xð Þð Þ:

Proofs

Logical implications are so important in Mathematics that we use some special sym-

bols, namely),( and,, to denote implications.

Thus, if we are discussing two statements P and Q, we write

P)Q to mean: ‘if P is true, then Q is true’ or ‘P implies Q’;

P(Q to mean: ‘P is implied by Q’, or ‘if Q is true, then P is true’ or ‘Q implies P’;

P,Q to mean: ‘P is true if and only if Q is true’;

this is equivalent to the two separate statements

P) Q and Q) P:

In this case, P and Q are equivalent statement.

The implications P)Q and Q)P are called converse implications.

In order to prove that an assertion ‘P)Q’ is true, we need to verify that, in all

situations where the statement P holds, then the statement Q also holds. To prove that an

assertion ‘P)Q’ is not true, all that we need do is to find one example of a situation

where the statement P holds but the statement Q does not. Such a situation is called a

counter-example to the assertion.

Sometimes we prove assertions P by simply checking all possible cases; generally,

though, we devise logical arguments that deal with all cases at the same time.

Sometimes our logical arguments appear slightly convoluted to non-mathematicians;

two examples of these are:

Proof by contradiction: In order to show that a statement P holds, we start by

assuming that P is false; we then look at the consequences of that assumption, and

identify some specific consequence that is untrue or contradictory. It then follows that

P must hold, after all.

Proof by contraposition: In order to show that an implication P)Q is true, it is

sufficient to prove that

if Q does not hold, then P does not hold.

Sometimes mathematical proofs are long and complicated, and involve the use of a

variety of approaches to prove the desired result.

Notice that not all approaches to proving a result are necessarily valid! Some

examples of such approaches are:

Proof by picture: Here you make a false claim in your argument because it appears

to be true in the particular diagram that you have drawn.

Proof by example: For instance, you prove that some property holds for n¼ 1 and

n¼ 2, and then assert that it therefore holds for all positive integers.

Sometimes we use the termextension to denote a functiong : C!D where A�C,B�D, and g(x)¼ f(x), for allx2A.

Thus, to prove that P,Q, weneed to prove that P)Q andQ)P.

This is sometimes calleddisproof by counter-example.

This approach is sometimescalled proof by exhaustion.

This fact is sometimes(humorously) called proof byperspiration.

356 Appendix 1

Proof by omission: For instance, you prove one or two special cases of a result, and

claim that the general proof is ‘similar’.

Proof by superiority: Here you claim that the result is ‘obvious’, rather than sit down

to construct a logical proof of it.

Proof by mumbo-jumbo: Here you write down a jumble of formulas and relevant

words, ending up with the claim that you have proved the result.

Principle of Mathematical Induction

This is a standard method of proving statements involving an integer n, generally a

positive integer.

Principle of Mathematical Induction

Let P(n) denote a statement involving a positive integer n. If the following two

conditions are satisfied:

1. the statement P(1) is true,

2. whenever the statement P(k) is true for a positive integer k, then the statement

P(kþ 1) is also true,

then the statement P(n) is true for all positive integers n.

Sometimes we need to use an equivalent version of the Principle.

(Second) Principle of Mathematical Induction

Let P(n) denote a statement involving a positive integer n. If the following two

conditions are satisfied:

1. the statement P(1) is true,

2. whenever the statements P(1), P(2), . . ., P(k) are true for a positive integer k, then

the statement P(kþ 1) is also true,

then the statement P(n) is true for all positive integers n.

We have stated the Principle when P(n) applies to all integers n� 1; analogous

results hold when P(n) applies to all integers n�N, whatever integer N may be.

We end with some useful results that can be proved using the Principle of

Mathematical Induction.

Sums

X

n

k¼1

1 ¼ n;X

n

k¼1

k ¼ n nþ 1ð Þ2

;

X

n

k¼1

k2 ¼ n nþ 1ð Þ 2nþ 1ð Þ6

;X

n

k¼1

k3 ¼ n nþ 1ð Þ2

� �2

;

X

n

k¼0

2k þ 1ð Þ ¼ nþ 1ð Þ2;

sin Aþ sin Aþ Bð Þ þ � � � þ sin Aþ n� 1ð ÞBð Þ

¼sin 1

2nB

� �

sin 12

B� � sin Aþ n� 1

2B

� �

; B 6¼ 0:

Arithmetic progression

aþ aþ dð Þ þ aþ 2dð Þ þ � � � þ aþ n� 1ð Þdð Þ ¼ n aþ n�12

d� �

:

Thus, in order to use thePrinciple, we have a two stagestrategy:

1. check P(1),

2. check that P(k))P(kþ 1).

Sets, functions and proofs 357

Geometric progression

aþ ar þ ar2 þ � � � þ arn�1 ¼ a 1�rn

1�r; r 6¼ 1:

Binomial Theorem

1þ xð Þn ¼X

n

k¼0

n

k

� �

xk ¼ 1þ nxþ n n� 1ð Þ2!

x2 þ � � � þ xn;

aþ bð Þn ¼X

n

k¼0

n

k

� �

an�kbk

¼ an þ nan�1bþ n n� 1ð Þ2!

an�2b2 þ � � � þ bn:

It follows that the sum of the

geometric seriesP

1

k¼0

ark, for

jrj< 1, is a1�r

.

Here,n

k

� �

¼ n!k! n�kð Þ! and

0!¼ 1.

358 Appendix 1

Appendix 2: Standard derivativesand primitives

f(x) f 0(x) Domain

k 0 R

x 1 R

xn, n2Z� {0} nxn�1R

x�, �2R �x��1 (0,1)

ax, a> 0 ax loge a R

xx xx (1þ loge x) (0,1)

sin x cos x R

cos x �sin x R

tan x sec2 x R � nþ 12

� �

p :n 2 Z� �

cosec x �cosec x cot x R � {np : n2Z}

sec x sec x tan x R � nþ 12

� �

p :n 2 Z� �

cot x �cosec2 x R � {np : n2Z}

sin�1 x 1ffiffiffiffiffiffiffiffi

1�x2p (�1, 1)

cos�1 x �1ffiffiffiffiffiffiffiffi

1�x2p (�1, 1)

tan�1 x 11þx2 R

ex exR

loge x 1x

(0,1)

sinh x cosh x R

cosh x sinh x R

tanh x sech2 x R

sinh�1 x 1ffiffiffiffiffiffiffiffi

1þx2p R

cosh�1 x 1ffiffiffiffiffiffiffiffi

x2�1p (1,1)

tanh�1 x 11�x2 (�1, 1)

359

f (x) Primitive F(x) Domain

xn, n2Z� {�1} xnþ1

nþ1R

x�, �2R � {�1} x�þ1

�þ1(0,1)

ax, a> 0 ax

logeaR

sin x �cos x R

cos x sin x R

tan x loge (sec x) � 12p; 1

2p

� �

ex exR

1x

loge x (0,1)1x

loge jxj (�1, 0)

loge x x loge x� x (0,1)

sinh x cosh x R

cosh x sinh x R

tanh x loge (cosh x) R

1a2�x2, a> 0 1

2aloge

aþxa�x

� �

(�a, a)1

a2þx2, a> 0 1a

tan�1 xa

� �

R

1ffiffiffiffiffiffiffiffiffi

a2�x2p , a> 0

sin�1 xa

� �

�cos�1 xa

� �

(

(�a, a)

(�a, a)


x2�a2p , a> 0 loge xþ


x2 � a2p� �

cosh�1 xa

� �

(

(a,1)

(a,1)


a2þx2p , a> 0 loge xþ


a2 þ x2p

� �

sinh�1 xa

� �

R

Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 � x2p

, a> 0 12

xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 � x2p

þ 12

a2 sin�1 xa

� �

(�a, a)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � a2p

, a> 0 12


x2 � a2p

� 12

a2 loge xþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � a2p� �

(a,1)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ x2p

, a> 0 12


a2 þ x2p

þ 12

a2 loge xþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ x2p

� �

R

eax sin bx, a,b 6¼ 0 eax

a2þb2 a sin bx� b cos bxð Þ R

eax cos bx, a,b 6¼ 0 eax

a2þb2 a cos bxþ b sin bxð Þ R

360 Appendix 2

Appendix 3: The first 1000 decimal places offfiffiffi

2p

, e and p

More than a million decimal places of these numbers, and various similar commonly

arising numbers, are easily available on the internet.

ffiffiffi

2p¼ 1.

4142135623 7309504880 1688724209 6980785696 7187537694 8073176679

7379907324 7846210703 8850387534 3276415727 3501384623 0912297024

9248360558 5073721264 4121497099 9358314132 2266592750 5592755799

9505011527 8206057147 0109559971 6059702745 3459686201 4728517418

6408891986 0955232923 0484308714 3214508397 6260362799 5251407989

6872533965 4633180882 9640620615 2583523950 5474575028 7759961729

8355752203 3753185701 1354374603 4084988471 6038689997 0699004815

0305440277 9031645424 7823068492 9369186215 8057846311 1596668713

0130156185 6898723723 5288509264 8612494977 1542183342 0428568606

0146824720 7714358548 7415565706 9677653720 2264854470 1585880162

0758474922 6572260020 8558446652 1458398893 9443709265 9180031138

8246468157 0826301005 9485870400 3186480342 1948972782 9064104507

2636881313 7398552561 1732204024 5091227700 2269411275 7362728049

5738108967 5040183698 6836845072 5799364729 0607629969 4138047565

4823728997 1803268024 7442062926 9124859052 1810044598 4215059112

0249441341 7285314781 0580360337 1077309182 8693147101 7111168391

6581726889 4197587165 8215212822 9518488472 . . .

e¼ 2.

7182818284 5904523536 0287471352 6624977572 4709369995 9574966967

6277240766 3035354759 4571382178 5251664274 2746639193 2003059921

8174135966 2904357290 0334295260 5956307381 3232862794 3490763233

8298807531 9525101901 1573834187 9307021540 8914993488 4167509244

7614606680 8226480016 8477411853 7423454424 3710753907 7744992069

5517027618 3860626133 1384583000 7520449338 2656029760 6737113200

7093287091 2744374704 7230696977 2093101416 9283681902 5515108657

4637721112 5238978442 5056953696 7707854499 6996794686 4454905987

9316368892 3009879312 7736178215 4249992295 7635148220 8269895193

6680331825 2886939849 6465105820 9392398294 8879332036 2509443117

3012381970 6841614039 7019837679 3206832823 7646480429 5311802328

7825098194 5581530175 6717361332 0698112509 9618188159 3041690351

5988885193 4580727386 6738589422 8792284998 9208680582 5749279610

4841984443 6346324496 8487560233 6248270419 7862320900 2160990235

3043699418 4914631409 3431738143 6405462531 5209618369 0888707016

7683964243 7814059271 4563549061 3031072085 1038375051 0115747704

1718986106 8739696552 1267154688 9570350354 . . .

Numerical Analysts enjoyadding more digits to suchdecimals, using ever-moresophisticated computingtechniques!

For most practical purposesffiffiffi

2p’ 1:414:

For most practical purposese ’ 2:718:

A useful fact sometimes isthat e3’ 20.

361

p¼ 3.

1415926535 8979323846 2643383279 5028841971 6939937510 5820974944

5923078164 0628620899 8628034825 3421170679 8214808651 3282306647

0938446095 5058223172 5359408128 4811174502 8410270193 8521105559

6446229489 5493038196 4428810975 6659334461 2847564823 3786783165

2712019091 4564856692 3460348610 4543266482 1339360726 0249141273

7245870066 0631558817 4881520920 9628292540 9171536436 7892590360

0113305305 4882046652 1384146951 9415116094 3305727036 5759591953

0921861173 8193261179 3105118548 0744623799 6274956735 1885752724

8912279381 8301194912 9833673362 4406566430 8602139494 6395224737

1907021798 6094370277 0539217176 2931767523 8467481846 7669405132

0005681271 4526356082 7785771342 7577896091 7363717872 1468440901

2249534301 4654958537 1050792279 6892589235 4201995611 2129021960

8640344181 5981362977 4771309960 5187072113 4999999837 2978049951

0597317328 1609631859 5024459455 3469083026 4252230825 3344685035

2619311881 7101000313 7838752886 5875332083 8142061717 7669147303

5982534904 2875546873 1159562863 8823537875 9375195778 1857780532

1712268066 1300192787 6611195909 2164201989 . . .

For most practical purposesp ’ 3:1416:

A useful fact sometimes isthat p2’ 10.

362 Appendix 3

Appendix 4: Solutions to the problems

Chapter 1

Section 1.1

1. �1 < � 1720< � 45

53< 0 < 45

53< 17

20< 1:

2. Let a and b be two distinct rational numbers, where a< b. Let c ¼ 12

aþ bð Þ; then

c is rational, and

c� a ¼ 1

2b� að Þ > 0; and

b� c ¼ 1

2b� að Þ > 0;

so that a< c< b.

3. 17¼ 0:142857142857 . . .; 2

13¼ 0:153846153846 . . ..

4. (a) Let x ¼ 0:231.

Multiplying both sides by 103, we obtain

1000x ¼ 231:231 ¼ 231þ x:

Hence

999x ¼ 231) x ¼ 231

999¼ 77

333:

(b) Let x ¼ 0:81.

Multiplying both sides by 102, we obtain

100x ¼ 81:81 ¼ 81þ x:

Hence

99x ¼ 81 ) x ¼ 81

99¼ 9

11:

Thus

2:281 ¼ 2þ 2

10þ 9

110¼ 251

110:

5. 1720¼ 0:85 and 45

53¼ 0:84 . . ., so that 45

53< 17

20.

6. Suppose that there exists a rational number x such that x3¼ 2. Then we can write

x ¼ pq. By cancelling, if necessary, we may assume that p and q have no common

factor. The equation x3¼ 2 now becomes

p3 ¼ 2q3:

Now, the cube of an odd number is odd, because

2k þ 1ð Þ3 ¼ 8k3 þ 12k2 þ 6k þ 1

¼ 2 4k3 þ 6k2 þ 3k� �

þ 1;

and so p must be even. Hence we can write p¼ 2r, say. Our equation now becomes

363

2rð Þ3¼ 2q3;

so that

q3 ¼ 4r3:

Hence q is also even, so that p and q do have a common factor 2, which contradicts

our earlier statement that p and q have no common factors.

Thus, no such number x exists.

7. We may take, for example, x¼ 0.34 and y¼ 0.34001000100001. . ..

8. Let a and b have decimal representations

a ¼ a0 � a1a2a3 . . . and b ¼ b0 � b1b2b3 . . .;

where we arrange that a does not end in recurring 9s, whereas b does not terminate

(this latter can be arranged by replacing a terminating representation by an equiva-

lent representation that ends in recurring 9s).

Since a< b, there must be some integer n such that

a0 ¼ b0; a1 ¼ b1; . . . ; an�1 ¼ bn�1; but an < bn:

Then x¼ a0 � a1a2a3 . . . an� 1bn is rational, and a< x< b as required.

Next, let c ¼ 12

xþ bð Þ, so that x< c< b. By repeating the same procedure, this

time to the interval between the numbers c and b, we can find a rational number y

with c< y< b.

We have thus constructed two distinct rational numbers between a and b, as

required.

Section 1.2

1. Rule 1 For any a, b2R , a� b, b� a� 0.

Rule 2 For any a, b, c2R , a� b, aþ c� bþ c.

Rule 3

� For any a, b2R and any c> 0, a� b, ac� bc;

� For any a, b2R and any c< 0, a� b, ac� bc.

Remark

Note that the following results are NOT true in general:

� For any a, b2R and any c� 0, a� b, ac� bc;

� For any a, b2R and any c� 0, a� b, ac� bc.

For, if c¼ 0, then we can make no assertion as to whether a� b or a� b from

the information that ac� bc or ac� bc. To see this, take in turn a¼ 2, b¼ 3 and

a¼ 2, b¼ 1.

Rule 4 (Reciprocal Rule) For any positive a, b2R , a � b, 1a� 1

b.

Rule 5 (Power Rule) For any non-negative a, b2R , and any p> 0, a� b, ap� bp.

2. (a) xþ 3> 5.

(b) 2� x< 0.

(c) 5xþ 2> 12.

(d) �1ð5xþ 2Þ >

�112

.

3. (a) Rearranging the inequality, we obtain

4x� x2 � 7

x2 � 1� 3, 4x� x2 � 7

x2 � 1� 3 � 0 , 4x� 4x2 � 4

x2 � 1� 0

, x2 � xþ 1

x2 � 1� 0 ,

x� 12

� �2þ 34

x2 � 1� 0:

Here a0, b0 are non-negativeintegers, and a1, b1, a2, b2, . . .are digits.

Note that x does not end inrecurring 9s, from the way inwhich it was constructed.

364 Appendix 4

Since x� 12

� �2þ 34> 0, for all x, the inequality holds if and only if x2� 1< 0.

Hence the solution set is

x :4x� x2 � 7

x2 � 1� 3

� �

¼ �1; 1ð Þ:

(b) Rearranging the inequality, we obtain

2x2 � xþ 1ð Þ2 , 2x2 � x2 þ 2xþ 1

, x2 � 2x� 1 � 0

, x� 1ð Þ2� 2:


x : 2x2 � xþ 1ð Þ2n o

¼ x : x� 1 � �ffiffiffi

2pn o

[ x : x� 1 �ffiffiffi

2pn o

¼ �1; 1�ffiffiffi

2p� i

[ 1þffiffiffi

2p

;1h �

:

4. We can obtain an equivalent inequality by squaring, provided that both sides are non-

negative.

Now,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

is defined when 2x2� 2� 0; that is, for x2� 1. So,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

is

defined and non-negative if x lies in (�1, �1][ [1,1).

Hence, for x� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

> x, 2x2 � 2 > x2

, x2 > 2:

So the part of the solution set in [1,1) isffiffiffi

2p

; 1� �

.

Suppose, next, that x��1. Thenffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

� 0 > x;

so that the whole of (�1, �1] lies in the solution set.

Hence the complete solution set is

x :ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2x2 � 2p

> xn o

¼ �1;�1ð � [ffiffiffi

2p

;1� �

:

5. (a) 2x2 � 13

< 5, �5 < 2x2 � 13 < 5 ðby Rule 6Þ, 8 < 2x2 < 18

, 4 < x2 < 9

, 2 < xj j < 3:


x : 2x2 � 13

< 5 �

¼ �3;�2ð Þ [ 2; 3ð Þ:

(b) x� 1j j � 2 xþ 1j j , x� 1ð Þ2� 4 xþ 1ð Þ2

, x2 � 2xþ 1 � 4x2 þ 8xþ 4

, 0 � 3x2 þ 10xþ 3

, 0 � 3xþ 1ð Þ xþ 3ð Þ:Hence the solution set is

x : x� 1j j � 2 xþ 1j jf g ¼ �1;�3ð � [ � 1

3;1

�

:

Solutions to the problems 365

Section 1.3

1. (a) Suppose that aj j � 12.

The Triangle Inequality gives

aþ 1j j � aj j þ 1

� 1

2þ 1 ðsince aj j � 1

2Þ

¼ 3

2:

Hence

aþ 1j j � 3

2:

(b) Suppose that bj j < 12.

The ‘reverse form’ of the Triangle Inequality gives

b3 � 1

� jjb3j � 1j¼ bj j3�1

� 1� bj j3:Now

bj j < 1

2) bj j3< 1

8) 1� bj j3 > 7

8;

so, from the previous chain of inequalities

b3 � 1

>7

8:

2. Rearranging the inequality, we obtain

3n

n2 þ 2< 1, 3n < n2 þ 2

, 0 < n2 � 3nþ 2

, 0 < n� 1ð Þ n� 2ð Þ;and this final inequality certainly holds for n> 2.

3. The result is true for all those a and b for which aþ b� 0.

We now consider those a and b for which aþ b> 0. Rearranging the given

inequality, we obtain

aþ bffiffiffi

2p �


a2 þ b2p

, aþ bð Þ2

2� a2 þ b2

, a2 þ 2abþ b2 � 2a2 þ 2b2

, 0 � a2 � 2abþ b2

, 0 � a� bð Þ2;

and this final inequality certainly holds.

This completes the proof.

4. (a) Using the rules for rearranging inequalities, we obtain

1

2aþ 2

a

�

< a, 1

2� 2

a< a� 1

2a

, 1

a<

1

2a

, 2 < a2 ðsince a > 0Þ:

366 Appendix 4

Since the final inequality is true, the first inequality must also be true. Hence

1

2a þ 2

a

�

< a:

(b) In Example 3 and the subsequent remark, we saw that, if a 6¼ b, then

ab <a þ b

2

� 2

: ( )

Now let b ¼ 2a . Then a 6¼ b, since a > 2

a (as a2 > 2); so, by ( ), it follows that

a � 2

a <

1

2a þ 2

a

� � 2

;

which gives the required inequality.

Alternatively , use a direct argument after squaring the expression 12

a þ 2a

� �

:

5. Since c , d � 0, we can choose non-negative numbers a and b with c ¼ a2 and d ¼ b2 .

Substituting into the resultffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

� a þ b, for a, b � 0, of Example 4, we obtainffiffiffiffiffiffiffiffiffiffiffi

cþ dp

�ffiffiffi

cpþ

ffiffiffi

dp

:

6. In Problem 5 we saw thatffiffiffiffiffiffiffiffiffiffiffi

cþ dp

�ffiffiffi

cpþ

ffiffiffi

dp

; for numbers c; d � 0: ( )

Following the good advice in the margin note before the Problem, we use this!

For a, b, c� 0, we deduce from () that

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

aþ bþ cp

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

aþ bþ cð Þp

�ffiffiffi

apþ


bþ cp

ðby an application of ðÞÞ�

ffiffiffi

apþ

ffiffiffi

bpþ

ffiffiffi

cp

ðby a further application of ðÞÞ:

7. If we substitute x ¼ 1n

in the Binomial Theorem for (1þ x)n, when n� 2 we get

1þ 1

n

� n

¼ 1þ n1

n

�

þ n n� 1ð Þ2!

1

n

� 2

þ � � � þ 1

n

� n

� 1þ 1þ n� 1

2n

¼ 2þ 1

2� 1

2n¼ 5

2� 1

2n:

When n¼ 1, the inequality also holds. This completes the proof.

8. Let P(n) be the statement

PðnÞ : 4n > n4:

STEP 1 First we show that P(5) is true: 45� 54.

Since 45¼ 1024 and 54¼ 625, P(5) is certainly true.

STEP 2 We now assume that P(k) holds for some k� 5, and deduce that P(kþ 1) is

then true.

So, we are assuming that 4k> k4. Multiplying this inequality by 4 we get

4kþ1 > 4k4;

so it is therefore sufficient for our purposes to prove that 4k4� (kþ 1)4.

Now

4k4 � k þ 1ð Þ4, 4 � 1þ 1

k

� 4

: ( )

As k increases, the expression 1þ 1k

decreases. Since k� 5, 1þ 1k�

1þ 15¼ 6

5; it follows that

This assumption is just P(k).

Since P(kþ 1) is:4kþ 1> (kþ 1)4.

1þ 1

k

� 4

� 6

5

� 4

¼ 1296

625¼ 2:0736 < 4:

Thus the inequality () certainly holds for k� 5, and so it follows that

4kþ 1� (kþ 1)4 also holds for k� 5.

In other words: P(k) true for some k� 5)P(kþ 1) true.

It follows, by the Principle of Mathematical Induction, that 4n> n4,

for n� 5.

9. Substituting x ¼ �1ð2nÞ into Bernouilli’s Inequality (1þ x)n� 1þ nx (which we may

do since �1ð2nÞ � �1 for all natural numbers n), we obtain

1� 1

2n

� n

� 1� n� 1

2n

¼ 1

2:

If we take the nth root of both sides of this final inequality (which is permissible, by

the Power Rule), we find that

1� 1

2n� 1

21n

;

so that

21n � 1

1� 12n

¼ 2n

2n� 1

¼ 1þ 1

2n� 1:

10. If we apply the Cauchy–Schwarz Inequality, withffiffiffiffiffi

akp

in place of ak and 1ffiffiffiffi

akp in

place of bk, we have

a1 þ a2 þ � � � þ anð Þ 1

a1

þ 1

a2

þ � � � þ 1

an

�

� ffiffiffiffiffi

a1

p � 1ffiffiffiffiffi

a1p þ ffiffiffiffiffi

a2


a2p þ � � � þ ffiffiffiffiffi

an


anp

� 2

¼ 1þ 1þ � � � þ 1ð Þ2 ðwith n terms in the bracketÞ

¼ n2:

11. Applying Theorem 3 to the nþ 1 positive numbers 1; 1þ 1n; 1þ 1

n; . . . ; 1þ 1

n,

we obtain

1þ 1

n

� n�

1nþ1

� 1

nþ 1� nþ 1þ n� 1

n

�

¼ nþ 2

nþ 1

¼ 1þ 1

nþ 1;

taking the (nþ 1)th power of this last inequality, by the Power Rule, we deduce

that

1þ 1

n

� n

� 1þ 1

nþ 1

� nþ1

:

368 Appendix 4

Section 1.4

1. (a)

E1 is bounded above. For example, M¼ 1 is an upper bound of E1, since

x � 1; for all x 2 E1:

Also, max E1¼ 1, since 12E1.

(b)

E2 is bounded above. For example, M¼ 1 is an upper bound of E2, since

1� 1

n� 1; for n ¼ 1; 2; . . .:

However, E2 has no maximum element. If x2E2, then x ¼ 1� 1n

for some

positive integer n; and so there is another element of E2, such as 1� 1nþ1

, with

1� 1

n< 1� 1

nþ 1since

1

n>

1

nþ 1

�

:

Hence x is not a maximum element of E2.

(c)

The set E3 is not bounded above, and so it cannot have a maximum element. For

each number M, there is a positive integer n such that n2>M (for instance, take

n>M, which implies that n2> n>M).

Hence M cannot be an upper bound of E3.

2. (a) The set E1¼ (�1, 1] is not bounded below, and so it cannot have a minimum

element. For each number m, there is a (negative) number x such that x<m.

Since x2E1, m cannot be a lower bound of E1.

(b) E2 is bounded below by 0, since

1� 1

n� 0; for n ¼ 1; 2 . . .:

Also, 02E2, and so min E2¼ 0.

(c) E3 is bounded below by 1, since

n2 � 1; for n ¼ 1; 2 . . .:

Also, 12E3, and so min E3¼ 1.

3. ƒ is increasing on the interval [�3, �2). Since �3� x<�2 so that x22 (4, 9], we

have f xð Þ ¼ 1x2 2 1

9, 1

4

� �

. Hence ƒ is bounded above and bounded below.

Next, since f ð�3Þ ¼ 19

and 19

is a lower bound for ƒ on the interval [�3, �2), it

follows that ƒ has a minimum value of 19

on this interval.

Finally, 14

is an upper bound for ƒ on the interval [�3, �2) but there is no point x

in [�3, �2) for which f ðxÞ ¼ 14. So 1

4cannot be a maximum of f on the interval.

However, if y is any number in 19; 1

4

� �

, there is a number x > � 1ffiffi

yp in

� 1ffiffi

yp ;�2

� �

�3;�2½ Þ such that f xð Þ ¼ 1x2 > y, so that no number in 1

9, 1

4

� �

will

serve as a maximum of f on the interval. It follows that f has no maximum value

on [�3, �2).

4. (a) The set E1¼ (�1, 1] has a maximum element 1, and so

sup E1 ¼ max E1 ¼ 1:

(b) We know that E2 ¼ f1� 1n

: n ¼ 1, 2, . . .g is bounded above by 1, since

1� 1

n� 1; for n ¼ 1; 2 . . .:

To show that M¼ 1 is the least upper bound of E2, we prove that, if M0< 1, then

there is an element 1� 1n

of E2 such that

1� 1

n> M0:

However, since M0< 1, we have

1� 1

n> M0 , 1�M0 >

1

n

, n >1

1�M0ðsince 1�M0 > 0Þ:

Choosing a positive integer n so large that n > 11�M0, we obtain 1� 1

n> M0, as

required. Hence 1 is the least upper bound of E2.

(c) The set E3¼ {n2: n¼ 1, 2, . . .} is not bounded above, and so it cannot have a

least upper bound.

5. (a) The set E1¼ (1, 5] is bounded below by 1, since

1 � x; for all x 2 E1:

To show that m¼ 1 is the greatest lower bound of E1, we prove that, if m0> 1,

then there is an element x in E1 which is less than m0. Since m0> 1, there is a

number x of the form x¼ 1.00 . . . 01 such that

1 < x < m0;

and clearly x2E1.

Hence 1 is the greatest lower bound of E1.

(b) The set E2 ¼ f 1n2 : n ¼ 1, 2, . . .g is bounded below by 0, since

0 <1

n2; for n ¼ 1; 2; . . .:

To show that m¼ 0 is the greatest lower bound of E2, we prove that, if m0> 0,

then there is an element 1n2 in E2 such that 1

n2 < m0. Since m0> 0, we have

1

n2< m0 , n2 >

1

m0

, n >1ffiffiffiffiffi

m0p :

Choosing a positive integer n so large that n > 1ffiffiffiffi

m0p , we obtain 1

n2 < m0, as

required. Hence 0 is the greatest lower bound of E2.

6. ƒ is decreasing on the interval [1, 4). Since 1� x< 4 so that x22 [1, 16), we have

f xð Þ ¼ 1x2 2 1

16, 1

� �

. Hence ƒ is bounded above by 1 and bounded below by 116

.

Next, since f(1)¼ 1 and 1 is an upper bound for ƒ on the interval [1, 4), it follows

that ƒ has least upper bound 1 on [1, 4).

Finally, 116

is a lower bound for ƒ on the interval [1, 4) but there is no point x in

[1, 4) for which f ðxÞ ¼ 116

.

However, if y is any number in 116

, 1� �

, there is a number x < 1ffiffi

yp in [1, 4) such that

f xð Þ ¼ 1x2 < y, so that no number in 1

16, 1

� �

will serve as a lower bound of ƒ on [1, 4).

It follows that ƒ has 116

as its greatest lower bound on [1, 4).

370 Appendix 4

Chapter 2

Section 2.1

1. (a) (i) 4, 7, 10, 13, 16;

(ii) 13

, 19

, 127

, 181

, 1243

;

(iii) �1, 2, �3, 4, �5.

(b) (i) a1¼ 1, a2¼ 2, a3¼ 6, a4¼ 24, a5¼ 120;

(ii) a1¼ 2, a2¼ 2.25, a3¼ 2.37, a4¼ 2.44, a5¼ 2.49.

2. (a)

(b)

(c)

(d)

3. (a) n! is monotonic, because

an ¼ n! and anþ1 ¼ nþ 1ð Þ!;


so that

anþ1� an ¼ nþ 1ð Þ!� n!

¼ n� n! > 0; for n ¼ 1; 2; . . .:

Thus {n!} is increasing.

Alternatively, an> 0, for all n, and

anþ1

an

¼ nþ 1ð Þ!n!

¼ nþ 1 � 1; for n ¼ 1; 2; . . .;

so that {n!} is increasing.

(b) {2�n} is monotonic, because

an ¼ 2�n and anþ1 ¼ 2� nþ1ð Þ;

so that

anþ1 � an ¼1

2nþ1� 1

2n

¼ 1

2n

1

2� 1

�

< 0; for n ¼ 1; 2; . . .:

Thus {2�n} is decreasing.

Alternatively, an> 0, for all n, and

anþ1

an

¼ 2n

2nþ1¼ 1

2< 1; for n ¼ 1; 2; . . .;

so that {2�n} is decreasing.

(c) nþ 1n

�

is monotonic, because

an ¼ nþ 1

nand anþ1 ¼ nþ 1þ 1

nþ 1;

so that

anþ1 � an ¼ nþ 1þ 1

nþ 1

�

� nþ 1

n

�

¼ n nþ 1ð Þ � 1

n nþ 1ð Þ > 0; for n ¼ 1; 2; . . .:

Thus nþ 1n

�

is increasing.

4. (a) TRUE: 2n> 1000, for n> 9, since {2n} is increasing and 210¼ 1024.

(b) FALSE: All the terms a1, a3, a5, . . . are negative.

(c) TRUE: 1n< 0:025, for all n > 1

0:025¼ 40:

(d) TRUE: an> 0 for all n, and anþ1

an¼ 1

4nþ1

n

� �4:

Now

1

4

nþ 1

n

� 4

� 1, nþ 1

n

� 4

� 4 , 1þ 1

n� 4

14

, 1

n�

ffiffiffi

2p� 1 , n � 1

ffiffiffi

2p� 1’ 2:414:

Soanþ1

an

� 1; for n > 2:

Hence

anþ1 � an; for n > 2;

so that

n4

4n

�

is eventually decreasing.

372 Appendix 4

Section 2.2

1. (a) 1n< 1

100, n > 100, by the Reciprocal Rule (for positive numbers). Hence we

may take X¼ 100.

(b) 1n< 3

1000, n > 1000

3¼ 333:3333 . . ., by the Reciprocal Rule (for positive num-

bers). Hence we may take X¼ 333.

2. (a) �1ð Þn

n2

<1

100, 1

n2<

1

100

, n2 > 100

, n > 10:

Hence we may take X¼ 10.

(b) �1ð Þn

n2

<3

1000, 1

n2<

3

1000

, n2 >1000

3

, n >


1000

3

r

’ 18:25:

Hence we may take X¼ 18.

3. (a) The sequencen

12n� 1

o

is a null sequence.

To prove this, we want to show that:


1

2n� 1

< "; for all n > X: ( )

We know that

1

2n� 1

< ", 1

2n� 1< " ðsince 2n� 1 > 0Þ

, 2n� 1 >1

"

, n > 12

1þ 1

"

�

;

it follows that () holds if we take X ¼ 12

�

1þ 1"

�

. Hencen

12n�1

o

is null.

(b) The sequence�1ð Þn10

n o

is not a null sequence.

To prove this, we must find a positive value of " such that the sequence does

not eventually lie in the horizontal strip on the sequence diagram from �" up

to ". We can take " ¼ 120

, as the following sequence diagram shows:

Any value for X greater than100 will also be valid.

Any value for X greater than333 will also be valid.

Any value for X greater than 10will also be valid.

Any value for X greater than 18will also be valid.

There is NO value of X such that the following statement holds

�1ð Þn

10

<1

20; for all n > X:

(c) The sequence�1ð Þnn4þ1

n o

is a null sequence.

To prove this, we want to show that:


�1ð Þn

n4 þ 1

< "; for all n > X: ( )

We know that

�1ð Þn

n4 þ 1

< ", 1

n4 þ 1< "

, n4 þ 1 >1

"

, n4 >1

"� 1:

Now, if "� 1, then 1" � 1 � 0, so that

n4 >1

"� 1; for n ¼ 1; 2; . . .;

hence () holds with X¼ 1.

On the other hand, if 0<"< 1, then 1" � 1 > 0, so that

n4 >1

"� 1, n >

1

"� 1

� 14

;

hence () holds with X ¼ 1" � 1� �1

4.

Thus () holds in either case. Hence�1ð Þn

n4 þ 1

n o

is null.

4. (a) We know thatn

12n�1

o

is null, and so 1

2n�1ð Þ3n o

is also null, by the Power Rule.

(b) The sequences 1n

�

andn

12n�1

o

are null, so 6ffiffi

n5p

n o

and 5

2n�1ð Þ7n o

are also null, by

the Power Rule and Multiple Rule.

Hence the sequence 6ffiffi

n5p þ 5

2n�1ð Þ7n o

is null, by the Sum Rule.

(c) The sequences 1n

�

andn

12n�1

o

are null, so 1n4

�

and 1

2n�1ð Þ13

� �

are also null, by

the Power Rule.

Hence 1

3n4 2n�1ð Þ13

� �

is also null, by the Product Rule and Multiple Rule.

5. Here, using the Hint, we have

n1

2

� n

¼ n� 1

2n

� n� 1

n2¼ 1

n:

Thus, the sequence 1n

�

dominates the sequence n 12

� �n �

, and is itself null. It

follows, from the Squeeze Rule, that the sequence n 12

� �n �

is null.

6. (a) We guess that 1n2þn

n o

is dominated by 1n

�

.

To check this, we have to show that

1

n2 þ n� 1

n; for n ¼ 1; 2; . . .;

this certainly holds, because

n2 þ n � n; for n ¼ 1; 2; . . .:

Since 1n

�

is null, we deduce that 1n2þn

n o

is null, by the Squeeze Rule.

374 Appendix 4

(b) We guess that�1ð Þnn!

n o

is dominated by 1n

�

.


�1ð Þn

n!

� 1

n; for n ¼ 1; 2; . . . ;


�1ð Þn

n!

¼ 1

n!; for n ¼ 1; 2; . . .;

and

n! � n; for n ¼ 1; 2; . . .:

Since 1n

�

is null, we deduce that �1ð Þn

n!

�


(c) We guess that sin n2

n2 þ 2n

n o

is dominated by 1n2

�

.


sin n2

n2 þ 2n

� 1

n2; for n ¼ 1; 2; . . .;


sin n2

� 1; for n ¼ 1; 2; . . .;

and

n2 þ 2n � n2; for n ¼ 1; 2; . . .:

Since 1n2

�

is null, we deduce that sin n2

n2 þ 2n

n o


Section 2.3

1. (a)

The sequence appears to converge to 1.

(b) Since bn ¼ an � 1 ¼ nþ 1n� 1 ¼ 1

n, it follows that bnf g ¼ 1

n

�

is a null sequence.

2. (a)


Here an � 1 ¼ n2 � 1n2 þ 1� 1 ¼ �2

n2 þ 1. We know that �2

n2 þ 1

n o

is a null sequence, so it

follows that {an� 1} is also a null sequence.

Hence {an} converges to 1.

(b)

Here an � 12¼ n3þ �1ð Þn

2n3 � 12¼ �1ð Þn

2n3 . We know that�1ð Þn2n3

n o

is a null sequence, so it

follows that an � 12

�


Hence {an} converges to 12.

3. (a) The dominant term is n3, so we write an as

an ¼n3 þ 2n2 þ 3

2n3 þ 1

¼1þ 2

nþ 3

n3

2þ 1n3

:

Since 1n

�

and 1n3

�

are basic null sequences

limn!1

an ¼1þ 0þ 0

2þ 0¼ 1

2;


(b) The dominant term is 3n, so we write an as

an ¼n2 þ 2n

3n þ n3

¼n2

3n þ 23

� �n

1þ n3

3n

:

Since n2

3n

n o

, 23

� �n �

and n3

3n

n o


limn!1

an ¼0þ 0

1þ 0¼ 0;


(c) The dominant term is n!, so we write an as

an ¼n!þ �1ð Þn

2n þ 3n!

¼1þ �1ð Þn

n!2n

n! þ 3:

Since�1ð Þnn!

n o

and 2n

n!

�


limn!1

an ¼1þ 0

0þ 3¼ 1

3;


376 Appendix 4

4. (a) We know that

n1n � 1þ


2

n� 1

r

, n � 1þffiffiffiffiffiffiffiffiffiffiffi

2

n� 1

r

!n

:

Using the hint, with x ¼ffiffiffiffiffiffi

2n�1

q

, we obtain

1þffiffiffiffiffiffiffiffiffiffiffi

2

n� 1

r

!n

� n n� 1ð Þ2!


2

n� 1

r

!2

¼ n n� 1ð Þ2

� 2

n� 1¼ n;

as required.

(b) Since n� 1, we have n1n � 1. Combining this inequality with that in part (a), we

obtain

1 � n1n � 1þ


2

n� 1

r

; for n � 2:

Now,ffiffiffiffiffiffiffiffi

2n� 1

qn o1

n¼2is a null sequence, by the Power Rule, so that

limn!1

1þffiffiffiffiffiffiffiffiffiffiffi

2

n� 1

r

!

¼ 1:

Hence, by the Squeeze Rule, limn!1

n1n

� �

¼ 1:

Section 2.4

1. (a) {1þ (� 1)n} is bounded, since the terms 1þ (�1)n take only the values 0 and 2.

Hence

1þ �1ð Þnj j � 2; for n ¼ 1; 2; . . .:

(b) {(� 1)n n} is unbounded. Given any number K, there is a positive integer n such

that

�1ð Þnnj j > K;

for instance, n ¼ Kj j þ 1.

(c) 2nþ 1n

�

is bounded, since 2nþ 1n¼ 2þ 1

nso that

2nþ 1

n

¼ 2þ 1

n� 3; for n ¼ 1; 2; . . .:

(d) 1� 1n

� �n �

is bounded, since

0 � 1� 1

n� 1; for n ¼ 1; 2; . . .;

so that

1� 1

n

� n

¼ 1� 1

n

� n

� 1; for n ¼ 1; 2; . . .:

2. (a)ffiffiffi

npf g is unbounded, and hence divergent, by Corollary 1.

(b) n2 þ nn2 þ 1

n o

is convergent (with limit 1), and hence bounded, by Theorem 1. In fact

n2 þ n

n2 þ 1¼

1þ 1n

1þ 1n2

� 1þ 1

n� 2; for n ¼ 1; 2; . . .:

(c) {(� 1)nn2} is unbounded, and hence divergent, by Corollary 1.

(d) The terms of the sequence n �1ð Þn �

are

1; 2;1

3; 4;

1

5; 6; . . .;

so the sequence is unbounded: given any number K, there is an even positive

integer 2n such that 2n>K. Hence the sequence is divergent, by Corollary 1.


3. (a) Each term of 2n

n

�

is positive, and n2n

�

is a basic null sequence. Hence 2n

n!1,

by the Reciprocal Rule.

(b) First, note that

2n � n100 ¼ 2n 1� n100

2n

�

; for n ¼ 1; 2; . . .;

and that n100

2n

n o

is a basic null sequence. It follows that n100

2n is eventually less

than 1, so that 2n� n100 is eventually positive.

Also

limn!1

1

2n � n100¼ lim

n!1

12n

1� n100

2n

¼ 0

1� 0¼ 0;


Hence 2n� n100!1, by the Reciprocal Rule.

(c) We know that 2n

n!1, by part (a), and that

2n

nþ 5n100 � 2n

n; for n ¼ 1; 2; . . .:

Hence 2n

nþ 5n100 !1, by the Squeeze Rule.

Remark

You could have used the Reciprocal Rule or the Sum and Multiple Rules.

(d) Each term of the sequence 2n þ n2

n10 þ n

n o

is positive, and

limn!1

2n þ n2

n10 þ n

� �1

¼ limn!1

n10 þ n

2n þ n2

¼ limn!1

n10

2n þ n2n

1þ n2

2n

¼ 0þ 0

1þ 0¼ 0;


Hence 2n þ n2

n10 þ n!1, by the Reciprocal Rule.

4. (a) (i) a2¼ 4, a4¼ 16, a6¼ 36, a8¼ 64, a10¼ 100;

(ii) a3¼ 9, a7¼ 49, a11¼ 121, a15¼ 225, a19¼ 361;

(iii) a1¼ 1, a4¼ 16, a9¼ 81, a16¼ 256, a25¼ 625.

(b) a1 ¼ 1; a3 ¼ 13; a5 ¼ 1

5;

a2¼ 2, a4¼ 4, a6¼ 6.

5. (a) If an ¼ �1ð Þnþ 1n, for n¼ 1, 2, . . ., then

a2k ¼ 1þ 1

2kand a2k�1 ¼ �1þ 1

2k � 1;

so that

limk!1

a2k ¼ 1; whereas limk!1

a2k�1 ¼ �1:

Hence {an} is divergent, by the First Subsequence Rule.

(b) If an ¼ 13n� 1

3n� �

, for n¼ 1, 2, . . ., then

a3k ¼ 0 and a3kþ1 ¼1

3; for k ¼ 1; 2; . . .;

so that

limk!1

a3k ¼ 0; whereas limk!1

a3kþ1 ¼ 13:

Hence {an} is divergent, by the First Subsequence Rule.

378 Appendix 4

(c) If an ¼ n sin 12np

� �

, for n¼ 1, 2, . . ., then

a1 ¼ 1; a2 ¼ 0; a3 ¼ �3; a4 ¼ 0; a5 ¼ 5; a6 ¼ 0; . . .:

Now

a4kþ1 ¼ 4k þ 1ð Þ sin 2kpþ 1

2p

�

¼ 4k þ 1; for k ¼ 1; 2; . . .;

so that a4kþ1!1.

Hence {an} is divergent, by the Second Subsequence Rule.

Section 2.5

1. Suppose that the sequence {an} is decreasing and bounded below, so that it is

necessarily convergent. We show that the limit of {an} is the number

m¼ inf {an : n¼ 1, 2, . . .}.

To see this, let ‘ denote limn!1

an. By the Limit Inequality Rule, since an�m we know

that ‘�m. Now assume that in fact ‘>m.

Since ‘>m, it follows that ‘ > 12ð‘þ mÞ > m. It follows, from the definition of

greatest lower bound, that there then exists some integer X such that aX <12‘þ mð Þ;

and so, since {an} is decreasing, that

an <1

2‘þ mð Þ; for all n > X:

We deduce from the Limit Inequality Rule that ‘ � 12‘þ mð Þ. We may rearrange this

inequality as 2‘� ‘þm, or ‘�m.

This contradicts our assumption that ‘>m. It follows that, in fact, ‘¼m.

2. We use the ideas in the discussion prior to the problem.

(a) Let a1 > 3. Then, from equation (5) in the discussion before the statement of the

problem

a2 � a1 ¼1

4a1 � 1ð Þ a1 � 3ð Þ

> 0;

so that a2> a1. This suggests that, in general, it might be that anþ1> an in the

case that a1> 3; we check this, using Mathematical Induction.

Let P(n) be the statement: If a1> 3, then anþ 1> an.

We have just seen that the statement P(1) is true.

Now assume that the statement P(k) is true; that is, that akþ1> ak for some

k � 1. It follows, from equation ( 5) in the discussion before the statement of the

problem, that

akþ1 � ak ¼1

4ak � 1ð Þ ak � 3ð Þ

> 0;

so that akþ 1> ak; in other words, the statement P(kþ 1) is true.

This proves that P(n) holds for all n� 1.

Thus, if a1> 3, the sequence {an} is increasing.

It follows, from the Monotone Convergence Theorem, that either {an} con-

verges to some (finite) limit ‘ or tends to infinity as n!1.

But the only possible limits of the sequence are 1 and 3, so that since a1> 3

and the sequence is increasing, clearly {an} cannot tend to a limit. It follows thatan!1 as n!1.

(b) Let 0� a1 < 1. Then, from equation ( 5) in the discussion before the statement of

the problem,

We shall show that thisassumption leads to acontradiction.

By letting n!1.

For we cannot have ‘<m and‘�m!

We need a discussion of thistype in order to identify theresult that we wish to verifyusing Mathematical Induction.

a2 � a1 ¼1

4a1 � 1ð Þ a1 � 3ð Þ

> 0;

so that a2> a1. This suggests that, in general, it might be that anþ1> an in the

case that 0� a1< 1; we check this, using Mathematical Induction.

Let P(n) be the statement: If 0� a1< 1, then anþ1> an.

We have just seen that the statement P(1) is true.

Now assume that the statement P(k) is true; that is, that akþ1> ak for some

k � 1. It follows, from equation (5) in the discussion before the statement of the

problem, that

akþ1 � ak ¼1

4ak � 3ð Þ ak � 1ð Þ

> 0;

so that akþ 1> ak; in other words, the statement P(kþ 1) is true.

This proves that P(n) holds for all n� 1.

Thus, if 0� a1< 1, the sequence {an} is increasing.

It follows, from the Monotone Convergence Theorem, that either {an} con-

verges to some (finite) limit ‘ or tends to infinity as n!1.

Now, we can use equation (3) before the problem, with n¼ 1, to see that the

assumption a1< 1 implies that

a2 � 1 ¼ 1

4a2

1 � 1� �

< 0;

so that a2< 1. We can then prove, by Mathematical Induction, using an argu-

ment similar to that in part (a), that, if a1< 1, then an< 1 for all n� 1.

But the only possible limits of the sequence are 1 and 3, so that since an< 1 for

all n� 1, clearly {an} cannot tend to 3 or to infinity. It follows that an! 1 as

n!1.

(c) Let a1< 0. Since anþ1 ¼ 14

a2n þ 3

� �

, it is clear that the behaviour of the sequence

as n!1 depends only on the magnitude of a1 and not on its sign.

It follows from this observation that:

if �1< a1< 0, then an! 1 as n!1;

if a1¼�1, then an! 1 as n!1;

if �3< a1<�1, then an! 1 as n!1;

if a1¼�3, then an! 3 as n!1;

if a1<�3, then an!1 as n!1.

3. (a) The Binomial Theorem gives

1þ x

n

� �n

¼ 1þ nx

n

� �

þ n n� 1ð Þ2!

x

n

� �2

þ � � � þ x

n

� �n

¼ 1þ xþ 1

2!1� 1

n

�

x2 þ � � � þ xn

nn:

The general term in this expansion is

n n� 1ð Þ . . . n� k þ 1ð Þk!

x

n

� �k

¼ 1

k!1� 1

n

�

1� 2

n

�

. . . 1� k � 1

n

�

xk:

If k and x are fixed, this general term increases as n increases. Hence the

sequence 1þ xn

� �n �

is increasing if x> 0.

(b) By the Binomial Theorem

1þ 1

n

� k

� 1þ k1

n

�

¼ 1þ k

n; for k ¼ 1; 2; . . .: ( )

380 Appendix 4

To prove that 1þ xn

� �n �

is bounded above, we choose an integer k� x and use

the inequality (), as follows

1þ x

n

� �n

� 1þ k

n

� n

� 1þ 1

n

� k !n

¼ 1þ 1

n

� n� k

� ek;

since the sequence 1þ 1n

� �n �

is increasing with limit e. Hence 1þ xn

� �n �

is

bounded above by ek.

(c) Since 1þ xn

� �n �

is increasing and bounded above, it must be convergent, by the

Monotone Convergence Theorem.

4. The first n terms of the sequence 1þ 1n

� �n �

are

2

1

� 1

;3

2

� 2

;4

3

� 3

; . . . ;nþ 1

n

� n

:

Taking the product of these terms, each of which is less than e, we obtain

213243 . . . nþ 1ð Þn

112233 . . . nn< en;

it follows, by cancellation, that

nþ 1ð Þn

n!< en:

Hence

n! >nþ 1ð Þn

en¼ nþ 1

e

� n

; for n ¼ 1; 2; . . .:

5. We use the formulas:

(a) Area ¼ 12� 1� 1� sin 1

3p

� �

¼ 12�

ffiffi

3p

2¼

ffiffi

3p

4.

(b) Area ¼ 12� 1� 1� sin 1

6p

� �

¼ 12� 1

2¼ 1

4.

(c) Area ¼ 12� 2 tan 1

6p

� �

� 1 ¼ 12� 2� 1

ffiffi

3p ¼ 1

ffiffi

3p .

(d) Area ¼ 12� 2 tan 1

12p

� �

� 1 ¼ tan 112p

� �

.

To determine tan 112p

� �

, we use the formula

tan1

6p

�

¼2 tan 1

12p

� �

1� tan2 112p

� � :

Since tan 16p

� �

¼ 1ffiffi

3p , we obtain

tan2 1

12p

�

þ 2ffiffiffi

3p

tan 112p

� �

� 1 ¼ 0;

so that

tan1

12p

�

¼�2

ffiffiffi

3p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2ffiffiffi

3p� �2þ4

q

2

¼ �ffiffiffi

3p� 2:

Since tan 112p

� �

> 0, we must have tan 112p

� �

¼ 2�ffiffiffi

3p

; it follows that

Area ¼ 2�ffiffiffi

3p

:

Chapter 3

Section 3.1

1. (a) Using the formula for summing a geometric series, with a ¼ r ¼ � 13, we obtain

sn ¼ � 1

3

�

þ � 1

3

� 2

þ � 1

3

� 3

þ � � � þ � 1

3

� n

¼ � 1

3�

1� � 13

� �n

1� � 13

� �

¼ � 1

41� � 1

3

� n�

:

Since � 13

� �n �

is a basic null sequence

limn!1

sn ¼ �1

4;

and so

X

1

n¼1

� 1

3

� n

is convergent;with sum� 1

4:

(b) Here

sn ¼ 1þ �1ð Þ1þ �1ð Þ2þ � � � þ �1ð Þn�1

¼ 1� 1þ 1� 1þ � � � þ �1ð Þn�1;

so that

sn ¼1; n odd;0; n even:

�

Hence s2kþ1! 1 and s2k! 0 as k!1, so that {sn} is divergent, by the First

Subsequence Rule. Thus the seriesP

1

n¼0

�1ð Þn is divergent.

(c) Using the formula for summing a geometric series with a¼ 2 and r ¼ 12,

we obtain

sn ¼ 2þ 1þ 1

2

�

þ 1

2

� 2

þ � � � þ 1

2

� n�3

¼ 2�1� 1

2

� �n

1� 12

� �

¼ 4 1� 1

2

� n�

:

Since 12

� �n �

is a basic null sequence, limn!1

sn ¼ 4, and so

X

1

n¼�1

1

2

� n

is convergent;with sum 4:

2. (a) We interpret 0.111. . . as

1

101þ 1

102þ 1

103þ � � �:

382 Appendix 4

This is a geometric series with a ¼ 110

and r ¼ 110

. Since 110< 1, the series is

convergent with sum

a

1� r¼

110

1� 110

¼ 1

9;

hence

0:111 . . . ¼ 1

9:

(b) We interpret 0.86363. . . as

8

10þ 1

10

63

1001þ 63

1002þ � � �

�

:

The series in the bracket is a geometric series with a ¼ 63100

and r ¼ 1100

. Since1

100< 1, this series is convergent with sum

a

1� r¼

63100

1� 1100

¼ 63

99¼ 7

11;

hence

0:86363 . . . ¼ 8

10þ 7

110¼ 19

22:

(c) We interpret 0.999. . . as

9

101þ 9

102þ 9

103þ � � �:

This is a geometric series with a ¼ 910

and r ¼ 110

. Since 110< 1, this series is

convergent with sum

a

1� r¼

910

1� 110

¼ 1;

hence

0:999 . . . ¼ 1:

3. s1 ¼1

1� 2¼ 1

2;

s2 ¼1

1� 2þ 1

2� 3¼ 1

2þ 1

6¼ 2

3;

s3 ¼1

1� 2þ 1

2� 3þ 1

3� 4¼ 2

3þ 1

12¼ 3

4;

s4 ¼1

1� 2þ 1

2� 3þ 1

3� 4þ 1

4� 5¼ 3

4þ 1

20¼ 4

5:

4. Since

1

n nþ 2ð Þ ¼1

2

1

n� 1

nþ 2

�

; for n ¼ 1; 2; . . .;

we have

X

1

n¼1

1

n nþ 2ð Þ ¼X

1

n¼1

1

2

1

n� 1

nþ 2

�

;

and so

sn¼1

1�3þ 1

2�4þ 1

3�5þ 1

4�6þ �� þ 1

n nþ2ð Þ

¼ 1

21�1

3

�

þ 1

2�1

4

�

þ 1

3�1

5

�

þ 1

4�1

6

�

þ�� þ 1

n� 1

nþ2

� � �

:

Most of the terms in alternate brackets cancel out, leaving

sn ¼1

21þ 1

2� 1

nþ 1� 1

nþ 2

�

¼ 3

4� 1

2 nþ 1ð Þ �1

2 nþ 2ð Þ :

Since 1nþ 1

n o

and 1nþ 2

n o

are null sequences, limn!1

sn ¼ 34, and so

X

1

n¼1

1

n nþ 2ð Þ is convergent, with sum3

4:

5. The seriesP

1

n¼1

34

� �nis a geometric series, with a ¼ r ¼ 3

4. Hence, it is convergent, with

sum34

1�34

¼ 3.

The seriesP

1

n¼1

1n nþ 1ð Þ is convergent, with sum 1 (cf. Sub-section 3.1.3). Hence, by

the Combination Rules

X

1

n¼1

3

4

� n

� 2

n nþ 1ð Þ

�

is convergent, with sum 3� 2� 1ð Þ ¼ 1:

6. By the Combination Rules for sequences

limn!1

n2

2n2 þ 1¼ lim

n!1

1

2þ 1n2

¼ 1

2;

so that the sequence n2

2n2 þ 1

n o

is not null.

Hence, by the Non-null Test,P

1

n¼1

n2

2n2 þ 1is divergent.

Section 3.2

1. Let sn ¼ 1þ 122 þ 1

32 þ � � � þ 1n2 and tn ¼ 1þ 1

2þ 1

3þ � � � þ 1

n. The values of these

partial sums are as listed below:

n 1 2 3 4 5 6 7 8

sn 1 1.25 1.36 1.42 1.46 1.49 1.51 1.53

tn 1 1.50 1.83 2.08 2.28 2.45 2.59 2.72

2. (a) We use the Comparison Test. Since

n3 þ n � n3; for n ¼ 1; 2; . . .;

384 Appendix 4

we have1

n3 þ n� 1

n3; for n ¼ 1; 2; . . .:

SinceP

1

n¼1

1n3 is convergent, we deduce, from the Comparison Test, that

X

1

n¼1

1

n3 þ nis convergent:

(b) Let

an ¼1

nþffiffiffi

np and bn ¼

1

n; for n ¼ 1; 2; . . .;

so that

limn!1

an

bn

¼ limn!1

n

nþffiffiffi

np

¼ limn!1

1

1þ 1ffiffi

np

¼ 1 6¼ 0:

SinceP

1

n¼1

1n

is divergent, we deduce, from the Limit Comparison Test, that

X

1

n¼1

1

nþffiffiffi

np is divergent:

(c) Let

an ¼nþ 4

2n3 � nþ 1and bn ¼

1

n2; for n ¼ 1; 2; . . .;

so that

limn!1

an

bn

¼ limn!1

n3 þ 4n2

2n3 � nþ 1

¼ limn!1

1þ 4n

2� 1n2 þ 1

n3

¼ 1

26¼ 0:

SinceP

1

n¼1

1n2 is convergent, we deduce, from the Limit Comparison Test, that

X

1

n¼1

nþ 4

2n3 � nþ 1is convergent:

(d) We use the Comparison Test. Since

0 � cos2 2nð Þ � 1; for n ¼ 1; 2; . . .;

we have

0 � cos2 2nð Þn3

� 1

n3; for n ¼ 1; 2; . . .:

SinceP

1

n¼1

1n3 is convergent, we deduce, from the Comparison Test, that

X

1

n¼1

cos2 2nð Þn3

is convergent:

3. (a) Let an ¼ n3

n!, for n¼ 1, 2, . . ., so that

anþ1

an

¼ nþ 1ð Þ3

nþ 1ð Þ!

!

� n!

n3

�

¼ nþ 1ð Þ2

n3

¼ n2 þ 2nþ 1

n3¼ 1

nþ 2

n2þ 1

n3:


Hence, by the Combination Rules for sequences

limn!1

anþ1

an

¼ 0;

it follows, from the Ratio Test, thatP

1

n¼1

n3

n! is convergent.

(b) Let an ¼ n22n

n! , for n¼ 1, 2, . . ., so that

anþ1

an

¼ nþ 1ð Þ22nþ1

nþ 1ð Þ!

!

� n!

n22n

�

¼ 2 nþ 1ð Þn2

¼ 21

nþ 1

n2

�

:

Hence, by the Combination Rules for sequences

limn!1

anþ1

an

¼ 0;

it follows, from the Ratio Test, thatP

1

n¼1

n22n

n! is convergent.

(c) Let an ¼ 2nð Þ!nn , for n¼ 1, 2, . . ., so that

anþ1

an

¼ 2nþ 2ð Þ!nþ 1ð Þnþ1

!

� nn

2nð Þ!

�

¼ 2nþ 2ð Þ 2nþ 1ð Þnn

nþ 1ð Þnþ1

¼ 2 2nþ 1ð Þnn

nþ 1ð Þn

¼ 4nþ 2

1þ 1n

� �n :

We know that limn!1

1þ 1n

� �n¼ e and that 14nþ 2

n o

is null, so that1þ1

nð Þn4nþ 2

� �

is null,

by the Product Rule for sequences. It follows, from the Reciprocal Rule for

sequences, that

anþ1

an

!1 as n!1:

It follows, from the Ratio Test, thatP

1

n¼1

2nð Þ!nn is divergent.

Remark

Notice that

2nð Þ!nn� 2n

n

�

� 2n� 1

n

�

� � � � � nþ 1

n

�

� 1;

it follows, from the Non-null Test, thatP

1

n¼1

2nð Þ!nn is divergent.

4. (a) Let an ¼ 1n loge n

, n¼ 2, . . .. Then an is positive; and, since {nlogen} is an increas-

ing sequence, anf g ¼ 1n loge n

n o

is a decreasing sequence.


bn ¼ 2n � 1

2n loge 2nð Þ

¼ 1

n loge 2¼ 1

loge 2� 1

n:

386 Appendix 4

SinceP

1

n¼2

1n

is a basic divergent series, it follows by the Multiple Rule thatP

1

n¼2

bn must be divergent. Hence, by the Condensation Test, the original series

P

1

n¼2

1n loge n

must also be divergent.

(b) Let an ¼ 1

n loge nð Þ2, n¼ 2, . . .. Then an is positive; and, since {n(loge n)2} is an

increasing sequence, anf g ¼ 1

n loge nð Þ2n o



bn ¼ 2n � 1

2n loge 2nð Þð Þ2

¼ 1

n loge 2ð Þ2¼ 1

loge 2ð Þ2� 1

n2:

SinceP

1

n¼2

1n2 is a basic convergent series, it follows by the Multiple Rule that

P

1

n¼2

bn must be convergent. Hence by the Condensation Test, the original series

P

1

n¼2

1

n loge nð Þ2 must also be convergent.

Section 3.3

1. (a) Let an ¼ �1ð Þnþ1n

n3 þ 1, for n¼ 1, 2, . . . , so that

anj j ¼n

n3 þ 1; for n ¼ 1; 2; . . .:

Now

n

n3 þ 1� n

n3¼ 1

n2; for n ¼ 1; 2; . . .;

andP

1

n¼1

1n2 is a basic convergent series. Hence, by the Comparison Test

X

1

n¼1

n

n3 þ 1is convergent:

It follows, from the Absolute Convergence Test, that

X

1

n¼1

�1ð Þnþ1n

n3 þ 1is convergent:

(b) Let an ¼ cos n2n , for n¼ 1, 2, . . .; then

anj j �1

2n; for n ¼ 1; 2; . . .;

since cos nj j � 1.

NowP

1

n¼1

12n is a basic convergent series (in fact, a convergent geometric

series). Hence, by the Comparison Test

X

1

n¼1

cos n

2nis absolutely convergent:

It follows, from the Absolute Convergence Test, that

X

1

n¼1

cos n

2nis convergent:


2. The series1

2þ 1

4� 1

8þ 1

16þ 1

32� 1

64þ � � �

is absolutely convergent, and hence convergent, by the Comparison Test, since the

seriesP

1

n¼1

12n is a convergent geometric series.

By the infinite form of the Triangle Inequality, we have

1

2þ 1

4� 1

8þ 1

16þ 1

32� 1

64þ � � �

� 1

2þ 1

4þ 1

8þ 1

16þ 1

32þ 1

64þ � � �

¼ 1:

It follows that the sum of the original series lies in the interval [�1, 1].

3. (a) The sequence�1ð Þnþ1

n3

n o

is of the form

�1ð Þnþ1bn

�

, where

bn ¼1

n3; for n ¼ 1; 2; . . .:

Now:

1. 1n3 � 0, for n¼ 1, 2, . . .;

2. 1n3

�


3. 1n3

�

is decreasing, because {n3} is increasing.


1

n¼1

�1ð Þnþ1

n3 is convergent.

(b) The sequence �1ð Þnþ1 nnþ2

n o

is is not a null sequence, since

limn!1

n

nþ 2¼ 1;

and so the odd subsequence tends to 1.

Hence, by the Non-null Test

X

1

n¼1

�1ð Þnþ1 n

nþ 2is divergent:

(c) The sequence�1ð Þnþ1

n13þn

12

� �

is of the form �1ð Þnþ1bn

n o

, where

bn ¼1

n13 þ n

12

; for n ¼ 1; 2; . . .:

Now:

1. 1

n13 þ n

12

� 0, for n¼ 1, 2, . . .;

2. 1

n13 þ n

12

� �

is a null sequence, by the Squeeze Rule, since

0 � 1

n13 þ n

12

� 1

n12

;

where 1

n12

n o


3. 1

n13 þ n

12

� �

is decreasing, because n13 þ n

12

n o

is increasing.


1

n¼1

�1ð Þnþ1

n13 þ n

12

is convergent.

4. Let sn and tn denote the nth partial sums of the series

1� 1

2� 1

4þ 1

3� 1

6� 1

8þ 1

5� 1

10� 1

12þ � � � and 1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � ;

respectively. Denote by Hn the nth partial sumP

n

k¼1

1k¼ 1þ 1

2þ � � � þ 1

nof the harmonic

series.

388 Appendix 4

Th e t e r m s o f t he s e r i e sP

1

n ¼1

an come ‘in a natural way’ in t hr ees, so it seems

sensible to loo k at the p arti al sum s s3n

s3 n ¼ 1 � 1

2 � 1

4 þ 1

3 � 1

6 � 1

8 þ 1

5 � 1

10 � 1

12 þ �� þ 1

2n � 1 � 1

4n � 2 � 1

4n

¼ 1 � 1

2 � 1

4

�

þ 1

3 � 1

6 � 1

8

�

þ 1

5 � 1

10 � 1

12

�

þ�� þ 1

2 n � 1 � 1

4 n � 2 � 1

4 n

�

¼ 1 þ 1

3 þ 1

5 þ �� þ 1

2n � 1

�

� 1

2 þ 1

4 þ 1

6 þ 1

8 þ �� þ 1

4n � 2 þ 1

4n

�

¼ 1 þ 1

2 þ 1

3 þ 1

4 þ �� þ 1

2n � 1 þ 1

2n

�

� 1

2 þ 1

4 þ �� þ 1

2n

�

� 1

21 þ 1

2 þ 1

3 þ �� þ 1

2 n

�

¼ H2n �1

2 Hn �

1

2 H 2n

¼ 1

2H2n � Hnð Þ:

But as we saw in Example 2

t2n ¼ H2 n � H n ;

so that

s3n ¼1

2 t2 n :

By assumption, tn ! log e 2 and so t2n ! log e 2 as n !1. It follows, by the Multiple

Rule for sequences, that

s3n !1

2loge 2 as n !1:

Finally

s3n �1 ¼ s3n þ1

4n ! 1

2loge 2 as n !1;

and

s3n �2 ¼ s3n þ1

4n � 2 þ 1

4n ! 1

2loge 2 as n !1:

It follows from these three results that sn ! 12

loge 2 as n !1.

5. We follow the pattern of the solution to Example 3, by finding a subsequence of partial

sums of the seriesP

1

n¼1

�1ð Þnþ1

n¼1� 1

2þ 1

3� 1

4þ 1

5� 1

6þ � � � that are greater than 2, 3, . . .

in turn and ensuring that the other partial sums are not much less than these values.

First, note that all terms of the seriesP

1

n¼1

�1ð Þnþ1

nare at most 1 in modulus.

Next, since the ‘positive’ part of the seriesP

1

n¼1

�1ð Þnþ1

nhas partial sums that tend to

1, we start to construct the desired rearranged series as follows. Take enough of the

‘positive’ terms 1; 13; 1

5; . . .; 1

2N1 � 1so that the sum 1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1is greater

than 2, choosing N1 so that it is the first integer such that this sum is greater than 2.

Then these N1 terms will form the first N1 terms in our desired rearranged series.

Next, take one ‘negative’ term 12

and consider the expression

1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1

�

� 1

2

�

:

It is certainly less than the partial sum tN1¼ 1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1of the rearranged

series, for which tN1> 2. But, since all terms are at most 1 in magnitude, it follows

that tN1þ1 > 1.


Now add in some more ‘positive’ terms 12N1þ1

; 12N1þ3

; . . .; 12N2�1

so that the sum

1þ 1

3þ 1

5þ � � � þ 1

2N1 � 1

�

� 1

2

�

þ 1

2N1 þ 1þ 1

2N1 þ 3þ � � � þ 1

2N2 � 1

�

is greater than 3, choosing N2 so that it is the first available integer such that this sum

is greater than 3. Then these N2þ 1 terms will form the first N2þ 1 terms in our

desired rearranged series.

We then add in just one ‘negative’ term; this makes the next partial sum of the

rearranged series satisfy tN2þ2 > 2. Then we add in sufficient positive terms to make

the partial sum tN3þ2 > 4. And so on indefinitely.

In each set of two steps in this process we must use at least one of the ‘positive’

terms and one of the ‘negative’ terms, so that eventually all the ‘positive’ terms and

all the ‘negative’ terms of the original series will be taken exactly once in the new

series, which we denote byP

1

n¼1

bn. So certainly the seriesP

1

n¼1

bn is a rearrangement of

the original seriesP

1

n¼1

�1ð Þnþ1

n.

From our construction, we have ensured that:

for all n>Nk, for any k, the nth partial sums ofP

1

n¼1

bn are >k.

It follows, then, that the nth partial sums ofP

1

n¼1

bn tend to infinity as n!1.

6. (a) The sequence 12

n �

is not null; hence, by the Non-null Test, the seriesP

1

n¼1

12

n is

divergent.

ThusP

1

n¼1

12

n is neither convergent nor absolutely convergent.

(b) We have

5nþ 2n

3n¼ 5n

1

3

� n

þ 2

3

� n

; for n ¼ 1; 2; . . .:

Now,P

1

n¼1

n 13

� �nand

P

1

n¼1

23

� �nare both basic convergent series; hence, by the

Combination Rules for series

X

1

n¼1

5nþ 2n

3nis convergent ðand absolutely convergentÞ:

(c) The sequence fang ¼n

32n3 � 1

o

is null (so the Non-null Test is inappropriate),

and contains only positive terms.

The Ratio Test fails, since

limn!1

anþ1

an

¼ limn!1

2n3 � 1

2 nþ 1ð Þ3�1¼ 1:

Instead, we use the Limit Comparison Test, with

bn ¼1

n3; for n ¼ 1; 2; . . .:

Then

limn!1

an

bn

¼ limn!1

32n3�1

1n3

¼ limn!1

3n3

2n3 � 1

¼ 3

26¼ 0:

Since all terms are non-negative.

390 Appendix 4

SinceP

1

n¼1

1n3 is a basic convergent series, we deduce, from the Limit Comparison

Test, thatX

1

n¼1

3

2n3 � 1is convergent:

Since the terms in the series are non-negative, it follows that it is also absolutely

convergent.

(d) The sequence�1ð Þnþ1

n13

n o

is null, butP

1

n¼1

1

n13

is a basic divergent series. HenceP

1

n¼1

�1ð Þnþ1

n13

is not absolutely convergent.

However,�1ð Þnþ1

n13

n o


n o

, where

bn ¼1

n13

; for n ¼ 1; 2; . . . :

Now:

1. 1

n13

� 0, for n¼ 1, 2, . . .;

2. 1

n13

n o

is a null sequence;

3. 1

n13

n o

is decreasing, because

n13

�

is increasing.


1

n¼1

�1ð Þnþ1

n13

is convergent.

(e) The sequence an ¼ �1ð Þnþ1n2

n2 þ 1; n ¼ 1; 2; . . ., is not null, since

limn!1

n2

n2 þ 1¼ lim

n!1

1

1þ 1n2

¼ 1;

so that the odd subsequence tends to 1.

Hence,P

1

n¼1

anj j andP

1

n¼1

an are divergent, by the Non-null Test. It follows that

X

1

n¼1

�1ð Þnþ1n2

n2 þ 1is neither convergent nor absolutely convergent:

(f) Let an ¼ �1ð Þnþ1n

n3þ5; n ¼ 1; 2; . . .; then

anj j ¼n

n3 þ 5; n ¼ 1; 2; . . .;

so that

anj j �n

n3¼ 1

n2; for n ¼ 1; 2; . . .:

It follows, by the Comparison Test, thatP

1

n¼1

anj j is convergent; that is, thatP

1

n¼1

an

is absolutely convergent.

Hence, by the Absolute Convergence Test,P

1

n¼1

an is convergent.

(g) Since

n6

2n

�

is a basic null sequence of positive terms, it follows, from the

Reciprocal Rule for sequences, that

2n

n6!1 as n!1:

Hence 2n

n6

�

is not a null sequence; it follows, from the Non-null Test, that

X

1

n¼1

2n

n6is divergent:

Thus,P

1

n¼1

2n

n6 is neither convergent nor absolutely convergent.

SinceP

1

n¼1

1n2 is a basic

convergent series.


(h) Let an ¼ �1ð Þnþ1n

n2þ2, n¼ 1, 2, . . ., so that

anj j ¼n

n2 þ 2; for n ¼ 1; 2; . . .:

Now we try the Limit Comparison Test, with

bn ¼1

n; for n ¼ 1; 2; . . .:

We obtain

limn!1

anj jbn

¼ limn!1

nn2þ2

1n

¼ limn!1

n2

n2 þ 2¼ 1 6¼ 0:

Thus,P

1

n¼1

anj j is divergent, sinceP

1

n¼1

1n

is divergent; it follows thatP

1

n¼1

an is not

absolutely convergent.

However,�1ð Þnþ1

n

n2 þ 2

n o


n o

, where

bn ¼n

n2 þ 2; for n ¼ 1; 2; . . .:

Now:

1. nn2 þ 2

� 0, for n¼ 1, 2, . . .;

2. nn2 þ 2

n o

is a null sequence, since

n

n2 þ 2¼

1n

� �

1þ 2n2

� �! 0 as n!1;

3. nn2 þ 2

n o

is decreasing, sincen

n2þ2n

o

is increasing, because

nþ 1ð Þ2þ2

nþ 1� n2 þ 2

n¼ nþ 1þ 2

nþ 1

�

� nþ 2

n

�

¼ 1� 2

n nþ 1ð Þ � 0:


1

n¼1

�1ð Þnþ1n

n2 þ 2is convergent.

(i) Let an ¼ 1

n loge nð Þ34

, n¼ 2, . . .. Then an is positive; and, since n loge nð Þ34

n o

is an

increasing sequence, anf g ¼n

1

n loge nð Þ34

o



bn ¼ 2n � 1

2n loge 2nð Þ34

¼ 1

n loge 2ð Þ34

¼ 1

loge 2ð Þ34

� 1

n34

:

SinceP

1

n¼2

1

n34

is a basic divergent series, it follows (for example, by the Ratio Test)

thatP

1

n¼2

bn must be divergent. Hence, by the Condensation Test, the original

seriesP

1

n¼2

1

n loge nð Þ34

must also be divergent. Hence the series is not convergent, and,

since its terms are non-negative, is not absolutely convergent.

392 Appendix 4

Section 3.4

1. Substituting x¼ 2 into the power series for ex, we find that the seventh partial sum of

the power series for e2 gives

e2 ’ 1þ 2

1!þ 22

2!þ 23

3!þ 24

4!þ 25

5!þ 26

6!

¼ 1þ 2þ 4

2þ 8

6þ 16

24þ 32

120þ 64

720

¼ 1þ 2þ 2þ 4

3þ 2

3þ 4

15þ 4

45

¼ 7þ 16

45¼ 7:355:

Hence this method gives as an estimate for e2 to 3 decimal places the number 7.355.

Remark

In fact, e2¼ 7.389, to 3 decimal places.

2. From equation ( 5) we know that

0 < e� sn <1

n� 1ð Þ!�1

n� 1;

so that we want n to satisfy the requirement that

1

n� 1ð Þ!�1

n� 1< 5� 10�11; or n� 1ð Þ!� n� 1ð Þ > 1

5� 1011 ¼ 2� 1010:

But 13!� 13 ’ 8.095� 1010, so n¼ 14 is sufficient.

In other words, 14 terms will suffice.

Chapter 4

Section 4.1

1. (a) Let {xn} be any sequence in R that converges to 2. Then, by the Combination

Rules for sequences, we have

f xnð Þ ¼ x3n � 2x2

n

! 23 � 2� 22 ¼ 8� 8 ¼ 0 as n!1:But f(0)¼ 0, so that f(xn)! f(0) as n!1. Thus f is continuous at 2.

(b) Consideration of the graph of f near 1 suggests that f is not continuous at 1.

So, let {xn} be any sequence in (0, 1) that tends to 1. Then f(xn)¼ 0, for all n,

so that

f xnð Þ ! 0 as n!1:But f(1)¼ 1, so that f(xn) 6! f(1) as n!1. Thus f is not continuous at 1.

Remark

If we had chosen {xn} to be any sequence in (1, 2) that tends to 1, we would

have found that f(xn)! f(1) as n!1. However this does NOT prove that f is

continuous at 1, since the definition of continuity requires that f(xn)! f(1) for

ALL sequences that tend to 1.

2. (a) Let c be any point in R .

Let {xn} be any sequence in R that converges to c. Then f(xn)¼ 1, for all n,

so that

f xnð Þ!1 as n!1:

For example, xn ¼ 1� 12n

.

But f(c)¼ 1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.

Since c is an arbitrary point of R , it follows that f is continuous on R .

(b) Let c be any point in R .

Let {xn} be any sequence in R that converges to c. Then f(xn)¼ xn, for all n,

so thatf xnð Þ ! c as n!1:

But f(c)¼ c, so that f(xn)! f(c) as n!1. Thus f is continuous at c.


3. The domain of f is the interval I¼ {x : x� 0} if n is even and R if n is odd.

Thus, we have to show that for each c in I:

for each sequence {xk} in I such that xk! c, then x1n

k ! c1n.

First, let c¼ 0. We have already seen (in the Power Rule for null sequences and the

subsequent Remark) that, for any null sequence {xk} in I, x1n

k

n o

is also a null

sequence. In other words, f(xk)! f(0) as k!1. Hence f is continuous at 0.

Next, let c 6¼ 0. We have to prove that if {xk� c} is a null sequence, then so is

x1n

k � c1n

n o

.

Now, using the hint, we have

x1n

k � c1n ¼ xk � c

xn�1

n

k þ xn�2

n

k c1n þ x

n�3n

k c2n þ � � � þ c

n�1n

:

By the result of part (b) of Exercise 3 on Section 2.3, we obtain

xn�1

n

k þ xn�2

n

k c1n þ x

n�3n

k c2n þ � � � þ c

n�1n

! cn�1

n þ cn�2

n c1n þ c

n�3n c

2n þ � � � þ c

n�1n as k!1

¼ ncn�1

n ;

since c 6¼ 0, we deduce that

x1n

k � c1n ! 0

ncn�1

n

as k!1

¼ 0:

In other words, x1n

k � c1n

n o

is null, as required. It follows that f is continuous at c.

4. (a) Let d be any point in R .

Let {xn} be any sequence in R that converges to d. Then f(xn)¼ c, for all n, so that

f xnð Þ ! c as n!1:But f(d)¼ c, so that f(xn)! f(d) as n!1. Thus f is continuous at d.

Since d is an arbitrary point of R , it follows that f is continuous on R .

(b) Let c be any point in R .

Let {xk} be any sequence in R that converges to c. Then f(xk)¼ xkn, for all k,

so that, by the Product Rule for sequences

f xkð Þ ! cn as k!1:But f(c)¼ cn, so that f (xk)! f (c) as k!1. Thus f is continuous at c.


(c) Let c be any point in R .

Let {xn} be any sequence in R that converges to c. Then, by Theorem 4 of

Sub-section 2.3.3, we have

f xnð Þ ¼ xnj j ! cj j as n!1:But f(c)¼ jcj, so that f(xn)! f(c) as n!1. Thus f is continuous at c.


394 Appendix 4

5. First, let c¼ 0. Consideration of the graph of f near 0 suggests that f is not continuous at 0.

So, let {xn} be any sequence in (0, 1) that tends to 0. Then f(xn)¼ 1, for all n, so that


Next, let c> 0. We will show that f is continuous at c.

So, let {xn} be any sequence that tends to c. Since c> 0, it follows that eventually

xn> 0; thus xn> 0 for all n>N, for a suitable choice of N. Then f(xn)¼ 1, for all

n>N, so that

f xnð Þ ! 1 as n!1:But f(c)¼ 1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.

Finally, let c< 0. We will show that f is continuous at c.

So, let {xn} be any sequence that tends to c. Since c< 0, it follows that eventually

xn< 0; thus xn< 0 for all n>N, for a suitable choice of N. Then f(xn)¼�1, for all

n>N, so that

f xnð Þ ! �1 as n!1:But f(c)¼�1, so that f(xn)! f(c) as n!1. Thus f is continuous at c.

It follows that f is continuous on R � {0}, and discontinuous at 0.

6. We have already seen, in Example 3 of Sub-section 4.1.1 , that the function

g : x 7! x12; for x � 0;

is continuous on [0, 1). Also, it follows from Problem 4 that the function

h : x 7! x3; for x 2 R ;


We may apply the Composition Rule to the functions g and h, since

g([0,1))R ; it follows that

h � g : x 7! x12

� �3

¼ x32; for x � 0;

is continuous on [0,1).

7. By the Sum and Product Rules, and by using the fact that the constant function and

the identity function are continuous on R , we see that

g : x 7! x2 þ xþ 1; x 2 R ;

is continuous on R . Since g(R) (0,1) and the square root function h is continuous

on R , it follows, by the Composition Rule, that the function

h � g : x 7!ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ xþ 1p

; x 2 R ;


Next, by the Sum, Multiple and Product Rules, and by using the fact that the constant

function and the identity function are continuous on R , we see that the functions

k : x 7! �5x and l : x 7!1þ x2; x 2 R ;

are continuous on R . Since l(x) 6¼ 0, it follows from the Product and Quotient Rules,

that

m : x 7! �5x

1þ x2; x 2 R ;


It then follows, by applying the Sum Rule to h� g and m, that the function

f xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ xþ 1p

� 5x

1þ x2; x 2 R ;


Equivalently, all polynomialsare continuous on R .


Equivalently, rational functionsare continuous on theirdomains.

8. (a) The graph of f suggests that we should find functions g and h that squeeze f near 0.

y = x sin 1x

y

x

So, we define g(x)¼� jxj, x2R , and h(x)¼ jxj, x2R . With these two chosen,

we check the conditions of the Squeeze Rule.

Now we know that

� 1 � sin1

x

�

� 1; for any x 6¼ 0:

It follows that

� xj j � x sin1

x

�

� xj j; for any x 6¼ 0;

so that

g xð Þ ¼ � xj j½ � � f xð Þ � xj j ¼½ � h xð Þ; for any x 2 R :

So condition 1 of the Squeeze Rule is satisfied.

Next, the functions f, g and h all take the value 0 at the point 0. Thus condition

2 of the Squeeze Rule is satisfied.

Finally, the functions g and h are both continuous at 0.

It then follows, from the Squeeze Rule, that f is continuous at 0.

(b) Consideration of the graph of f near 0 suggests that f is not continuous at 0.

y = sin 1x

1

–1

So, let {xn} be the sequence

xnf g ¼1

2nþ 12

� �

p

( )

:

Then xn! 0 as n!1, and

f xnð Þ ¼ sin 2nþ 1

2

�

p�

¼ sin1

2p

�

¼ 1

for all n, so that


396 Appendix 4

9. Since the constant function and the identity function are continuous on R , it follows

from the Combination Rules that the function

g xð Þ ¼ x2 þ 1; x 2 R ;


Next, since the sine function is continuous on R , it follows by the Composition

Rule, that

sin � g xð Þ ¼ sin x2 þ 1� �

; x 2 R ;


It then follows from the Multiple and Sum Rules, that the function

f xð Þ ¼ x2 þ 1þ 3 sin x2 þ 1� �

; x 2 R ;


10. The cosine function is continuous on R , so that, by the Multiple Rule, the function

x 7!p2

cos x; x 2 R ;


Then, since the sine function, also, is continuous on R , we deduce, from the

Composition Rule, that

f xð Þ ¼ sinp2

cos x� �

; x 2 R ;


11. Since the identity function is continuous on R , it follows, from the Combination

Rules, that the function

g xð Þ ¼ x5 � 5x2; x 2 R ;


Then, since the cosine function is continuous on R , we deduce, from the


h xð Þ ¼ cos x5 � 5x2� �

; x 2 R ;


Next, since the identity function is continuous on R , it follows, from the

Combination Rules, that the function

k xð Þ ¼ �x2; x 2 R ;


Then, since the exponential function is continuous on R , we deduce, from the


l xð Þ ¼ e�x2

; x 2 R ;


Finally, it follows, from the Combination Rules, that

f xð Þ ¼ cos x5 � 5x2� �

þ 7e�x2

; x 2 R ;


Section 4.2

1. Consider the function

f xð Þ ¼ cos x� x; x 2 0; 1½ �:We shall show that f has a zero c in (0, 1).





Certainly, f is continuous on [0, 1], and

f 0ð Þ ¼ cos 0� 0 ¼ 1 > 0;

f 1ð Þ ¼ cos 1� 1< 0:

Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that

f (c)¼ 0, and so such that

cos c ¼ c:

2. If f (0)¼ 0 or f (1)¼ 1, we can take c¼ 0 or c¼ 1, respectively.

Otherwise, we have f (0)> 0 and f (1)< 1, since the image of [0, 1] under f lies

in [0, 1].

We now define the auxiliary function

g xð Þ ¼ f xð Þ � x; x 2 0; 1½ �;and show that g has a zero c in (0, 1).

Certainly, g is continuous on [0, 1], and

g 0ð Þ ¼ f 0ð Þ � 0 > 0;

g 1ð Þ ¼ f 1ð Þ � 1< 0:

Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that

g(c)¼ 0, and so such that

f cð Þ ¼ c:

3. For the function f (x)¼ x5þ x� 1, x2 [0, 1], we have

f 0ð Þ ¼ �1 < 0; f 1ð Þ ¼ 1þ 1� 1 ¼ 1 > 0; and

f1

2

�

¼ 1

32þ 1

2� 1 ¼ � 15

32:

It follows, by the Intermediate Value Theorem applied to 12; 1

� �

, that f has a zero

in 12; 1

� �

.

Next, f 34

� �

¼ 2431024þ 3

4� 1 ¼ � 13

1024< 0. Thus, since f 3

4

� �

< 0 and f(1)> 0 it fol-

lows, by the Intermediate Value Theorem applied to 34; 1

� �

, that f has a zero in 34; 1

� �

.

Finally, f 78

� �

¼ 16;80732;768

� 18¼ 12;711

32;768> 0. Thus, since f 3

4

� �

< 0 and f 78

� �

> 0 it fol-

lows, by the Intermediate Value Theorem applied to 34; 7

8

� �

, that f has a zero in 34; 7

8

� �

.

This interval 34; 7

8

� �

is of length 18, as required.

4. We compile the following table of values for the continuous function p:

x �1 0 1 2

p(x) �3 1 �1 3

Since p(�1)< 0 and p(0)> 0, it follows, from the Intermediate Value Theorem, that

p has a zero in (�1, 0).

398 Appendix 4

Since p(0)> 0 and p(1)< 0, it follows, from the Intermediate Value Theorem,

that p has a zero in (0, 1).

Since p(1)< 0 and p(2)> 0, it follows, from the Intermediate Value Theorem,

that p has a zero in (1, 2).

5. For the polynomial p(x)¼ x5þ 3x4� x� 1, x2R , that is continuous on R , we have

M ¼ 1þmax 3j j; �1j j; �1j jf g ¼ 4;

so that, by the Zeros Localisation Theorem, all the zeros of p lie in (�4, 4).

We now compile a table of values of p(x), for x¼�4, �3, . . . , 4:

x �4 �3 �2 �1 0 1 2 3 4

p(x) �253 2 17 2 �1 2 77 482 1,787

By applying the Intermediate Value Theorem to p on each of the three intervals

[�4, �3], [�1, 0] and [0, 1], we deduce that p has a zero in each of the intervals

ð�4;�3Þ; ð�1; 0Þ and ð0; 1Þ:6. (a) First, we look separately at the two closed subintervals [�1, 0] and [0, 2], on

each of which f is strictly monotonic. The function f is continuous on each of

these subintervals, and so has a maximum and a minimum on each.

Since f is strictly decreasing on [�1, 0], it follows that

max f xð Þ : x 2 �1; 0½ �f g ¼ f �1ð Þ ¼ 1;

and this is attained in [�1, 0] only at � 1; also

min f xð Þ : x 2 �1; 0½ �f g ¼ f 0ð Þ ¼ 0;

and this is attained in [�1, 0] only at 0.

Since f is strictly increasing on [0, 2], it follows that

max f xð Þ : x 2 0; 2½ �f g ¼ f 2ð Þ ¼ 4;

and this is attained in [0, 2] only at 2; also

min f xð Þ : x 2 0; 2½ �f g ¼ f 0ð Þ ¼ 0;

and this is attained in [0, 2] only at 0.

Combining these results, we see that

max f xð Þ : x 2 �1; 2½ �f g ¼ 4;

and this is attained in [� 1, 2] only at 2; also

min f xð Þ : x 2 �1; 2½ �f g ¼ 0;

and this is attained in [�1, 2] only at 0.

(b) First, we look separately at the three closed subintervals 0; 12p

� �

, 12p; 3

2p

� �

and32p; 2p� �

on each of which f is strictly monotonic. The function f is continuous on

each of these subintervals, and so has a maximum and a minimum on each.

Since f is strictly increasing on 0; 12p

� �

, it follows that

max f xð Þ : x 2 0; 12p

� � �

¼ f 12p

� �

¼ 1;

and this is attained in 0; 12p

� �

only at 12p; also

min f xð Þ : x 2 0; 12p

� � �

¼ f 0ð Þ ¼ 0;

and this is attained in 0; 12p

� �

only at 0.

Since f is strictly decreasing on 12p; 3

2p

� �

, it follows that

max f xð Þ : x 2 12p; 3

2p

� � �

¼ f 12p

� �

¼ 1;

and this is attained in 12p; 3

2p

� �

only at 12p; also

min f xð Þ : x 2 12p; 3

2p

� � �

¼ f 32p

� �

¼ �1;

and this is attained in 12 p; 3

2 p

� �

only at 32p .

Next, since f is strictly increasing on 32 p; 2p� �

, it follows that

max f xð Þ : x 2 32 p; 2p� � �

¼ f 2 pð Þ ¼ 0;

and this is attained in 32 p; 2p� �

only at 0; also

min f xð Þ : x 2 32p ; 2p� � �

¼ f 32p

� �

¼ �1;

and this is attained in 32 p; 2p� �

only at 32p.

Combining these results, we see that

max f xð Þ : x 2 0; 2p½ �f g ¼ 1;

and this is attained in [0, 2p] only at 12p; also

min f xð Þ : x 2 0; 2p½ �f g ¼ �1;

and this is attained in [0, 2p] only at 32p.

Section 4.3

1. (a) If 0 � x1 < x 2, then 2x 1 < 2x 2 and x14< x2

4. Hence

x 41 þ 2x 1 þ 3 < x 42 þ 2x2 þ 3;

so that f is strictly increasing on R , and is thus one–one on R .

(b) If 0 < x1 < x 2 , then x 12< x2

2 and 1x1> 1

x2, so that � 1

x1< � 1

x2: Hence

x21 �

1

x1

< x22 �

1

x2

;

it follows that f is strictly increasing on (0,1), and is thus one–one on (0,1).

2. We perform the three steps of the strategy.

1. We showed, in Problem 1(b) above, that f is strictly increasing on (0, 1).

2. The function

x 7! x2 � 1

x¼ x3 � 1

x; x 2 R � 0f g;

is a rational function, and therefore is continuous on its domain R � {0}. In

particular, f is continuous on (0,1).

3. Choose the increasing sequence {n}, which tends to1, the right-hand end-point

of (0,1). Then

f nð Þ ¼ n2 � 1

n!1 as n!1;

by the Reciprocal Rule for sequences. Thus the right-hand end-point of

J ¼ f 0;1ð Þð Þ is1.

4. Choose the decreasing sequence 1n

�

, which tends to 0, the left-hand end-point of

(0,1). Then

f1

n

�

¼ 1

n2� n! �1 as n!1;

by the Reciprocal Rule for sequences. Thus the left-hand end-point of

J ¼ f 0;1ð Þð Þ is �1.Hence f has a continuous inverse f�1: R! (0, 1), by the Inverse Function

Rule.

3. (a) Since sin 14p

� �

¼ 1ffiffi

2p and 1

4p lies in �1

2p; 1

2p

� �

, we have sin�1 1ffiffi

2p� �

¼ 14p.

Since cos 13p

� �

¼ 12, we have cos 2

3p

� �

¼ �12, and 2

3p lies in [0, p], so that

cos�1 �12

� �

¼ 23p.

Since tan 13p

� �

¼ffiffiffi

3p

and 13p lies in �1

2p; 1

2p

� �

, we have tan�1ffiffiffi

3p� �

¼ 13p.

400 Appendix 4

(b) Following the Hint, we put y¼ sin�1x. Then

cos 2 sin�1 x� �

¼ cos 2yð Þ¼ 1� 2 sin2 y

¼ 1� 2x2;

since x¼ sin y.

4. Following the Hint, we put a¼ logex and b¼ logey. Then x¼ ea and y¼ eb, so that

loge xyð Þ ¼ loge eaeb� �

¼ loge eaþb� �

¼ aþ b

¼ loge xþ loge y:

5. Let y¼ cosh�1 x, where x� 1. Then

x ¼ cosh y ¼ 12

ey þ e�yð Þ;so that

e2y � 2xey þ 1 ¼ 0:

This is a quadratic equation in ey, and so

ey ¼ x�ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p

:

Both choices of þ or � give a positive expression on the right, but recall that y� 0

and so ey� 1. Since


x2 � 1p� �

� x�ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p� �

¼ 1;

we choose the þ sign, because xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p

� 1, whereas x�ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p

� 1.

Hence

y ¼ cosh�1 x ¼ loge xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p� �

:

(The value y ¼ loge x�ffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 1p� �

gives the negative solution of the equation

cosh y¼ x.)

Section 4.4

1. (a) For x> 0, f xð Þ ¼ x� ¼ e� loge x.

Now, the functions

x 7! loge x; x 2 0;1ð Þ; and x 7!ex; x 2 R ;

are continuous; hence, by the Multiple Rule and the Composition Rule, f is

continuous.

(b) For x> 0, f xð Þ ¼ xx ¼ ex loge x.

Now, the functions

x 7! loge x; x 2 0;1ð Þ; x 7! x; x 2 R ;

and x 7! ex; x 2 R ;

are continuous; hence, by the Product Rule and the Composition Rule, f is

continuous.

2. Since ax ¼ ex loge a and ay ¼ ey loge a, we have

axay ¼ ex loge aey loge a

¼ ex loge aþy loge a

¼ e xþyð Þ loge a

¼ axþy:


3. We use the inequality ex> 1þ x, for x> 0. Applying this inequality with x replaced

by xe

� �

� 1, we obtain

exeð Þ�1 > 1þ x

e� 1

� �

; for x > e;

¼ x

e¼ xe�1:

It follows that

exe > x;

so that, by Rule 5 with p¼ e

ex > xe:

Chapter 5

Section 5.1

1. (a) The function

f xð Þ ¼ x2 þ x

x

has domain R � {0}, so that f is defined on any punctured neighbourhood of 0.

Next, notice that f (x) ¼ xþ 1, for x 6¼ 0. It follows that, if {xn} lies in R � {0}

and xn! 0 as n!1, then f(xn)! 1 as n!1.

Hence

limx!0

f xð Þ ¼ 1:

(b) First, notice that

f xð Þ ¼ xj jx¼ 1; x > 0,

�1; x < 0.

�

The domain of f is R � {0}, so that f is defined on any punctured neighbour-

hood of 0.

However, the two null sequences 1n

�

and � 1n

�

both have non-zero terms,

but

limn!1

f1

n

�

¼ 1; whereas limn!1

f � 1

n

�

¼ �1:

Hence, limx!0

f xð Þ does not exist.

2. (a) The function f xð Þ ¼ffiffiffi

xp

is defined on [0, 1), which contains the point 2; it is

also continuous at 2.

Hence, by Theorem 2

limx!2

ffiffiffi

xp¼

ffiffiffi

2p

:

(b) The function f xð Þ ¼ffiffiffiffiffiffiffiffiffi

sin xp

is defined on [0, p], which contains the point p2; it is

also continuous at p2, by the Composition Rule for continuous functions.

Hence, by Theorem 2

limx!p

2


sin xp

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

sin� p

2

�

r

¼ 1:

(c) The function f xð Þ ¼ ex

1þ xis defined on (�1,1), which contains the point 1; it is

also continuous at 1, by the Composition Rule for continuous functions.

Hence, by Theorem 2

limx!1

ex

1þ x¼ 1

2e:

402 Appendix 4

3. (a) Since

limx!0

sin x

x¼ 1 and lim

x!0

1

2þ x¼ 1

2;

we deduce from the Product Rule for limits that

limx!0

sin x

2xþ x2¼ 1

2:

(b) Let f(x)¼ x2 and g xð Þ ¼ sin xx

; then

limx!0

x2 ¼ 0 and limx!0

sin x

x¼ 1:

Also, f(x)¼ x2 6¼ 0 in R � {0}; hence, by the Composition Rule

limx!0


sin x2ð Þx2

¼ 1:

(c) Let f xð Þ ¼ xsin x

and g xð Þ ¼ffiffiffi

xp

; then f is defined on the punctured neighbourhood

(�p, 0)[ (0, p) of 0, and, by the Quotient Rule for limits

limx!0

f xð Þ ¼ limx!0

sin x

x

� �1

¼ 1:

Also, g is defined on (0,1), and is continuous at 1; hence, by Theorem 2 and the

Composition Rule

limx!0


x

sin x

� �12¼ g 1ð Þ ¼ 1:

4. Here

ð f � gÞðxÞ ¼f ð0Þ; x ¼ 0,

f ð1þ xÞ; x 6¼ 0,

�

¼f ð0Þ; x ¼ 0,

f ð0Þ; x ¼ �1,

f ð1þ xÞ; x 6¼ 0, �1 (so that 1 + x 6¼ 1, 0)

8

>

<

>

:

¼0; x ¼ 0,

0; x ¼ �1,

2þ ð1þ xÞ; x 6¼ 0, �1,

8

<

:

¼0; x ¼ 0, �1,

3þ x; x 6¼ 0, �1.

�

In particular, ( f� g)(0)¼ 0.

Next, limx!0

g xð Þ ¼ 1, so that

f limx!0

g xð Þ�

¼ f 1ð Þ ¼ �2:

Finally, since f(g(x))¼ 3þ x on the punctured neighbourhood (�1, 0)[ (0, 1) of 0

limx!0

f g xð Þð Þ ¼ limx!0

3þ xð Þ ¼ 3:

5. (a) The inequalities

� x2 � x2 sin1

x

�

� x2; for x 6¼ 0;

show that condition 1 of the Squeeze Rule holds, with

g xð Þ ¼ �x2 and h xð Þ ¼ x2; x 2 R :

Since

limx!0

x2� �

¼ 0 and limx!0�x2� �

¼ 0;

it follows, from the Squeeze Rule, that

limx!0

x2 sin1

x

�

¼ 0:

(b) The inequalities

� xj j � x cos1

x

�

� xj j; for x 6¼ 0;

show that condition 1 of the Squeeze Rule holds, with

g xð Þ ¼ � xj j and h xð Þ ¼ xj j; x 2 R :

Since

limx!0� xj jð Þ ¼ 0 and lim

x!0xj jð Þ ¼ 0;

it follows, from the Squeeze Rule, that

limx!0

x cos1

x

�

¼ 0:

6. Let

f xð Þ ¼ sin1

x; �1� x < 0,

1þ x; 0 � x � 0.

(

Then, for n¼ 1, 2, . . ., we have

f � 1

np

�

¼ sin �npð Þ ¼ 0! 0 as n!1;

whereas

f � 1

2nþ 12

ð Þp

�

¼ sin � 2nþ 12

� �

p� �

¼ �1! �1 as n!1:

It follows that limx!0�

f xð Þ does not exist.

However, for any sequence {xn} in ( 0, 3) for which xn! 0, it follows, from the

fact that the function x 7! 1þ x is continuous on [0, 3), that

limx!0þ

f xð Þ ¼ f 0ð Þ ¼ 1:

7. (a) Since limx!0

sin xx¼ 1, we have, by Theorem 7, that

limx!0þ

sin x

x¼ 1:

Hence, by the Sum Rule

limx!0þ

sin x

xþ

ffiffiffi

xp�

¼ limx!0þ

sin x

xþ lim

x!0þ

ffiffiffi

xp

¼ 1þ 0

¼ 1:

(b) Let f xð Þ ¼ffiffiffi

xp

, x� 0, and g xð Þ ¼ sin xx

, x 6¼ 0. Then

limx!0þ

ffiffiffi

xp¼ 0 and lim

x!0

sin x

x¼ 1:

Also, f xð Þ ¼ffiffiffi

xp6¼ 0 in (0,1). Hence, by the Composition Rule for limits

limx!0þ

g f xð Þð Þ ¼ limx!0þ

sinffiffiffi

xpð Þffiffiffi

xp ¼ 1:

404 Appendix 4

8. Since 1þ x � ex � 11�x

, for jxj< 1, we have

x � ex � 1 � x

1� x; for xj j < 1; ( )

so that

1 � ex � 1

x� 1

1� x; for 0 < x < 1:

Then, since limx!0þ

1 ¼ 1 and limx!0þ

11�x¼ 1, it follows, from the Squeeze Rule, that

limx!0þ

ex � 1

x¼ 1:

Next, it follows, from the inequalities () above, that

1 � ex � 1

x� 1

1� x; for�1 < x < 0;

so that, by an argument similar to the one above, we have

limx!0�

ex � 1

x¼ 1:

Hence, by Theorem 7, we may combine these two one-sided limit results to obtain

limx!0

ex � 1

x¼ 1:

Section 5.2

1. (a) Let f (x)¼ jxj. Then f(x)> 0 for x 6¼ 0, and, since f is continuous at 0

limx!0

xj j ¼ 0:


1

f xð Þ ¼1

xj j ! 1 as x! 0:

(b) Let f(x)¼ x3� 1. Then f(x)> 0 for x2 (1,1), and, since f is continuous at 1

limx!1þ

x3 � 1� �

¼ 0:


1

f xð Þ ¼1

x3 � 1!1 as x! 1þ;

so that

� 1

f xð Þ ¼1

1� x3! �1 as x! 1þ:

(c) Let f xð Þ ¼ x3

sin x. Then f(x)> 0 for x 2 � 1

2p; 1

2p

� �

, and

limx!0

x3

sin x¼ lim

x!0

x2

sin xx

� �

¼limx!0

x2

� �

limx!0

sin xx

� �

¼ 0

1¼ 0;

by the Quotient Rule for limits.


1

f xð Þ ¼sin x

x3!1 as x! 0:

2. (a) Let

f xð Þ ¼ 1

x2; x 2 R � 0f g; and g xð Þ ¼ 1

x2; x 2 R � 0f g:

Then f(x)!1 as x! 0, and g(x)!1 as x! 0; but

f xð Þ � g xð Þ ¼ 0! 0 as x! 0:

(b) Let

f xð Þ ¼ 1

x2; x 2 R � 0f g; and g xð Þ ¼ x2; x 2 R � 0f g:

Then f(x)!1 as x!1, and g(x)! 0 as x!1; but

f xð Þg xð Þ ¼ 1! 1 as x! 0:

3. (a) By the Sum Rule for limits as x!1, we have

limx!1

2x3 þ x

x3¼ lim

x!12þ 1

x2

�

¼ limx!1

2ð Þ þ limx!1

1

x2

�

¼ 2þ 0 ¼ 2:

(b) Since � 1� sin x � 1 for x2R , we have

� 1

x� sin x

x� 1

x; for x 2 0;1ð Þ:

Also

g xð Þ ¼ � 1

x! 0 as x!1

and

h xð Þ ¼ 1

x! 0 as x!1:

Hence, by the Squeeze Rule for limits as x!1 ,

f xð Þ ¼ sin x

x! 0 as x!1:

4. (a) Let f(x)¼ log ex and g(x)¼ log ex, for x2 (0,1). Then

f xð Þ ! 1 as x!1 and g xð Þ ! 1 as x!1:Hence, by the Composition Rule for limits

g f xð Þð Þ ¼ log e log exð Þ ! 1 as x!1:

(b) Let f xð Þ ¼ 1x

and g xð Þ ¼ ex

x, for x2 (0,1). Then

g f xð Þð Þ ¼ xe1x; for x 2 0;1ð Þ:

Now

f xð Þ ! �1 as x! 0�:

In order to use the Composition rule for limits, we need to determine the

behaviour of g(x) as x!�1; that is, the behaviour of g(�x) as x!1.

406 Appendix 4

Now

g �xð Þ ¼ � e�x

x¼ � 1

xex

andxex !1 as x!1:

It follows, by the Reciprocal Rule, that

g �xð Þ ! 0 as x!1:Hence, by the Composition Rule for limits,

g f xð Þð Þ ¼ xe1x ! 0 as x! 0�:

5. Let

f xð Þ ¼ 1; for all x 2 R ;

and

gðxÞ ¼ 2; x ¼ 1,

3; x 6¼ 1.

�

Then

f xð Þ ! 1 as x!1 ðso that ‘ ¼ 1Þ;and

g xð Þ ! 3 as x! 1 ðso that m ¼ 3Þ;

however as x!1, we have

g f xð Þð Þ ¼ g 1ð Þ ¼ 2

! 2 6¼ 3ð¼ mÞ:

Section 5.3

1. (a) an � 0j j < 0:1

, �1ð Þnn2

< 0:1

, 1n2 < 0:1

, n2 > 10:

It follows that j an� 0j< 0.1 for all n>X, so long as X �ffiffiffiffiffi

10p

’ 3:16:

(b) an � 0j j < 0:01

, �1ð Þnn2

< 0:01

, 1n2 < 0:01

, n2 > 100

, n > 10:

It follows that jan� 0j< 0.01 for all n>X, so long as X� 10.

(c) an � 0j j < "

, �1ð Þnn2

< "

, 1n2 < "

, n2 > 1"

, n > 1ffiffi

"p :

It follows that jan� 0 j<" for all n>X, so long as X � 1ffiffi

"p .

2. (a) f xð Þ � 5j j < 0:1

, 2x� 2j j < 0:1

, x� 1j j < 0:05:

It follows that j f (x)� 5j< 0.1 whenever 0< jx� 1j<�, so long as �� 0.05.

(b) f xð Þ � 5j j < 0:01

, 2x� 2j j < 0:01

, x� 1j j < 0:005:

It follows that j f (x)� 5j< 0.01 whenever 0< jx� 1j<�, so long as �� 0.005.

(c) f xð Þ � 5j j < "

, 2x� 2j j < "

, x� 1j j <1

2":

It follows that j f (x)� 5j<" whenever 0< jx� 1j<�, so long as � � 12".

3. (a) The function f (x)¼ 5x� 2 is defined on every punctured neighbourhood of 3.

We must prove that:


5x� 2ð Þ � 13j j < "; for all x satisfying 0 < x� 3j j < �;

that is

x� 3j j < 1

5"; for all x satisfying 0 < x� 3j j < �: ( )

Choose � ¼ 15" ; then the statement () holds.

Hence

limx!3

5x� 2ð Þ ¼ 13:

(b) The function f (x )¼ 1� 7x3 is defined on every punctured neighbourhood of 0.

We must prove that:


1� 7x3� �

� 1


that is

7x3

< "; for all x satisfying 0 < xj j < �: ( )

Choose � ¼ffiffiffiffiffi

17"3

q

(so that 7�3¼ "); then the statement () holds.

Hence

limx!0

1� 7x3� �

¼ 1:

(c) The function f xð Þ ¼ x2 cos 1x3

� �

is defined on every punctured neighbourhood of 0.

We must prove that:


x2 cos1

x3

�

� 0


that is

x2 cos1

x3

�

< "; for all x satisfying 0 < xj j < �: ( )

408 Appendix 4

Choose � ¼ ffiffiffi

"p

. Then, since cos 1x3

� �

� 1 for all non-zero x, it follows that, for

0< jxj<�

x2 cos1

x3

�

� x2

< �2 ¼ ";so that the statement () holds.

Hence

limx!0

x2 cos1

x3

�

¼ 0:

(d) We consider the cases that x< 0 and x> 0 separately; that is, the two one-sided

limits.

Since jxj ¼� x when x< 0, we have

limx!0�

xþ xj jx¼ lim

x!0�

0

x¼ 0;

also, since jxj ¼ x when x> 0, we have

limx!0þ

xþ xj jx¼ lim

x!0þ

2x

x¼ lim

x!0þ2ð Þ ¼ 2:

Since the two one-sided limits are not equal, it follows that limx!0

xþ xj jx

does not

exist.

4. The function f(x)¼ x3, x2R , is defined on every punctured neighbourhood of 1.

We must prove that:


jx3� 1j<" for all x satisfying 0 < x� 1j j < �;

that is

x2 þ xþ 1

� x� 1j j < "; for all x satisfying 0 < x� 1j j < �: ( )

Choose � ¼ min 1; 17"f g, so that:

(i) for 0< jx� 1j<�, we have x2 (0, 2), and so

x2 þ xþ 1

¼ x2 þ xþ 1

< 22 þ 2þ 1 ¼ 7;

(ii) x� 1j j < 17":

It follows that

x2 þ xþ 1

� x� 1j j < 7� 1

7" ¼ "; for all x satisfying 0 < x� 1j j < �;

that is, the statement () holds.

Hence

limx!1

x3� �

¼ 1:

5. Since the limits for f and g exist, the functions f and g must be defined on some

punctured neighbourhoods (c� r1, c)[ ( c, cþ r1) and (c� r2, c)[ (c, cþ r2) of c, it

follows that the function fg is certainly defined on the punctured neighbourhood

(c� r, c)[ ( c, cþ r), where r¼min{r1, r2}.



f xð Þg xð Þð Þ � ‘mð Þj j < "; for all x satisfying 0 < x� cj j < �: ( )

In order to examine the first inequality in (), we write

f xð Þg xð Þ � ‘m as f xð Þ � ‘ð Þ g xð Þ � mð Þ þ m f xð Þ � ‘ð Þ þ ‘ g xð Þ � mð Þ;

so that

f xð Þg xð Þ � ‘mj j ¼ f xð Þ � ‘ð Þ g xð Þ �mð Þ þm f xð Þ � ‘ð Þ þ ‘ g xð Þ �mð Þj j� f xð Þ � ‘j j � g xð Þ �mj j þ mj j f xð Þ � ‘j j þ ‘j j g xð Þ �mj j: (1)

We know that, since limx!c

f xð Þ ¼ ‘, there is a positive number �1 such that

f xð Þ � ‘j j < "; for all x satisfying 0 < x� cj j < �1; (2)

and a positive number �2 such that

f xð Þ � ‘j j < 1; for all x satisfying 0 < x� cj j < �2; (3)

similarly, since limx!c

g xð Þ ¼ m, there is a positive number �3 such that

g xð Þ � mj j < "; for all x satisfying 0 < x� cj j < �3: (4)

We now choose �¼min {�1, �2, �3}. Then the statements (2), (3) and (4) hold for all

x satisfying 0< jx� cj<�, so that, from (1), we deduce

f xð Þg xð Þ � ‘mj j � f xð Þ � ‘j j � g xð Þ � mj j þ mj j f xð Þ � ‘j j þ ‘j j g xð Þ � mj j� 1� "þ mj j"þ ‘j j"¼ 1þ mj j þ ‘j jð Þ";

for all x satisfying 0< jx� cj<�.This is equivalent to the statement (*), in which the number " has been replaced

by K", where K¼ 1þ jmj þ j‘j. Since K is a constant, this is sufficient, by the

K" Lemma.

It follows that limx!c

f xð Þg xð Þ ¼ ‘m, as required.

6. Since limx!c

f xð Þ ¼ ‘ and ‘ 6¼ 0, we may take " ¼ 12‘j j in the definition of limit; it

follows that there is a positive number � such that

f xð Þ � ‘j j < 1

2‘j j for all x satisfying 0 < x� cj j < �;

in other words, for all x satisfying 0< jx� cj<�, we have

� 1

2‘j j < f xð Þ � ‘ < 1

2‘j j;

which we may rewrite in the form

‘� 1

2‘j j < f xð Þ < ‘þ 1

2‘j j: ( )

It follows, from (*), that, if ‘> 0, then

f xð Þ > ‘� 1

2‘j j

¼ ‘� 1

2‘ ¼ 1

2‘ > 0;

and, if ‘< 0, then

f xð Þ < ‘þ 1

2‘j j

¼ ‘� 1

2‘ ¼ 1

2‘ < 0:

In either case, we obtain that f(x) has the same sign as ‘ on the punctured neighbour-

hood {x: 0< jx� cj<�} of c.

410 Appendix 4

Section 5.4

1. The function f (x)¼ x3, x2R , is defined on R .

We must prove that:


x3 � 1

< "; for all x satisfying x� 1j j < �;

that is

x2 þ xþ 1

� x� 1j j < "; for all x satisfying x� 1j j < �: ( )

Choose � ¼ min 1; 17"

�

, so that

(i) for jx� 1j<�, we have x2 (0, 2), and so

x2 þ xþ 1

¼ x2 þ xþ 1

< 22 þ 2þ 1 ¼ 7;

(ii) x� 1j j < 17":

It follows that

x2 þ xþ 1

� x� 1j j < 7� 1

7" ¼ "; for all x satisfying x� 1j j < �;

that is, the statement (*) holds.

Hence f is continuous at 1.

2. The function f xð Þ ¼ffiffiffi

xp

is defined on [0,1), and f 4ð Þ ¼ffiffiffi

4p¼ 2.

We must prove that:

for each positive number ", there is a positive number � such thatffiffiffi

xp� 2

< "; for all x satisfying x� 4j j < �: ðÞNow

ffiffiffi

xp� 2

¼ x� 4ffiffiffi

xpþ 2

¼ x� 4j jffiffiffi

xpþ 2

� 1

2x� 4j j; since

ffiffiffi

xp� 0:

So, choose �¼ ". It follows that, for all x satisfying jx� 4j<�, we have

ffiffiffi

xp� 2

� 1

2x� 4j j

<1

2�

¼ 1

2"

< ":

That is, the statement () holds.

Hence f is continuous at 4.

3. The graph of f suggests that, while f is ‘well behaved’ to the left of 2, we should

examine the behaviour of f to the right of 2. If we choose " to be any positive number

less than 14, say, then there will always be points x (with x > 2) as close as we please to

2 where f (x) is not within a distance " of f (2)¼ 1.

So, take " ¼ 14. Then, if f were continuous at 2, there would be some positive

f xð Þ � 1j j < 1

4; for all x satisfying x� 2j j < �: ( )

Now let xn ¼ 2þ 1n, for n¼ 1, 2, . . .. Then

xn ¼ 2þ 1

n

�

2 2; 2þ �ð Þ; for all n > X; where X ¼ 1

�;

but

f xnð Þ � 1j j ¼ 2þ 1

n� 1

2

�

� 1

¼ 1

2þ 1

n

65 1

4¼ ":

Thus, with this choice of ", no value of � exists such that requirement (*) holds.

This proves that f is discontinuous at 2.

4. Since f is continuous at an interior point c of I and f (c) 6¼ 0, it follows, by Theorem 2

of Sub-section 5.4.1, that there is a neighbourhood N¼ ( c� r, cþ r) of c on which

f xð Þj j > 1

2f cð Þj j: (1)

Since f (c) 6¼ 0, it follows from the inequality (1) that

f xð Þj j > 1

2f cð Þj j

> 0; for all x with x� cj j < r:

In particular, f (x) 6¼ 0 on N, so that 1f

is defined on N.

To prove that 1f

is continuous at c, we must prove that,


1

f xð Þ �1

f cð Þ

< "; for all x satisfying x� cj j < �: ðÞ

Now

1

f xð Þ �1

f cð Þ

¼ f cð Þ � f xð Þf xð Þf cð Þ

¼ f xð Þ � f cð Þj jf xð Þj j � f cð Þj j :

(2)

Since f is continuous at c, we know that, for the given value of " in (*), there is a

positive number �1 such that

f xð Þ � f cð Þj j < "; for all x satisfying x� cj j < �1: (3)

Now choose �¼min{r, �1}, so that both (1) and (3) hold, for all x satisfying

jx� cj<�. It then follows, from (1), (2) and (3), that

1

f xð Þ �1

f cð Þ

¼ f xð Þ � f cð Þj jf xð Þj j � f cð Þj j

<"

12

f cð Þj j2

¼ 112

f cð Þj j2� ":

This is equivalent to the statement (*), in which the number " has been replaced by

K", where K ¼ 112 f cð Þj j2. Since K is a constant, this is sufficient, by the K" Lemma.

It follows that 1f

is continuous at c, as required.

412 Appendix 4

Section 5.5

1. Let " be any given positive number. We want to find a positive number � such that

f xð Þ � f cð Þj j < "; for all x and c in ½2; 3� satisfying x� cj j < �;

that is, that

x2 � c2

< "; for all x and c in ½2; 3� satisfying x� cj j < �: ( )

Now, since x2� c2¼ (x� c)(xþ c), we have

x2 � c2

¼ x� cj j � xþ cj j:But, for x and c in [2, 3], we know that

4 � xþ c � 6:

Hence, in order to prove the statement (*), it is sufficient to prove that there exists a

positive number � such that

6 x� cj j < "; for all x and c in ½2; 3� satisfying x� cj j < �: ( )In view of (**), take � ¼ 1

6" (note that this choice of � depends only on ", and

not at all on x or c). With this choice of �, it follows that if x, c2 [2, 3] and satisfy

jx� cj<�, then

6 x� cj j < 6�

¼ ":In other words, the statement (**) holds, and hence the statement (*) also holds.

Hence the choice � ¼ 16" serves our purpose.

Chapter 6

Section 6.1

1. (a) Let c be any point of R . Then, for any non-zero h, the difference quotient Q(h)

at c is


¼ cþ hð Þn� cn

h

¼ 1

hncn�1hþ 1

2n n� 1ð Þcn�2h2 þ � � � þ nchn�1 þ hn

�

n terms in the bracketð Þ

¼ ncn�1 þ 1

2n n� 1ð Þcn�2hþ � � � þ nchn�2 þ hn�1

! ncn�1 as h! 0:

It follows that f is differentiable at c, and that f 0(c)¼ ncn�1.

(b) Let f (x)¼ k, and let c be any point of R . Then, for any non-zero h, the difference

quotient Q(h) at c is


¼ k � k

h

¼ 0

! 0 as h! 0:

It follows that f is differentiable at c, and that f 0(c)¼ 0.

2. Let c be any point of R , with c 6¼ 0. Then, for any non-zero h, the difference quotient

Q(h) at c is


¼ 1

h

1

cþ h� 1

c

�

¼ c� cþ hð Þh cþ hð Þc

¼ �1

cþ hð Þc

! � 1

c2as h! 0:

It follows that f is differentiable at c, and that f 0 cð Þ ¼ � 1c2.

3. To prove that f (x)¼ x4, x2R , is differentiable at 1, with derivative f 0(1)¼ 4, we

have to show that:


x4 � 1

x� 1� 4


that is

x3 þ x2 þ x� 3

< "; for all x satisfying 0 < x� 1j j < �:

Now, x3þ x2þ x� 3¼ (x� 1)(x2þ 2xþ 3); thus, for 0<jx� 1j< 1, we have

x2 (0, 2), so that

x2 þ 2xþ 3

< 22 þ 2� 2þ 3 ¼ 11:

Next, choose � ¼ min 1; 111"

�

. With this choice of �, it follows that, for all x

satisfying 0<jx� 1j<�, we have

x3 þ x2 þ x� 3

¼ x� 1j j � x2 þ 2xþ 3

<1

11"� 11 ¼ ":

It follows that f is differentiable at 1, with derivative f 0(1)¼ 4.

4. (a) Let f xð Þ ¼ jxj12. The graph of f near 0 suggests that f is not differentiable at 0.

Then, for any non-zero h, the difference quotient Q(h) at 0 is


¼ hj j12� 0

h

¼ hj j12

h:

For h> 0, it follows that Q hð Þ ¼ 1

h12

. So, in particular, if we take h ¼ 1n2 for n2N ,

we find that

Q1

n2

�

¼ n!1 as n!1:

It follows that f is not differentiable at 0.

(b) Let f(x)¼ [x], the integer part of x. The graph of f near 1 suggests that f is not

differentiable at 1.

We use the fact that

x4�1¼ x2�1� �

x2þ1� �

¼ x�1ð Þ xþ1ð Þ x2þ1� �

¼ x�1ð Þ x3þx2þxþ1� �

:

414 Appendix 4

Then, for any h with �1< h< 0, the difference quotient Q(h) at 1 is


¼ 0� 1

h

¼ � 1

h:

It follows that

limh!0�

Q hð Þ ¼ limh!0�

� 1

h

�

¼ 1:Hence f is not differentiable at 1.

5.

(a) For 2> h> 0, the difference quotient for f at �2 is

Q hð Þ ¼ f �2þ hð Þ � f �2ð Þh

¼ � �2þ hð Þ2þ 4

h

¼ 4� h

! 4 as h! 0þ:

Hence f is differentiable on the right at �2, and fR0 (�2)¼ 4.

f is not differentiable at �2, since it is not defined to the left of 2.

(b) For �1< h< 1, the difference quotient for f at 0 is


¼�h2�0

h; h < 0;

h4�0h; h > 0;

(

¼�h; h < 0;

h3; h > 0:

�

Thus

limh!0�

Q hð Þ ¼ 0 and limh!0þ

Q hð Þ ¼ 0;

so that fL0 (0) and fR

0 (0) both exist and equal 0.

It follows that f is differentiable at 0, and f 0(0)¼ 0.

(c) The difference quotient for f at 1 is


¼1þhð Þ4�1

h; �1 < h < 0,

1þhð Þ3�1

h; 0 < h < 1,

8

<

:

¼ 4þ 6hþ 4h2 þ h3; �1 < h < 0,

3þ 3hþ h2; 0 < h < 1.

�

Thus

limh!0�

Q hð Þ ¼ 4 and limh!0þ

Q hð Þ ¼ 3;

so that fL0 (1)¼ 4 and fR

0 (1)¼ 3.

It follows that f is not differentiable at 1.

(d) The difference quotient for f at 2 is


¼2þhð Þ3�8

h; �1 < h < 0,

0�8h; 0 < h < 1,

(

¼12þ 6hþ h2; �1 < h < 0,

� 8h; 0 < h < 1.

(

Since limh!0þ

Q hð Þ does not exist, it follows that f is not differentiable at 2.

However f is differentiable on the left at 2, and fL0 (2)¼ 12.

6. For n2N

f1

2nþ 12

� �

p

!

¼ sin 2nþ 1

2

�

p

¼ 1! 1 as n!1:Hence, although 1

2nþ12ð Þp! 0 as n!1

f1

2nþ 12

� �

p

!

6! 0

¼ f 0ð Þ:It follows that f is not continuous at 0; and so, by Corollary 1, f is not differentiable at 0.

7. For any non-zero h, the difference quotient Q(h) at 0 is


¼h2 sin 1

h

� �

h

¼ h sin1

h

�

! 0 as h! 0:

It follows that f is differentiable at 0, and that f 0(0)¼ 0.

Now, for x 6¼ 0, f 0 xð Þ ¼ 2x sin 1x� cos 1

x. Hence, if n2N

f 01

2np

�

¼ 1

npsin 2npð Þ � cos 2npð Þ

¼ 0� 1

¼ �1; as n!1:Since f 0(0) 6¼�1, it follows that f 0 is not continuous at 0.

416 Appendix 4

8. Let c be any point in R . At c, the difference quotient for f is

Q hð Þ ¼ cos cþ hð Þ � cos c

h

¼ cos c cos h� sin c sin h� cos c

h

¼ cos c� cos h� 1

h� sin c� sin h

h

! cos c� 0� sin c� 1; as h! 0;

¼ � sin c:

It follows that f is differentiable at c, and f 0(c)¼� sin c. Since c is an arbitrary point

of R , it follows that f is differentiable on R .

9. For c> 0, the different quotient for f at c is

Q hð Þ ¼ cþ hð Þ3� c3

h

¼ 1

h3c2hþ 3ch2 þ h3� �

¼ 3c2 þ 3chþ h2 ! 3c2 as h! 0;

so that f is differentiable at c and f 0(c)¼ 3c2.

Next, for c< 0, the different quotient for f at c is

Q hð Þ ¼ cþ hð Þ2� c2

h

¼ 1

h2chþ h2� �

¼ 2cþ h! 2c as h! 0;

so that f is differentiable at c and f 0(c)¼ 2c.

Finally, the difference quotient for f at 0 is


¼h2�0

h; h < 0,

h3�0h; h > 0,

(

¼h; h < 0,

h2; h > 0,

�

! 0; as h! 0:

It follows that f is differentiable at 0, and that f 0(0)¼ 0.

Therefore, the function f 0 is given by the following formula

f 0 xð Þ ¼2x; x < 0,

0; x ¼ 0,

3x2; x > 0.

(

It follows that f 0 is continuous at 0, since limx!0�

f 0 xð Þ ¼ limx!0þ

f 0 xð Þ ¼ f 0 0ð Þ ¼ 0:

Section 6.2

1. (a) f 0 xð Þ ¼ 7x6 � 8x3 þ 9x2 � 5; x 2 R :

(b) f 0 xð Þ ¼ x3 � 1ð Þ2x� x2 þ 1ð Þ3x2

x3 � 1ð Þ2

¼ �x4 � 3x2 � 2x

x3 � 1ð Þ2; x 2 R � 1f g:

We assume that h is sufficientlysmall that jhj< c, so thatcþ h> 0; and hence that we areusing the correct value for f atcþ h.

We assume that h is sufficientlysmall that jhj<� c, so thatcþ h< 0; and hence that we areusing the correct value for f atcþ h.

(c) f 0 xð Þ ¼ 2 cos2 x� 2 sin2 x

¼ 2 cos 2x; x 2 R :

(d) f 0 xð Þ ¼ 3þ sin x� 2 cos xð Þex � ex cos xþ 2 sin xð Þ3þ sin x� 2 cos xð Þ2

¼ ex 3� sin x� 3 cos xð Þ3þ sin x� 2 cos xð Þ2

; x 2 R :

2. f 0 xð Þ ¼ e2x þ 2xe2x ¼ e2x 1þ 2xð Þ;f 00 xð Þ ¼ 2e2x 1þ 2xð Þ þ 2e2x ¼ e2x 4þ 4xð Þ;f 000 xð Þ ¼ 2e2x 4þ 4xð Þ þ 4e2x ¼ e2x 12þ 8xð Þ:

3. In each case, we use the Quotient Rule and the derivatives of sine and cosine.

(a) Here f xð Þ ¼ sin xcos x

, so that

f 0 xð Þ ¼ cos x cos x� sin x �sin xð Þcos2 x

¼ 1

cos2 x

¼ sec2 x;

on the domain of f.

(b) Here f xð Þ ¼ 1sin x

, so that

f 0 xð Þ ¼ � cos x

sin2 x

¼ �cosec x cot x;

on the domain of f.

(c) Here f xð Þ ¼ 1cos x

, so that

f 0 xð Þ ¼ sin x

cos2 x

¼ sec x tan x;

on the domain of f.

(d) Here f xð Þ ¼ cos xsin x

, so that

f 0 xð Þ ¼ sin x � sin xð Þ � cos x cos x

sin2 x

¼ �1

sin2 x

¼ �cosec2x;

on the domain of f.

4. (a) Here f xð Þ ¼ 12

ex � e�xð Þ; x 2 R ; so that

f 0 xð Þ ¼ 1

2ex þ e�xð Þ

¼ cosh x; x 2 R :

(b) Here f xð Þ ¼ 12

ex þ e�xð Þ; x 2 R ; so that

f 0 xð Þ ¼ 1

2ex � e�xð Þ

¼ sinh x; x 2 R :

(c) Here

f xð Þ ¼ sinh x

cosh x; x 2 R ;

418 Appendix 4

so that

f 0 xð Þ ¼ cosh x cosh x� sinh x sinh x

cosh2 x

¼ 1

cosh2 x

¼ sech2x; x 2 R :

5. (a) Here

f xð Þ ¼ sinh x2� �

; x 2 R ;

so that

f 0 xð Þ ¼ 2x cosh x2� �

; x 2 R :

(b) Here

f xð Þ ¼ sin sinh 2xð Þð Þ; x 2 R ;

so that

f 0 xð Þ ¼ cos sinh 2xð Þð Þ2 cosh 2xð Þ; x 2 R :

(c) Here

f xð Þ ¼ sincos 2x

x2

�

; x 2 0;1ð Þ;

so that

f 0 xð Þ ¼ coscos 2x

x2

�

� �x22 sin 2x� 2x cos 2x

x4

�

; x 2 0;1ð Þ:

6. (a) The function

f xð Þ ¼ cos x; x 2 0; pð Þ;is continuous and strictly decreasing on (0, p), and

f 0; pð Þð Þ ¼ �1; 1ð Þ:Also, f is differentiable on (0, p), and its derivative f 0(x)¼�sin x is non-zero

there.

So f satisfies the conditions of the Inverse Function Rule.

Hence f�1¼ cos�1 is differentiable on (�1, 1); and, if y¼ f (x), then

f�1� �0

yð Þ ¼ 1

f 0 xð Þ ¼ �1

sin x:

Since sin x> 0 on (0, p), and sin2 xþ cos2x¼ 1, it follows that

sin x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� cos2 xp

¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

1� y2p

:

Hence

f�1� �0

yð Þ ¼ �1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1� y2p :


f�1� �0

xð Þ ¼ �1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1� x2p ; x 2 �1; 1ð Þ:

(b) The function

f xð Þ ¼ sinh x; x 2 R ;

is continuous and strictly increasing on R , and

f Rð Þ ¼ R :


Also, f is differentiable on R , and its derivative f 0(x)¼ cosh x is non-zero there.

So f satisfies the conditions of the Inverse Function Rule.

Hence f�1¼ sinh�1 is differentiable on R ; and, if y¼ f (x), then

f�1� �0

yð Þ ¼ 1

f 0 xð Þ ¼1

cosh x:

Since cosh x> 0 on R , and cosh2 x¼ 1þ sinh2 x, it follows that

cosh x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ sinh2 xp

¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ y2p

:

Hence

f�1� �0

yð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ y2p :


f�1� �0

xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ x2p ; x 2 R :

7. If x1< x2, then x15 < x2

5; it follows that f (x1)< f (x2), so that f is strictly increasing

on R .

Since f is a polynomial function, it is continuous and differentiable on R . Also

f 0 xð Þ ¼ 5x4 þ 1 6¼ 0 on R :

Thus, f satisfies the conditions of the Inverse Function Theorem.

Now, f (0)¼�1, f (1)¼ 1 and f (�1)¼�3. Hence, by the Inverse Function Rule

f�1� �0 �1ð Þ ¼ 1

f 0 0ð Þ ¼ 1;

f�1� �0

1ð Þ ¼ 1

f 0 1ð Þ ¼1

6;

and

f�1� �0 �3ð Þ ¼ 1

f 0 �1ð Þ ¼1

6:

8. By definition,

f xð Þ ¼ xx ¼ exp x loge xð Þ:The functions x 7! x and x 7! loge x are differentiable on (0, 1), and exp is differ-

entiable on R . It follows, by the Product Rule and the Composition Rule, that f is

differentiable on (0,1), and that

f 0 xð Þ ¼ exp x loge xð Þ � loge xþ x� 1

x

�

¼ xx loge xþ 1ð Þ:

Section 6.3

1. Since f is a polynomial function, f is continuous on [�1, 2] and differentiable on

(�1, 2); also

f 0 xð Þ ¼ x3 � x2 ¼ x2 x� 1ð Þ:Thus f 0 vanishes at 0 and 1.

First, we consider the behaviour of f near 0. For x2 (�1, 1), for example

f xð Þ ¼ 1

4x3 x� 4

3

�

has the opposite sign to that of x, since x� 43< 0; thus

420 Appendix 4

f xð Þ > 0 for x 2 �1; 0ð Þand

f xð Þ < 0 for x 2 0; 1ð Þ:

Since f (0)¼ 0, it follows that 0 is not a local extremum of f.

Next, we consider the behaviour of f near 1. Now

f xð Þ � f 1ð Þ ¼ 1

4x4 � 1

3x3

�

� 1

4� 1

3

�

¼ 1

4x4 � 1� �

� 1

3x3 � 1� �

¼ 1

4x� 1ð Þ x3 þ x2 þ xþ 1

� �

� 1

3x� 1ð Þ x2 þ xþ 1

� �

¼ 1

12x� 1ð Þ 3x3 � x2 � x� 1

� �

¼ 1

12x� 1ð Þ2 3x2 þ 2xþ 1

� �

:

It follows that, for x2 (0, 2), for example

f xð Þ � f 1ð Þ � 0;

so that f has a local minimum at 1, with value f 1ð Þ ¼ � 112

.

2. Since the sine and cosine functions are continuous and differentiable on R , so also is f.

Now

f 0 xð Þ ¼ 2 sin x cos x� sin x ¼ sin x 2 cos x� 1ð Þ;

thus f 0 vanishes in 0; 12p

� �

only when cos x ¼ 12; that is, when x ¼ 1

3p.

Since f (0)¼ 1, f 12p

� �

¼ 1 and

f1

3p

�

¼ 1

2

ffiffiffi

3p� 2

þ 1

2¼ 3

4þ 1

2¼ 5

4;

it follows that, on 0; 12p

� �

:

the minimum of f is 1, and occurs when x¼ 0 and x ¼ 12p;

the maximum of f is 54, and occurs when x ¼ 1

3p.

3. Since f is a polynomial function, f is continuous on [1, 3] and differentiable on (1, 3).

Also, f (1)¼ 2 and f (3)¼ 2, so that f (1)¼ f (3). Thus f satisfies the conditions of

Rolle’s Theorem on [1, 3].

Now

f 0 xð Þ ¼ 4x3 � 12x2 þ 6x ¼ 2x 2x2 � 6xþ 3� �

;

so that f 0(x)¼ 0 when x¼ 0 and x ¼ 12

3�ffiffiffi

3p� �

(the roots of the quadratic equation

2x2� 6xþ 3¼ 0). Thus the only value of x in (1, 3) such that f 0(x)¼ 0 is

x ¼ 12

3þffiffiffi

3p

� �

, so we must take c ¼ 12

3þffiffiffi

3p

� �

’ 2:37.

4. (a) NO: f is not defined at 12p.

(b) NO: f is not differentiable at 1, and f (0) 6¼ f (2).

(c) YES: All the conditions are satisfied.

(d) NO: f 0ð Þ 6¼ f 12p

� �

.

Section 6.4

1. (a) Since f is a polynomial function, f is continuous on [�2, 2] and differentiable on

(�2, 2). Thus f satisfies the conditions of the Mean Value Theorem on [�2, 2].

Nowf 0 xð Þ ¼ 3x2 þ 2;

so that c satisfies the conclusion of the theorem when

3c2 þ 2 ¼ f 2ð Þ � f �2ð Þ2� �2ð Þ ¼

12� �12ð Þ2� �2ð Þ

¼ 6;

that is, when 3c2¼ 4 so that c ¼ �ffiffi

43

q

’ �1:15.

Thus there are two possible values of c.

(b) The function exp is continuous on [0, 3] and differentiable on (0, 3). Thus f

satisfies the conditions of the Mean Value Theorem on [0, 3].

Now, f 0(x)¼ ex, so that c satisfies the conclusion of the theorem when

ec ¼ f 3ð Þ � f 0ð Þ3� 0

¼ 1

3e3 � 1� �

;

that is, when c ¼ loge13

e3 � 1ð Þ� �

’ 1:85.

2. (a) Here f 0 xð Þ ¼ 4x13 � 4 ¼ 4 x

13 � 1

� �

, so that f 0(x)> 0 on (1, 1). Hence, by the

Increasing-Decreasing Theorem, f is strictly increasing on [1,1).

(b) Here f 0 xð Þ ¼ 1� 1x, so that f 0(x)< 0 on (0, 1). Hence, by the Increasing-

Decreasing Theorem, f is strictly decreasing on (0, 1].

3. (a) Let

f xð Þ ¼ sin�1 xþ cos�1 x; x 2 �1; 1½ �:

Then f is continuous on [�1, 1] and differentiable on (�1, 1).

Now

f 0 xð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1� x2p � 1


1� x2p ¼ 0; for x 2 �1; 1ð Þ:

It follows, from Corollary 1, that

sin�1 xþ cos�1 x ¼ c; for x 2 �1; 1½ �;for some constant c.

Putting x¼ 0, we obtain 0þ 12p ¼ c, so that c ¼ 1

2p. Hence

sin�1 xþ cos�1 x ¼ 1

2p; x 2 �1; 1½ �:

(b) Let

f xð Þ ¼ tan�1 xþ tan�1 1

x; x 2 0;1ð Þ:

Then f is continuous and differentiable on (0,1).

Now

f 0 xð Þ ¼ 1

1þ x2�

1x2

1þ 1x

� �2¼ 0; for x 2 0;1ð Þ:

It follows, from Corollary 1, that

tan�1 xþ tan�1 1

x¼ c; for x 2 0; 1ð Þ;

for some constant c.

Putting x¼ 1, we obtain 14pþ 1

4p ¼ c, so that c ¼ 1

2p. Hence

tan�1 xþ tan�1 1

x¼ 1

2p; x 2 0; 1ð Þ:

422 Appendix 4

4. We have

f 0 xð Þ ¼ 3x2 � 6x ¼ 3x x� 2ð Þ; for x 2 R :

Thus f 0(x)¼ 0 when x¼ 0 and x¼ 2, so that c¼ 0, 2.

Now

f 00 xð Þ ¼ 6x� 6; for x 2 R ;

so that f 0 0(0)¼�6< 0 and f 0 0(2)¼ 6> 0. It follows, from the Second Derivative

Test, that f has a local maximum that occurs at 0 and a local minimum that occurs at 2.

The value of the local maximum is f (0)¼ 1, and the value of the local minimum is

f (2)¼�3.

5. (a) Let

f xð Þ ¼ x� sin x; x 2 0;1

2p

� �

:

Then f is continuous on 0; 12p

� �

and differentiable on 0; 12p

� �

.

Now

f 0 xð Þ ¼ 1� cos x > 0; for x 2 0;1

2p

�

;

and f (0)¼ 0.

It follows that f (x)� 0 for x 2 0; 12p

� �

, so that

x � sin x; for x 2 0;1

2p

� �

:

(b) Let

f xð Þ ¼ 2

3xþ 1

3� x

23; x 2 0; 1½ �:

Then f is continuous on [0, 1] and differentiable on (0, 1).

Now

f 0 xð Þ ¼ 2

3� 2

3x�

13

¼ 2

31� x�

13

� �

< 0; for x 2 0; 1ð Þ;

and f 1ð Þ ¼ 23þ 1

3� 1 ¼ 0.

It follows that f (x)� 0 for x2 [0, 1], so that

2

3xþ 1

3� x

23; for x 2 0; 1½ �:

Section 6.5

1. Here

f xð Þ ¼ x3 þ x2 sin x and g xð Þ ¼ x cos x� sin x on 0; p½ �:Since polynomial functions and the sine and cosine functions are continuous and

differentiable on R , it follows, by the Combination Rules, that f and g are continuous

on [0, p] and differentiable on (0, p). It follows that Cauchy’s Mean Value Theorem

applies to the functions f and g on [0, p].

Now, g(0)¼ 0� 1� 0¼ 0 and g(p)¼ pcosp� sinp¼ � p, so that g(p) 6¼ g(0).

Also

f 0 xð Þ ¼ 3x2 þ 2x sin xþ x2 cos x

¼ x2 3þ cos xð Þ þ 2x sin x

and

g0 xð Þ ¼ cos x� x sin x� cos x

¼ �x sin x

on (0, p). In particular, g0(x) 6¼ 0 on (0, p).

It therefore follows from Cauchy’s Mean Value Theorem that there exists at

least one point c in (0, p) for which

c2 3þ cos cð Þ þ 2c sin c

�c sin c¼ p3ð Þ � 0ð Þ�pð Þ � 0ð Þ ;

so thatc 3þ cos cð Þ þ 2 sin c

� sin c¼ �p2:

By cross-multiplying, we obtain

c 3þ cos cð Þ þ 2 sin c ¼ p2 sin c;

so that

3c ¼ p2 � 2� �

sin c� c cos c:

Hence the equation 3x¼ (p2� 2)sinx� xcosx has at least one root in (0, p).

2. (a) Letf xð Þ ¼ sinh 2x and g xð Þ ¼ sin 3x; for x 2 R :


f 0ð Þ ¼ g 0ð Þ ¼ 0;


Now

f 0 xð Þ ¼ 2 cosh 2x and g0 xð Þ ¼ 3 cos 3x;

so that


2 cosh 2x

3 cos 3x:

It follows, from l’Hopital’s Rule, that the desired limit limx!0

f xð Þg xð Þ exists and equals

limx!0


x!0

2 cosh 2x

3 cos 3x;


But, by the Combination Rules for continuous functions, the function

x 7! 2 cosh 2x

3 cos 3x

is continuous at 0, so that

limx!0


f 0 0ð Þg0 0ð Þ

¼ 2

3:

It follows, from l’Hopital’s Rule, that the original limit exists, and that its

value is 23.

(b) Let

f xð Þ ¼ 1þ xð Þ15� 1� xð Þ

15

and

g xð Þ ¼ 1þ 2xð Þ25� 1� 2xð Þ

25;

for x 2 � 12; 1

2

� �

:

424 Appendix 4

Then f and g are differentiable on � 12; 1

2

� �

, and

f 0ð Þ ¼ g 0ð Þ ¼ 0;


Now

f 0 xð Þ ¼ 1

51þ xð Þ�

45 þ 1

51� xð Þ�

45

and

g0 xð Þ ¼ 4

51þ 2xð Þ�

35 þ 4

51� 2xð Þ�

35;

so that


15

1þ xð Þ�45 þ 1

51� xð Þ�

45

45

1þ 2xð Þ�35 þ 4

51� 2xð Þ�

35

:



limx!0


x!0

15

1þ xð Þ�45 þ 1

51� xð Þ�

45

45

1þ 2xð Þ�35 þ 4

51� 2xð Þ�

35

;


But, by the Combination Rules and the Power Rule for continuous functions,

the functions f 0 and g0 are continuous at 0; hence, by the Quotient Rule

limx!0


f 0 0ð Þg0 0ð Þ

¼15þ 1

545þ 4

5

¼ 1

4:

It follows, from l’Hopital’s Rule, that the original limit exists, and equals 14.

(c) Let

f xð Þ ¼ sin x2 þ sin x2� ��

and g xð Þ ¼ 1� cos 4x; for x 2 R :


f 0ð Þ ¼ g 0ð Þ ¼ 0;


Now

f 0 xð Þ ¼ cos x2 þ sin x2� ��

� 2xþ 2x cos x2� ��

and

g0 xð Þ ¼ 4 sin 4x;

so that


cos x2 þ sin x2ð Þð Þ � 2xþ 2x cos x2ð Þð Þ4 sin 4x

:

It follows, from l’Hopital’s Rule, that the limit limx!0


limx!0

f 0 xð Þg0 xð Þ ; ( )

provided that this last limit () exists.

Next, f 0(0)¼ 0 and g0(0)¼ 0, and f 0 and g0 are differentiable on R . Thus f 0 and

g0 satisfy the conditions of l’Hopital’s Rule at 0.

It follows, from l’Hopital’s Rule, that the limit limx!0

f 0 xð Þg0 xð Þ exists and equals


limx!0

f 00 xð Þg00 xð Þ ; ()

provided that this last limit () exists.

Now

f 00 xð Þ ¼ �sin x2 þ sin x2� ��

� 2xþ 2x cos x2� �� 2

þ cos x2 þ sin x2� ��

� 2þ 2 cos x2� �

� 4x2 sin x2� ��

and

g00 xð Þ ¼ 16 cos 4x:

But, by the Combination Rules for continuous functions, the functions f 0 0 and g0 0

are continuous at 0, so that

limx!0


f 00 0ð Þg00 0ð Þ

¼ 4

16¼ 1

4:

It follows, from l’Hopital’s Rule, that the limit () exists, and that its value is 14.

It then follows, from a second application of l’Hopital’s Rule, that the original

limit exists, and that its value is 14.

(d) Let

f xð Þ ¼ sin x� x cos x and g xð Þ ¼ x3; for x 2 R :


f 0ð Þ ¼ g 0ð Þ ¼ 0;


Now

f 0 xð Þ ¼ x sin x and g0 xð Þ ¼ 3x2;

so that


x sin x

3x2

¼ 1

3� sin x

x:



limx!0


x!0

1

3� sin x

x

�

¼ 1

3� lim

x!0

sin x

x

�

;


But limx!0

sin xx

� �

¼ 1, so that the limit limx!0

f 0 xð Þg0 xð Þ does exist and equals 1

3. It follows,

from l’Hopital’s Rule, that the original limit also exists, and that its value is 13.

Chapter 7

Section 7.1

1. In this case, the function f is continuous, except at 12

and 1.

On the three subintervals 0; 13

� �

, 13; 3

4

� �

and 34; 1

� �

in P, we have

426 Appendix 4

m1 ¼ f 0ð Þ ¼ 0; M1 ¼ f1

3

�

¼ 2

3; �x1 ¼

1

3� 0 ¼ 1

3;

m2 ¼ f1

2

�

¼ 0; M2 ¼ f3

4

�

¼ 3

2; �x2 ¼

3

4� 1

3¼ 5

12;

m3 ¼ f 1ð Þ ¼ 1; M3 ¼ 2�

¼ limx!1�

f xð Þ�

; �x3 ¼ 1� 3

4¼ 1

4:

It then follows, from the definitions of L( f, P) and U( f, P), that

L f ;Pð Þ ¼X

3

i¼1

mi�xi ¼ m1�x1 þ m2�x2 þ m3�x3

¼ 0� 1

3þ 0� 5

12þ 1� 1

4

¼ 0þ 0þ 1

4¼ 1

4;

U f ;Pð Þ ¼X

3

i¼1

Mi�xi ¼ M1�x1 þM2�x2 þM3�x3

¼ 2

3� 1

3þ 3

2� 5

12þ 2� 1

4

¼ 2

9þ 5

8þ 1

2¼ 97

72:

2. In this case, the function f is increasing and continuous on [0, 1]. Thus, on each

subinterval in [0, 1], the infimum of f is the value of f at the left end-point of

the subinterval and the supremum of f is the value of f at the right end-point of the

subinterval.

Hence, on the ith subinterval i�1n; i

n

� �

in Pn, for 1� i� n, we have

mi ¼ fi� 1

n

�

¼ i� 1

n

� 2

; Mi ¼ fi

n

�

¼ i

n

� 2

; and

�xi ¼i

n� i� 1

n¼ 1

n:

It then follows, from the definitions of L( f, Pn) and U( f, Pn), that

L f ;Pnð Þ ¼X

n

i¼1

mi�xi ¼X

n

i¼1

i� 1

n

� 2

� 1

n

¼ 1

n3

X

n

i¼1

i2 � 2iþ 1� �

¼ 1

n3

n nþ 1ð Þ 2nþ 1ð Þ6

� 2n nþ 1ð Þ

2þ n

� �

¼ 1

n2

nþ 1ð Þ 2nþ 1ð Þ6

� nþ 1ð Þ þ 1

� �

¼ 1

6n22n2 þ 3nþ 1� �

� 6 nþ 1ð Þ þ 6 �

¼ 1

6n22n2 � 3nþ 1 �

¼ 1

3� 1

2nþ 1

6n2;

U f ;Pnð Þ ¼X

n

i¼1

Mi�xi ¼X

n

i¼1

i

n

� 2

� 1

n

¼ 1

n3

X

n

i¼1

i2

¼ 1

n3� n nþ 1ð Þ 2nþ 1ð Þ

6

¼ 2n2 þ 3nþ 1

6n2¼ 1

3þ 1

2nþ 1

6n2:

x134

12

0

1

2

y = f (x)

y

13

x

1

1

y

y = f (x)


It follows that

limn!1

L f ; Pnð Þ ¼ limn!1

1

3 � 1

2n þ 1

6n2

�

¼ 1

3

and

limn!1

U f ; Pnð Þ ¼ limn!1

1

3 þ 1

2n þ 1

6n2

�

¼ 1

3:

3. By definition of least upper bound, we know that f xð Þ � supx 2J

f , for x 2 J . It follows,

from the fact that I J , that

f xð Þ � supx 2J

f ; for x 2 I :

Hence supx2 J

f is an upper bound for f on I, so that supx2 I

f � supx 2J

f .

4. In Problem 2, we showed that for the function f ( x ) ¼ x 2 , x 2 [0, 1], and the partition

Pn ¼ 0 ; 1n

� �

; 1n ; 2

n

� �

; . . .; i�1n; i

n

� �

; . . .; 1 � 1n ; 1

� � �

of [0, 1]

limn!1

L f ; Pnð Þ ¼ 1

3and lim

n!1U f ; Pnð Þ ¼ 1

3:

It follows thatZ

�

1

0

f � 1

3and

Z

�1

0

f � 1

3:

ButZ

�

1

0

f �Z

�1

0

f ; by part ðbÞ of Theorem 4:


�

1

0 f ¼

R

�1

0f ¼ 1

3, so that f is integrable on [0, 1] and

R 1

0 f ¼ 1

3 .

5. Here the function f ( x ) = k , x 2 [0, 1], is the constant function on [0, 1].

Hence, on the i th subinterval i�1n; i

n

� �

in Pn , for 1 � i � n, we have

mi ¼ k ; M i ¼ k ; and � x i ¼i

n � i � 1

n¼ 1

n:

It then follows, from the definitions of L( f , Pn) and U ( f , Pn ), that

L f ; Pnð Þ ¼X

n

i¼1

mi � x i ¼X

n

i ¼1

k � 1

n

¼ kX

n

i¼1

1

n ¼ k ;

U f ; Pnð Þ ¼X

n

i¼ 1

Mi � x i ¼X

n

i¼1

k � 1

n

¼ kX

n

i ¼1

1

n ¼ k :

It follows thatZ

�

1

0

f � k and

Z

�1

0

f � k :

ButZ

�

1

0

f �Z

�1

0

f ; by part ðbÞ of Theorem 4:


�1

0f ¼

R

�1

0f ¼ k, so that f is integrable on [0, 1] and

R 1

0f ¼ k:

428 Appendix 4

6. (a) Let Pn ¼ x0; x1½ �; x1; x2½ �; . . .; xi�1; xi½ �; . . .; xn�1; xn½ �f g be a standard partition of

[0, 1].

The function f is constant on the interval [0, 1), so that, for this partition Pn,

we have

mi ¼ �2 for all i;

Mi ¼�2; 1 � i � n� 1;

3; i ¼ n;

�

�xi ¼1

n:

It follows that

L f ;Pnð Þ ¼X

n

i¼1

mi�xi ¼X

n

i¼1

�2ð Þ � 1

n

¼ �2X

n

i¼1

1

n¼ �2;

and

U f ;Pnð Þ ¼X

n

i¼1

Mi�xi ¼X

n�1

i¼1

Mi�xi þMn�xn

¼X

n�1

i¼1

�2ð Þ� 1

nþ 3� 1

n

¼ �2X

n�1

i¼1

1

nþ 3

n

¼ �2n� 1

nþ 3

n¼ �2þ 5

n:

Thus, in particular

U f ;Pnð Þ � L f ;Pnð Þ ¼ 5

n:

Now let " be any given positive number. It follows that, if we choose n such that5n< " (that is, if we choose n > 5

" ), then

U f ;Pnð Þ � L f ;Pnð Þ ¼ 5

n

�

< ":

It follows, from Riemann’s Criterion for integrability, that f is integrable on [0, 1].

Alternatively, let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be any

partition of [0, 1] . Then, since Mi ¼ mi for 1� i� n�1, we have


n

i¼1

Mi � mið Þ�xi ¼ Mn � mnð Þ�xn

¼ 3� �2ð Þð Þ�xn ¼ 5�xn: ()

Now let " be any given positive number, and choose P to be any partition of [0, 1]

such its mesh, jjPjj, is less than "5; then, in particular, we have �xn <

"5. It then

follows, from the inequality (), that

U f ;Pð Þ � L f ;Pð Þ ¼ 5�xn

< 5� "5¼ ":

It follows, from Riemann’s Criterion for integrability, that f is integrable

on [0, 1].

3

2

1

1 x

y

–1

–2

(b) Let P¼ {[x0, x1], [x1, x2], . . ., [xi�1, xi], . . ., [xn�1, xn]} be any partition of [0, 1].

xi–1 xi

1

y

10 x

y = 1, x ∈

y = 0, x ∉

Then, for the partition P, we have

mi ¼ 0 and Mi ¼ 1;

for all i.

It follows that

L f ;Pð Þ ¼X

n

i¼1

mi�xi ¼X

n

i¼1

0� �xi ¼ 0;

and

U f ;Pð Þ ¼X

n

i¼1

Mi�xi ¼X

n

i¼1

1� �xi

¼X

n

i¼1

�xi ¼ 1:

Thus, for ALL partitions P of [0, 1], we have

U f ;Pð Þ � L f ;Pð Þ ¼ 1:

In particular, it follows that, for any positive " for which 0<"< 1, there is NO

partition P of [0, 1] for which

U f ;Pð Þ � L f ;Pð Þ < ":

It follows, from Riemann’s Criterion for integrability, that f is not integrable

on [0, 1].

Section 7.2

1. First, we want to use the Strategy to prove that

infx2I


f xð Þf g:

Since f is bounded on I, infx2I

f xð Þf g exists. In particular

f xð Þ � infx2I

f xð Þf g; for all x 2 I:

It follows that

k þ f xð Þ � k þ infx2I

f xð Þf g; for all x 2 I;

so that k þ infx2I

f xð Þf g is a lower bound for kþ f (x) on I.

Next, let m0 be any number greater than k þ infx2I

f xð Þf g. It follows that

m0 � k > infx2I

f xð Þf g;

430 Appendix 4

so that, in particular, there is some number x in I such that

m0 � k > f xð Þ:We may reformulate this fact as: there is some number x in I such that

m0 > k þ f xð Þ;so that m0 is not a lower bound for kþ f (x) on I.

It follows that the greatest lower bound for kþ f (x) on I is k þ infx2I

f xð Þf g; in other

wordsinfx2I


f xð Þf g:

Next, we want to use the Strategy to prove that

supx2I


f xð Þf g:

Since f is bounded on I, supx2I

f xð Þf g exists. In particular

f xð Þ � supx2I

f xð Þf g; for all x 2 I:

It follows that

k þ f xð Þ � k þ supx2I

f xð Þf g; for all x 2 I;

so that k þ supx2I

f xð Þf g is an upper bound for kþ f (x) on I.

Next, let M0 be any number less than k þ supx2I

f xð Þf g. It follows that

M0 � k < supx2I

f xð Þf g;

so that, in particular, there is some number x in I such that

M0 � k < f xð Þ:We may reformulate this fact as: there is some number x in I such that

M0 < k þ f xð Þ;so that M0 is not an upper bound for kþ f (x) on I.

It follows that the least upper bound for kþ f (x) on I is k þ supx2I

f xð Þf g; in other

wordssupx2I


f xð Þf g:

2. The function f is decreasing on the interval (�2, 0], so that

infx2ð�2;0�

f ¼ f 0ð Þ ¼ 0 and supx2ð�2;0�

f ¼ limx!�2

þf xð Þ ¼ 4;

f is increasing on the interval [0, 3], so that

infx2 0;3½ �

f ¼ f 0ð Þ ¼ 0 and supx2 0;3½ �

f ¼ f 3ð Þ ¼ 9:

Since I¼ (�2, 0] [ [0, 3], it follows that

infI

f ¼ f 0ð Þ ¼ 0 and supI

f ¼ maxn

f 3ð Þ; limx!�2

þf xð Þ

o

¼ 9:

Since 0� f (x)� 9 and 0� f (y)� 9, so that�9 �� f (y)� 0, we have

� 9 � f xð Þ � f yð Þ � 9:

We will prove that infx;y2I

f xð Þ � f yð Þf g ¼ �9 and supx;y2I

f xð Þ � f yð Þf g ¼ 9.

First, since f (x)� f (y) ��9, �9 is a lower bound for f (x)� f (y), for x, y2 I. To

prove that �9 is the greatest lower bound, we now need to prove that:

for each positive number ", there are X and Y in I = (�2, 3] for which

f Xð Þ � f Yð Þ < �9þ ":

Now, by the definition of infimum and supremum on I, we know that, since 12" > 0,

there exist X and Y in (�2, 3] such that

f Xð Þ < 12" and f Yð Þ > 9� 1

2":


f Xð Þ � f Yð Þ < 12"

� �

þ �9þ 12"

� �

¼ �9þ ":

It follows that infx;y2I

f xð Þ � f yð Þf g ¼ �9.

Finally, since f (x)� f (y)� 9, 9 is an upper bound for f (x)� f (y), for x, y2 I. To

prove that 9 is the least upper bound, we now need to prove that:

for each positive number ", there are X and Y in I¼ (�2, 3] for which

f Xð Þ � f Yð Þ > 9� ":

Now, by the definition of infimum and supremum on I, we know that, since 12" > 0,

there exist X and Y in (�2, 3] such that

f Xð Þ > 9� 12" and f Yð Þ < 1

2":


f Xð Þ � f Yð Þ > 9� 12"

� �

þ �12"

� �

¼ 9� ":

It follows that supx;y2I

f xð Þ � f yð Þf g ¼ 9.

3. (a) For any x in I, f xð Þ � supI

f , so that, since l> 0

lf xð Þ � l supI

f ; for all x 2 I:

Thus

supx2I

lf xð Þð Þ � l supI

f :

To prove that supx2I

lf xð Þð Þ ¼ l�

supI

f�

, we now need to prove that:


lf Xð Þ > l�

supI

f�

� ": ( )

Now, since "> 0 and l> 0, we have "l > 0. It follows, from the definition of

supremum of f on I, that there exists an X in I for which

f Xð Þ > supI

f � "l:

Multiplying both sides by the positive number l, we obtain the desired result ().(b) For any x in I, f xð Þ � inf

If so that

� f xð Þ � � infI

f ; for all x 2 I:

Thus

supx2I

�f xð Þð Þ � � infI

f :

To prove that supx2I

�f xð Þð Þ ¼ � infI

f , we now need to prove that:


� f Xð Þ > � infI

f � ": ( )

For example, for any number Xin (�2, 3] with Xj j � 1

2

ffiffiffi

"p

, wehave

f Xð Þ ¼ X2 � 1

4";

we could, for instance, chooseX ¼ 1

2

ffiffiffi

"p

.

432 Appendix 4

Now, since "> 0 it follows, from the definition of infimum of f on I, that there

exists an X in I for which

f Xð Þ < infI

f þ ";so that

� f Xð Þ > � infI

f � ":

This is precisely the desired result ().

Section 7.3

1. (a) F0 xð Þ ¼ xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 4p� ��1

� 1þ 1

2x2 � 4� ��1

2 2xð Þ�

¼ xþffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 � 4p� ��1

� x2 � 4� ��1

2� x2 � 4� �

12þx

� �

¼ x2 � 4� ��1

2:

(b) F0 xð Þ ¼ 1

2


4� x2p

þ 1

2x� 1

24� x2� ��1

2 �2xð Þ þ 2 1� 1

4x2

� �121

2

¼ 4� x2� ��1

21

24� x2� �

� 1

2x2 þ 2

�

¼ 4� x2� ��1

2 4� x2� �

¼ 4� x2� �

12:

2. (a) From the Fundamental Theorem of Calculus and the Table of Standard

Primitives, we obtainZ 4

0

x2 þ 9� �

12dx ¼ 1

2x x2 þ 9� �

12 þ 9

2loge xþ x2 þ 9

� �

12

� �

� �4

0

¼ 10þ 9

2loge 9� 9

2loge 3

¼ 10þ 9

2loge 3:

(b) From the Fundamental Theorem of Calculus and the Table of Standard

Primitives, we obtainZ e

1

loge xdx ¼ x loge x� x½ �e1

¼ e� eð Þ � 0� 1ð Þ¼ 1:

3. Using the Table of Standard Primitives and the Combination Rules, we obtain the

following primitives:

(a) F xð Þ ¼ 4 x loge x� xð Þ � tan�1 x

2

� �

;

(b) F xð Þ ¼ 23

loge sec 3xð Þ þ 15e2x 2 sin x� cos xð Þ:

4. (a) Here we use integration by parts. Let

f xð Þ ¼ loge x and g0 xð Þ ¼ x13;

so that

f 0 xð Þ ¼ 1

xand g xð Þ ¼ 3

4x

43:

It follows thatZ

x13 loge x ¼ 3

4x

43 loge x �

Z

x �13

4 x

43dx

¼ 3

4x

43 loge x � 3

4

Z

x13dx

¼ 3

4x

43 loge x � 9

16x

43:

(b) Here we use integration by parts, twice. On each occasion we differentiate

the power function and integrate the trigonometric function.

HenceZ p

2

0

x 2 cos xdx ¼ x 2 sin x� �

p2

0�Z p

2

0

2x sin xdx

¼ 1

4p 2 � 2

Z p2

0

x sin xdx ;

andZ p

2

0

x sin xdx ¼ x � �cos xð Þ½ �p2

0 �Z p

2

0

� cos xð Þdx

¼ 0 þ sin x½ �p2

0

¼ 1:

It follows thatZ p

2

0

x 2 cos xdx ¼ 1

4p 2 � 2:

5. (a) We follow the strategy just before Example 3, with x in place of t and u in

place of x .

Let u ¼ g( x ) ¼ 2sin 3x , x 2R . Then

du

dx ¼ 6 cos 3 x; so that du ¼ 6 cos 3xdx:

Hence Z

sin 2 sin 3xð Þ cos 3xdx ¼Z

1

6sin udu

¼ �1

6cos u

¼ �1

6cos 2 sin 3xð Þ:

(b) We follow the strategy just before Example 3, with x in place of t and u in place of x.

Let u ¼ g xð Þ ¼ e x , x 2R . The function g is one–one on R . Then

du

dx ¼ ex ; so that du ¼ exdx;

also

when x ¼ 0; then u ¼ 1;

when x ¼ 1; then u ¼ e:

HenceZ 1

0

ex

1þ exð Þ2dx ¼

Z e

1

du

1þ uð Þ2

¼ �1

1þ u

� �e

1

¼ � 1

1þ eþ 1

2

¼ e� 1

2 1þ eð Þ :

434 Appendix 4

6. (a) We follow the strategy just before Example 4.

Let t ¼ g ðxÞ ¼ ðx � 1Þ12; x 2 ð1;1Þ. The function g is one–one on (1, 1).

Then t2 ¼ x � 1, so that x ¼ t2 þ 1. It follows that

dx

dt¼ 2t ; so that dx ¼ 2 tdt :

HenceZ

dx

3 x � 1ð Þ32þ x x� 1ð Þ

12

¼Z

2 tdt

3t 3 þ t 2 þ 1ð Þt

¼Z

2dt

4t 2 þ 1

¼ tan�1 2tð Þ

¼ tan�1 2 x � 1ð Þ12

� �

:

(b) We follow the strategy just before Example 4.

Let t ¼ g xð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ exp

, x2 [0,1). The function g is one–one on [0,1).

Then t2¼ 1þ ex, so that ex¼ t2�1 and x¼ loge (t2�1). It follows that:

dx

dt¼ 2t

t2 � 1; so that dx ¼ 2t

t2 � 1dt;

also:

when x¼ 0, then t ¼ffiffiffi

2p

,

when x¼ loge 3, then t¼ 2.

HenceZ loge 3

0

exffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ exp

dx ¼Z 2

ffiffi

2p t2 � 1� �

� t � 2t

t2 � 1dt

¼Z 2

ffiffi

2p 2t2dt

¼ 2

3t3

� �2

ffiffi

2p

¼ 16� 4ffiffiffi

2p

3:

Section 7.4

1. Since

x sin1

x10

�

� x; for x 2 1; 3½ �;

it follows, from part (a) of Theorem 2, thatZ 3

1

x sin1

x10

�

dx �Z 3

1

xdx

¼ 1

2x2

� �3

1

¼ 1

29� 1ð Þ¼ 4:

2. Since

ex2� 1

1� x2

� 1

1� 1

4

¼ 4

3; for x 2 0;

1

2

� �

;


it follows, from part (b) of Theorem 2, thatZ 1

2

0

ex2

dx � 4

3

1

2� 0

�

¼ 2

3:

(Remark We could obtain a smaller upper estimate for the integral by applying

part (a) of Theorem 2 to the inequality ex2� 11�x2.)

Next, since

ex2�1þ x2 � 1; for x 2 0;1

2

� �

;

it follows, from part (b) of Theorem 2, thatZ 1

2

0

ex2

dx � 1� 1

2� 0

�

¼ 1

2:

3. (a) Since

sin1

x

�

� 1; for x 2 1; 4½ �;

and

2þ cos1

x

�

� 1; for x 2 1; 4½ �;

it follows that

sin 1x

� �

2þ cos 1x

� �

� 1; for x 2 1; 4½ �:

It follows, from Theorem 3, thatZ 4

1

sin 1x

� �

2þ cos 1x

� �

� 1� 4� 1ð Þ

¼ 3:

(b) Since

tan x � 0 for x 2 0;p4

h i

;

and

3� sin x2� �

� 2; for x 2 0;p4

h i

;

it follows that

tan x

3� sin x2ð Þ �1

2tan x; for x 2 0;

p4

h i

:

It follows, from Theorems 2 and 3, thatZ p

4

0

tan x

3� sin x2ð Þ dx

�Z p

4

0

tan x

3� sin x2ð Þ

dx

�Z p

4

0

1

2tan xdx

¼ 1

2loge sec xð Þ

� �p4

0

¼ 1

2loge sec

p4

� �

� loge 1� �

¼ 1

2loge

ffiffiffi

2p¼ 1

4loge 2:

436 Appendix 4

4. (a) I0 ¼Z 1

0

exdx

¼ ex½ �10 ¼ e� 1:

(b) Using integration by parts, we obtain

In ¼Z 1

0

exxndx

¼ exxn½ �10 �Z 1

0

exnxn�1dx

¼ e� nIn�1:

(c) Using the result of part (b), with n¼ 1, 2, 3 and 4 in turn, we obtain

I1 ¼ e� I0 ¼ e� e� 1ð Þ ¼ 1;

I2 ¼ e� 2I1 ¼ e� 2;

I3 ¼ e� 3I2 ¼ e� 3 e� 2ð Þ ¼ 6� 2e;

I4 ¼ e� 4I3 ¼ e� 4 6� 2eð Þ ¼ 9e� 24:

5. (a) a1 ¼2

1� 23¼ 4

3;

a2 ¼2

1� 23� 43� 45¼ 64

45;

a3 ¼2

1� 23� 43� 45� 65� 67¼ 256

175;

b1 ¼1!ð Þ222

2!ffiffiffi

1p ¼ 2;

b2 ¼2!ð Þ224

4!ffiffiffi

2p ¼ 4

3

ffiffiffi

2p

;

b3 ¼3!ð Þ226

6!ffiffiffi

3p ¼ 16

15

ffiffiffi

3p

:

(b) b21 ¼ 4 and 3a1 ¼ 4;

b22 ¼

32

9and

5

2a2 ¼

32

9;

b23 ¼

256

75and

7

3a3 ¼

256

75:

(c) We know that

b2n ¼

n!ð Þ424n

2nð Þ!ð Þ2n:

We now express an in a similar way, tackling the numerator and denominator

separately.

The numerator is a product of 2n even numbers; so, taking a factor 2 from

each term, we obtain

2:2:4:4: . . . : 2nð Þ: 2nð Þ ¼ 22n 1:1:2:2: . . . :n:nð Þ¼ 22n n!ð Þ2:

The denominator of an cannot be tackled in quite the same way, as all its factors

are odd. However, we can relate it to factorials by introducing the missing

even factors:


1:3:3:5:5: . . . : 2n� 1ð Þ: 2nþ 1ð Þ ¼ 1:2:2:3:3:4:4: . . . : 2n� 1ð Þ: 2nð Þ: 2nð Þ: 2nþ 1ð Þ2:2: 4:4: . . . : 2nð Þ: 2nð Þ

¼ 2nð Þ!ð Þ2 2nþ 1ð Þ22n n!ð Þ2

:

It follows that

an ¼22n n!ð Þ22nð Þ!ð Þ2 2nþ1ð Þ

22n n!ð Þ2

¼ 24n n!ð Þ4

2nð Þ!ð Þ2 2nþ 1ð Þ¼ b2

n

n

2nþ 1;

so that

b2n ¼

2nþ 1

nan:

6. Let

f xð Þ ¼ 1

xp; for x 2 1;1½ Þ;

where p> 0 and p 6¼ 1.

Then f is positive and decreasing, and

f xð Þ ! 0 as x!1:Also,

Z n

1

f ¼Z n

1

dx

xp

¼ x1�p

1� p

� �n

1

¼ n1�p � 1

1� p: ( )

Now, if p> 1, then 1� p< 0, so thatZ n

1

f ¼ 1

p� 1� n1�p

p� 1<

1

p� 1:

Since the setR n

1f : n 2 N

�

is bounded above, it follows, from the Maclaurin

Integral Test, that the series converges.

Finally, if 0< p< 1, then 1� p> 0, so that

n1�p !1 as n!1:It follows, from the Maclaurin Integral Test, that the series diverges.

7. Let t¼ g(x)¼ loge x, x2 (1,1). The function g is one–one on (1,1).

Then x¼ et; also

dt

dx¼ 1

xand so dt ¼ dx

x:

HenceZ

dx

x loge xð Þ2¼Z

dt

t2

¼ � 1

t¼ � 1

loge x:

438 Appendix 4

Now let

f xð Þ ¼ 1

x logexð Þ2; x 2 2;1½ Þ:

Then f is positive and decreasing on [2,1), and

f xð Þ ! 0 as x!1:Also

Z n

2

f ¼Z n

2

dx

x loge xð Þ2

¼ � 1

loge x

� �n

2

¼ 1

loge 2� 1

loge n

� 1

loge 2:

Since the setR n

2f : n 2N

�

is bounded above, it follows, from the Maclaurin

Integral Test, that the series converges.

8. Let t¼ g(x)¼ loge x, x2 (1,1). The function g is one–one on (1,1).

Then x¼ et; also

dt

dx¼ 1

xand so dt ¼ dx

x:

HenceZ

dx

x loge x¼Z

dt

t

¼ loge t ¼ loge loge xð Þ:Now let

f xð Þ ¼ 1

x loge x; x 2 2;1½ Þ:

Then f is positive and decreasing on [2,1), and

f xð Þ ! 0 as x!1:Also

Z n

2

f ¼Z n

2

dx

x loge x

¼ loge loge xð Þ½ �n2¼ loge loge nð Þ � loge loge 2ð Þ! 1 as n!1:

Hence, by the Maclaurin Integral Test, the series diverges.

9. Let f xð Þ ¼ 11þx2, x2 [0, 1].

Then f is positive and decreasing on [0, 1]; it follows, from the strategy, that

1

n

X

n

i¼1

fi

n

�

!Z 1

0

f :

Now

1

n

X

n

i¼1

fi

n

�

¼ 1

n

X

n

i¼1

1

1þ in

� �2¼ n

X

n

i¼1

1

n2 þ i2

¼ n1

n2 þ 12þ 1

n2 þ 22þ � � � þ 1

n2 þ n2

�

;


andZ 1

0

f ¼Z 1

0

1

1þ x2dx

¼ tan�1 x� �1

0

¼ tan�1 1� tan�1 0 ¼ 1

4p:

It follows that

n1

n2 þ 12þ 1

n2 þ 22þ � � � þ 1

n2 þ n2

�

! 1

4p as n!1:

10. Let f xð Þ ¼ 1ffiffi

xp , x2 [1,1).

Then f is positive and decreasing on [1,1), and f (x)! 0 as x!1.

Now

X

n

i¼1

f ið Þ ¼X

n

i¼1

1ffiffi

ip

¼ 1þ 1ffiffiffi

2p þ 1

ffiffiffi

3p þ � � � þ 1

ffiffiffi

np ;

andR n

1f ¼

R n

1x�

12dx

¼ 2x12

h in

1

¼ 2n12 � 2!1 as n!1:

It follows, from the strategy, that

1þ 1ffiffiffi

2p þ 1

ffiffiffi

3p þ � � � þ 1

ffiffiffi

np � 2n

12 � 2 as n !1;

and so, using the remark before the problem, that

1þ 1ffiffiffi

2p þ 1

ffiffiffi

3p þ � � � þ 1

ffiffiffi

np � 2n

12 as n!1:

Section 7.5

1. 20! ’ 2:4329� 1018;

30! ’ 2:6525� 1032;

40! ’ 8:1592� 1047;

50! ’ 3:0414� 1064;

60! ’ 8:3210� 1081:

2. First, f2(n) � f5(n); that is, sin 1n2

� �

� 1n2 as n!1.

We know that

sin x

x! 1 as x! 0;

since the sequence 1n2

�

tends to 0 as n!1, it follows that

sin 1n2

� �

1n2

! 1 as n!1:

In other words, sin 1n2

� �

� 1n2 as n!1.

Second, f3(n) � f7(n); that is, 1� cos 1n

� �

� 12n2 as n!1.

We can use the formula cos x ¼ 1� 2 sin2 12

x� �

to obtain

440 Appendix 4

1 � cos x

x2¼

2 sin2 12 x

� �

x 2

¼ 1

2 �

sin 12

x� �

12

x�

sin 12

x� �

12

x:

Then, since 12

x ! 0 as x ! 0, we deduce, from the fact that sin xx! 1 as x ! 0, that

1 � cos x

x2! 1

2as x ! 0 :

Then, since the sequence 1n

�

tends to 0 as n !1, it follows that

1 � cos 1n

� �

1n 2

! 1

2as n !1;

so that

1 � cos 1n

� �

12 n2

! 1 as n !1;

in other words, 1 � cos 1n

� �

� 12n 2 as n !1.

3. Using the Hint, let f (n) ¼ n2 and g( n) ¼ n.

Then

f nð Þð Þ1n ¼ n2

� �

1n

¼ n1n

� �2

! 1ð Þ2 ¼ 1 as n !1;and

g nð Þð Þ1n ¼ nð Þ

1n

! 1 as n !1;

so thatf nð Þð Þ

1n

g nð Þð Þ1n! 1 as n !1. In other words, f nð Þð Þ

1n� g nð Þð Þ

1n as n !1.

However

f nð Þg nð Þ ¼

n2

n

¼ n 6! 1 as n !1;so that f ( n) 6� g(n) as n !1.

4. First, using a calculator, we find that the value of the expressionffiffiffiffiffiffiffiffi

2p np

ne

� �nwhen

n ¼ 5 is approximately 118.019.

Now 5! ¼ 120, so that the error in the Stirling’s Formula approximation is about

120 � 118.019 ¼ 1.981. The percentage error in this approximation is thus about

1 :981

120� 100 ’ 1: 651%:

This approximation is therefore not within 1% of the exact value.

5. Using Stirling’s Formula, we obtain

nn

n!� nn


2pnp

ne

� �n

¼ en

ffiffiffiffiffiffi

2pnp ;

it follows from Example 1 that

nn

n!

� 1n

� e

p1

2n 2nð Þ1

2n

as n!1:


We have seen previously that, for any positive number a

a1n ! 1 and n

1n ! 1; as n!1:

It follows that

p1

2n ! 1 and 2nð Þ1

2n ! 1; as n!1:It follows that

nn

n!

� 1n

� e as n!1;

that is

nn

n!

� 1n

! e as n!1:

6. (a) 300

150

!

1

2300¼ 300!

150!� 150!� 2300

’ffiffiffiffiffiffiffiffiffiffi

600pp

300e

� �300

300p 150e

� �3002300

¼ 1

30

ffiffiffi

6

p

r

¼ 0:046 ðto two significant figuresÞ:(b)

300!

100!ð Þ3� 1

3300’


600pp

300e

� �300

200pð Þ32 100

e

� �3003300

¼ffiffiffi

3p

200p

¼ 0:0028 ðto two significant figuresÞ:

7. Now

4n

2n

�

¼ 4nð Þ!2nð Þ!ð Þ2

and2n

n

�

¼ 2nð Þ!n!ð Þ2

;

so that

4n

2n

�

2n

n

� ¼ 4nð Þ! n!ð Þ2

2nð Þ!ð Þ3:

It follows, from Stirling’s Formula, that

�

4n

2n

�

2n

n

�ffiffiffiffiffiffiffiffi

8pnp

4ne

� �4n2pn n

e

� �2n

4pnð Þ32 2n

e

� �6n

¼ 44n

ffiffiffi

2p� 26n

¼ 22n

ffiffiffi

2p :

Hence l ¼ 1�ffiffiffi

2p

.

442 Appendix 4

Chapter 8

Section 8.1

1. The tangent approximation to f at a is

f xð Þ ’ f að Þ þ f 0 að Þ x� að Þ:

(a) f xð Þ ¼ ex; f 2ð Þ ¼ e2;

f 0 xð Þ ¼ ex; f 0 2ð Þ ¼ e2:


ex ’ e2 þ e2 x� 2ð Þ:

(b) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0:


cos x ’ 1þ 0 x� 0ð Þ ¼ 1:

2. (a) f xð Þ ¼ ex; f 2ð Þ ¼ e2;

f 0 xð Þ ¼ ex; f 0 2ð Þ ¼ e2;

f 00 xð Þ ¼ ex; f 00 2ð Þ ¼ e2;

f 000 xð Þ ¼ ex; f 000 2ð Þ ¼ e2:

Hence

T1 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!

x� 2ð Þ

¼ e2 þ e2 x� 2ð Þ;

T2 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!

x� 2ð Þ þ f 00 2ð Þ2!

x� 2ð Þ2

¼ e2 þ e2 x� 2ð Þ þ 1

2e2 x� 2ð Þ2;

T3 xð Þ ¼ f 2ð Þ þ f 0 2ð Þ1!

x� 2ð Þ þ f 00 2ð Þ2!

x� 2ð Þ2þ f 000 2ð Þ3!

x� 2ð Þ3

¼ e2 þ e2 x� 2ð Þ þ 1

2e2 x� 2ð Þ2þ 1

6e2 x� 2ð Þ3:

(b) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;

f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0:

Hence

T1 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!

x ¼ 1;

T2 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!

x þ f 00 0ð Þ2!

x2 ¼ 1� 1

2x2;

T3 xð Þ ¼ f 0ð Þ þ f 0 0ð Þ1!

x þ f 00 0ð Þ2!

x2 þ f 000 0ð Þ3!

x3 ¼ 1� 1

2x2:


3. The Taylor polynomial of degree 4 for f at a is

T4 xð Þ ¼ f að Þ þ f 0 að Þ1!

x� að Þ þ f 00 að Þ2!

x� að Þ2þ f 000 að Þ3!

x� að Þ3þ f 4ð Þ að Þ4!

x� að Þ4:

(a) f xð Þ ¼ 7� 6xþ 5x2 þ x3; f 1ð Þ ¼ 7;

f 0 xð Þ ¼ �6þ 10xþ 3x2; f 0 1ð Þ ¼ 7;

f 00 xð Þ ¼ 10þ 6x; f 00 1ð Þ ¼ 16;

f 000 xð Þ ¼ 6; f 000 1ð Þ ¼ 6;

f 4ð Þ xð Þ ¼ 0; f 4ð Þ 1ð Þ ¼ 0:

Hence

T4 xð Þ ¼ 7þ 7 x� 1ð Þ þ 8 x� 1ð Þ2þ x� 1ð Þ3:

(b) f xð Þ ¼ 1� xð Þ�1; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ 1� xð Þ�2; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ 2 1� xð Þ�3; f 00 0ð Þ ¼ 2;

f 000 xð Þ ¼ 3! 1� xð Þ�4; f 000 0ð Þ ¼ 3!;

f 4ð Þ xð Þ ¼ 4! 1� xð Þ�5; f 4ð Þ 0ð Þ ¼ 4!:

Hence

T4 xð Þ ¼ 1þ xþ x2 þ x3 þ x4:

(c) f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ 1þ xð Þ�1; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ � 1þ xð Þ�2; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ 2 1þ xð Þ�3; f 000 0ð Þ ¼ 2;

f 4ð Þ xð Þ ¼ �3! 1þ xð Þ�4; f 4ð Þ 0ð Þ ¼ �3!:

Hence

T4 xð Þ ¼ x� 1

2x2 þ 1

3x3 � 1

4x4:

(d) f xð Þ ¼ sin x; fp4

� �

¼ 1ffiffiffi

2p ;

f 0 xð Þ ¼ cos x; f 0p4

� �

¼ 1ffiffiffi

2p ;

f 00 xð Þ ¼ � sin x; f 00p4

� �

¼ � 1ffiffiffi

2p ;

f 000 xð Þ ¼ � cos x; f 000p4

� �

¼ � 1ffiffiffi

2p ;

f 4ð Þ xð Þ ¼ sin x; f 4ð Þ p4

� �

¼ 1ffiffiffi

2p :

Hence

T4 xð Þ ¼ 1ffiffiffi

2p 1þ x� p

4

� �

� 1

2x� p

4

� �2

� 1

6x� p

4

� �3

þ 1

24x� p

4

� �4�

:

444 Appendix 4

(e)f xð Þ ¼ 1þ 1

2x� 1

2x2 � 1

6x3 þ 1

4x4; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ 1

2� x� 1

2x2 þ x3; f 0 0ð Þ ¼ 1

2;

f 00 xð Þ ¼ �1� xþ 3x2; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ �1þ 3!x; f 000 0ð Þ ¼ �1;

f 4ð Þ xð Þ ¼ 3!; f 4ð Þ 0ð Þ ¼ 3!:

Hence

T4 xð Þ ¼ 1þ 1

2x� 1

2x2 � 1

6x3 þ 1

4x4:

4. f xð Þ ¼ tan x; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ sec2 x; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ 2 sec2 x tan x; f 00 0ð Þ ¼ 0;

f 000 xð Þ ¼ 4 sec2 x tan2 xþ 2 sec4 x; f 000 0ð Þ ¼ 2:

Hence

T3 xð Þ ¼ xþ 1

3x3;

so that

T3 0:1ð Þ ¼ 0:1þ 1

3� 0:001 ¼ 0:1003:

Since (by calculator) tan 0.1¼ 0.10033467. . ., the percentage error involved is

about

0:10033467� 0:10033333

0:10033467� 100 ’ 0:001%:

5. (a) f xð Þ ¼ 1� xð Þ�1; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ 1� xð Þ�2; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ 2� 1� xð Þ�3; f 00 0ð Þ ¼ 2;

..

.

f nð Þ xð Þ ¼ n!ð Þ � 1� xð Þ�n�1; f nð Þ 0ð Þ ¼ n!:

Hence

Tn xð Þ ¼ f 0ð Þ þ f 0 0ð Þxþ 1

2!f 00 0ð Þx2 þ � � � þ 1

n!f nð Þ 0ð Þxn

¼ 1þ xþ x2 þ � � � þ xn:

(b) f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ 1þ xð Þ�1; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ �1ð Þ � 1þ xð Þ�2; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ �1ð Þ �2ð Þ � 1þ xð Þ�3; f 000 0ð Þ ¼ 2;

..

.

f nð Þ xð Þ ¼ �1ð Þnþ1n� 1ð Þ!� 1þ xð Þ�n; f nð Þ 0ð Þ ¼ �1ð Þnþ1

n� 1ð Þ!:Hence


2!f 00 0ð Þx2 þ � � � þ 1

n!f nð Þ 0ð Þxn

¼ x� 1

2x2 þ 1

3x3 � � � � þ �1ð Þnþ1

nxn:


(c) f xð Þ ¼ ex; f 0ð Þ ¼ 1;

and, for each positive integer k

f kð Þ xð Þ ¼ ex; f kð Þ 0ð Þ ¼ 1:

Hence


2!f 00 0ð Þx2 þ � � � þ 1

n!f nð Þ 0ð Þxn

¼ 1þ xþ x2

2!þ x3

3!þ � � � þ xn

n!:

(d) f xð Þ ¼ sin x; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ cos x; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ �sin x; f 00 0ð Þ ¼ 0;

f 000 xð Þ ¼ �cos x; f 000 0ð Þ ¼ �1;

f 4ð Þ xð Þ ¼ sin x; f 4ð Þ 0ð Þ ¼ 0;

and, in general, for each positive integer k

f ðkÞð0Þ ¼0; if k is even,

1; if k � 1 (mod 4),

�1; if k � 3 (mod 4).

8

<

:

Hence

Tn xð Þ ¼ x� x3

3!þ x5

5!þ � � � þ 0 or 1 or�1ð Þ x

n

n!;

where the coefficient of xk isf kð Þ 0ð Þ

k! , and the value of f (k)(0) is as specified above.

(e) f xð Þ ¼ cos x; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;

f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;

f 4ð Þ xð Þ ¼ cos x; f 4ð Þ 0ð Þ ¼ 1;

and, in general, for each positive integer k

f kð Þ 0ð Þ ¼0; if k is odd;1; if k � 0 mod 4ð Þ;�1; if k � 2 mod 4ð Þ:

8

<

:

Hence

Tn xð Þ ¼ 1� x2

2!þ x4

4!� � � � þ 0 or 1 or� 1ð Þ x

n

n!;

where the coefficient of xk isf kð Þ 0ð Þ

k! , and the value of f (k)(0) is as specified

above.

Section 8.2

1. Here

f xð Þ ¼ 1

1� x; f 0 xð Þ ¼ 1

1� xð Þ2and f 00 xð Þ ¼ 2

1� xð Þ3:

Hence

R1 xð Þ ¼ f 00 cð Þ2!

x2 ¼ x2

1� cð Þ3:

446 Appendix 4

To calculate the value of c, we use Taylor’s Theorem, with n¼ 1

f xð Þ ¼ f að Þ þ f 0 að Þ x� að Þ þ R1 xð Þ:This gives

f3

4

�

¼ f 0ð Þ þ f 0 0ð Þ 34þ R1

3

4

�

;

that is

4 ¼ 1þ 3

4þ

34

� �2

1� cð Þ3;

and so

9

4¼

34

� �2

1� cð Þ3:

It follows that 1� cð Þ3¼ 14; so that 1� c ¼ 1

4

� �13 ’ 0:630. Thus c ’ 0:370:

2. When f is a polynomial of degree n or less

f nþ1ð Þ cð Þ ¼ 0; so that Rn xð Þ ¼ 0:

3. f xð Þ ¼ cos x; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;

f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;

f 4ð Þ xð Þ ¼ cos x:

Hence, by Taylor’s Theorem with a¼ 0, f (x)¼ cos x and n¼ 3, we have

cos x ¼ 1� 1

2x2 þ R3 xð Þ;

where

R3 xð Þ ¼ f 4ð Þ cð Þ4!

x4:

Thus

R3 xð Þj j � cos cj j24

x4

� 1

24x4:

4. f xð Þ ¼ loge 1þ xð Þ; f 0ð Þ ¼ 0;

f 0 xð Þ ¼ 1

1þ x; f 0 0ð Þ ¼ 1;

f 00 xð Þ ¼ �1

1þ xð Þ2; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ 2

1þ xð Þ3:

Hence, by Taylor’s Theorem with x¼ 0.02

loge 1:02 ¼ 0þ 1� 0:02ð Þ � 1

2� 0:022

�

þ R2 0:02ð Þ

¼ 0:0198þ R2 0:02ð Þ:Now, for c2 (0, 0.02)

f 000 cð Þj j ¼ 2

1þ cð Þ3

� 2:


Hence, by the Remainder Estimate with M¼ 2, we obtain

jR2 0:02ð Þj � 2

3!� 0:023 ¼ 0:0000026:

It follows that

loge 1:02 ¼ 0:0198 ðto four decimal placesÞ:5. For any positive integer n

f nð Þ xð Þ ¼ � cos x or � sin x;

hence

f nð Þ cð Þ

� 1; for all c 2 R ;

so that M¼ 1.

It follows that the remainder term in the Remainder Estimate satisfies the

inequality

Rn 0:2ð Þj j � 1

nþ 1ð Þ! 0:2ð Þnþ1:

Now

0:2ð Þ4

4!¼ 0:00006 and

0:2ð Þ5

5!¼ 0:0000026 . . .;

so we should try n¼ 4 in the Remainder Estimate, to be safe.

Here

f xð Þ ¼ cos x; f 0ð Þ ¼ 1;

f 0 xð Þ ¼ �sin x; f 0 0ð Þ ¼ 0;

f 00 xð Þ ¼ �cos x; f 00 0ð Þ ¼ �1;

f 000 xð Þ ¼ sin x; f 000 0ð Þ ¼ 0;

f 4ð Þ xð Þ ¼ cos x; f 4ð Þ 0ð Þ ¼ 1

f 5ð Þ xð Þ ¼ �sin x:

Hence, by Taylor’s Theorem with f (x)¼ cos x, a¼ 0, x¼ 0.2 and n¼ 4, we have

f xð Þ ¼ 1� 1

2x2 þ 1

24x4 þ R4 xð Þ;

in other words

cos 0:2 ¼ 1� 1

20:2ð Þ2þ 1

240:2ð Þ4þR4 0:2ð Þ

¼ 1� 0:02þ 0:00006þ R4 0:2ð Þ¼ 0:9801 ðrounded to four decimal placesÞ:

6. Using the same function f as in Problem 5, we find that

f pð Þ ¼ �1; f 0 pð Þ ¼ 0; f 00 pð Þ ¼ 1;

f 000 pð Þ ¼ 0; f 4ð Þ pð Þ ¼ �1:

It follows that

T4 xð Þ ¼ �1þ 1

2x� pð Þ2� 1

24x� pð Þ4:

The difference between T4(x) and f (x) is

R4 xð Þ ¼ f 5ð Þ cð Þ5!

x� pð Þ5

¼ � sin c

120x� pð Þ5;

448 Appendix 4

for some number c between p and x. Hence, for x 2 34p; 5

4p

� �

, we have

R4 xð Þj j � 1

120

p4

� �5

¼ 0:00249 . . . < 3� 10�3:

Section 8.3

1. (a) Applying the Ratio Test with an¼ 2nþ 4n, we obtain

anþ1

an

¼ 2nþ1 þ 4nþ1

2n þ 4n¼ 2þ 2nþ2

1þ 2n

¼ 212

� �nþ212

� �nþ1

! 2� 2 ¼ 4 as n!1:Hence the series has radius of convergence 1

4:

(b) Applying the Ratio Test with an ¼ ðn!Þ2ð2nÞ!, we obtain

anþ1

an

¼ nþ 1ð Þ nþ 1ð Þ2nþ 1ð Þ 2nþ 2ð Þ ¼

1þ 1n

� �

1þ 1n

� �

2þ 1n

� �

2þ 2n

� �

! 1

4as n!1:

Hence the series has radius of convergence 4.

(c) Applying the Ratio Test with an¼ nþ 2�n, we obtain

anþ1

an

¼ nþ 1þ 2�n�1

nþ 2�n¼

1þ 1nþ 1

n2nþ1

1þ 1n2n

! 1 as n!1:Hence the series has radius of convergence 1.

(d) Applying the Ratio Test with an ¼ 1

n!ð Þ1n; we obtain

anþ1

an

¼ n!ð Þ1n

nþ 1ð Þ!ð Þ1

nþ1

�ffiffiffiffiffiffiffiffi

2pnp� �

1n n

e

� �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2p nþ 1ð Þp� �

1nþ1 nþ1

e

� �

¼ffiffiffiffiffiffi

2pp� �

1n

ffiffiffiffiffiffi

2pp� �

1nþ1

��

n1n

�12

ð nþ 1ð Þ1

nþ1Þ12

� n

nþ 1

! 1 as n!1:(Here we have used Stirling’s Formula to estimate n! and (nþ 1)!.)


2. (a) Applying the Ratio Test with an¼ n, we obtain

anþ1

an

¼ nþ 1

n! 1 as n!1:


Thus the series converges if x2 (�1, 1), and diverges if jx j> 1.

When x¼�1, the sequence {nxn} is non-null; hence, by the Non-null Test, the

series diverges.

Hence the interval of convergence of the series is (�1, 1).

(b) Applying the Ratio Test with an ¼ 1n3n, we obtain

anþ1

an

¼ n3n

nþ 1ð Þ3nþ1¼ 1

1þ 1n

� �

3

! 1

3as n!1:

Thus the series converges if x2 (�3, 3), and diverges if jx j> 3.

When x¼ 3, the series isP

1

n¼1

1n, which is divergent.

When x¼�3, the series isP

1

n¼1

�1ð Þnn

, which is convergent.

Hence the interval of convergence of the series is [�3, 3).

3. Applying the Ratio Test with an ¼ � ��1ð Þ: ... : ��nþ1ð Þn! , we obtain

anþ1

an

¼ �� n

nþ 1

¼�n� 1

1þ 1n

! 1 as n!1:


4. From the Hint, we know that tan�1 x is the sum function of the series

X

1

n¼0

�1ð Þn

2nþ 1x2nþ1 ¼ x� x3

3þ x5

5� x7

7þ � � �; for xj j < 1: ( )

From the Hint, we also know that the radius of convergence of this power series is 1.

It follows that, by applying Abel’s Limit Theorem to the power series (), we may

obtain the following

limx!1�

tan�1 x� �

¼X

1

n¼0

�1ð Þn

2nþ 1;

provided that this last series is convergent; it is indeed convergent, by the Alternating

Test for series.

Since the function x 7! tan�1 x is continuous at 1, it follows that

X

1

n¼0

�1ð Þn

2nþ 1¼ lim

x!1�tan�1 x� �

¼ tan�1 1 ¼ p4:

Section 8.4

1. (a) We know that

ex ¼ 1þ xþ x2

2!þ x3

3!þ � � � þ xn

n!þ � � �; for x 2 R ;

and so

e�x ¼ 1� xþ x2

2!� x3

3!þ � � � þ �1ð Þnxn

n!þ � � �; for x 2 R :

Hence, by the Sum and Multiple Rules for power series

sinh x ¼ 1

2ex � e�xð Þ

¼ xþ x3

3!þ � � � þ x2nþ1

2nþ 1ð Þ!þ � � �; for x 2 R :

This series converges for all x.

For x 7! tan�1 x is continuouson its domain R .

450 Appendix 4

(b) We know that

loge 1� xð Þ ¼ �x� x2

2� x3

3� � � � � xn

n� � � �; for xj j < 1;

and1

1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � �; for xj j < 1:


loge 1� xð Þþ 2

1� x

¼ �x� x2

2� x3

3� � � � � xn

n� � � �

�

þ 2þ 2xþ 2x2 þ 2x3 þ � � � þ 2xn þ � � ��

¼ 2þ xþ 3

2x2 þ 5

3x3 þ � � � þ 2n� 1

nxn þ � � �; for xj j < 1:

The radius of convergence of this series is 1.

2. (a) We know that

sinh x ¼ xþ x3

3!þ � � � þ x2nþ1

2nþ 1ð Þ! þ � � �; for x 2 R ;

and that

sin x ¼ x� x3

3!þ � � � þ �1ð Þnx2nþ1

2nþ 1ð Þ! þ � � �; for x 2 R :


sinh xþ sin x

¼ xþ x3

3!þ � � � þ x2nþ1

2nþ 1ð Þ!þ � � ��

þ x� x3

3!þ � � � þ �1ð Þnx2nþ1

2nþ 1ð Þ! þ � � ��

¼ 2 xþ x5

5!þ � � � þ x4nþ1

4nþ 1ð Þ!þ � � ��

; for x 2 R :

This series converges for all x.

(b) We know that

loge 1þ xð Þ ¼ x� x2

2þ x3

3� � � � þ �1ð Þnþ1xn

nþ � � �; for xj j < 1;

and

loge 1� xð Þ ¼ �x� x2

2� x3

3� � � � � xn

n� � � �; for xj j < 1:


loge

1þ x

1� x

�

¼ loge 1þ xð Þ � loge 1� xð Þ

¼ x� x2

2þ x3

3� � � � þ �1ð Þnþ1xn

nþ � � �

�

� �x� x2

2� x3

3� � � � � xn

n� � � �

�

¼ 2 xþ x3

3þ � � � þ x2nþ1

2nþ 1þ � � �

�

; for xj j < 1:


(c) We know that

1

1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1:

It follows, by replacing x by � 2x 2, that

1

1 þ 2x 2 ¼ 1 þ �2x 2

� �

þ �2x2� �2þ �2x 2

� �3þ � � � þ �2x 2� �nþ � � �

¼ 1 � 2x 2 þ 22 x 4 � 23 x 6 þ � � � þ �1ð Þn2n x 2 n þ � � �;

for j2 x2 j< 1; that is, for xj j < 1ffiffi

2p .

The radius of convergence of this series is 1ffiffi

2p .

3. (a) We know that

loge 1 þ xð Þ ¼ x � x2

2þ x3

3� � � � þ �1ð Þnþ 1x n

nþ � � � ; for xj j < 1:

Hence, by the Product Rule

1 þ xð Þ loge 1 þ xð Þ

¼ x � x 2

2þ x 3

3� � � � þ �1ð Þn þ1x n

nþ � � �

�

þ x 2 � x 3

2þ x 4

3� � � � þ �1ð Þn xn

n � 1 þ � � �

�

¼ x þ x2

2� x 3

6� � � � þ �1ð Þn x n

n n� 1ð Þ þ � � � ; for xj j < 1:


(b) We know that

1

1� x¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1;

and (from Example 1) that

1þ x

1� xð Þ2¼ 1þ 3xþ 5x2 þ � � � þ 2nþ 1ð Þxn þ � � � ; for xj j < 1:

Hence, by the Product Rule, we obtain

1þ x

1� xð Þ3

¼ 1

1� x� 1þ x

1� xð Þ2

¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � ��

� 1þ 3xþ 5x2 þ � � � þ 2nþ 1ð Þxn þ � � ��

¼ 1þ 3þ 1ð Þxþ 5þ 3þ 1ð Þx2 þ 7þ 5þ 3þ 1ð Þx3 þ � � �þ 2nþ 1ð Þ þ 2n� 1ð Þ þ � � � þ 1ð Þxn þ � � �

¼ 1þ 4xþ 9x2 þ 16x3 þ � � � þ nþ 1ð Þ2xn þ � � � ; for xj j < 1:

(We can prove, by Mathematical Induction, that

1þ 3þ 5þ � � � þ 2nþ 1ð Þ ¼X

n

k¼0

2k þ 1ð Þ ¼ nþ 1ð Þ2; for n 2 N :Þ


4. (a) We know that

1� xð Þ�1¼ 1þ xþ x2 þ x3 þ � � � þ xn þ � � � ; for xj j < 1:

Hence, by the Differentiation Rule, we obtain

1� xð Þ�2¼ 1þ 2xþ 3x2 þ 4x3 þ � � � þ nxn�1 þ � � � ; for xj j < 1:


452 Appendix 4

(b) If we differentiate the series in part (a ), by the Differentiation Rule for power

series, we obtain

2 1� xð Þ�3 ¼ 2 þ 6x þ 12x 2 þ � � � þ n n� 1ð Þxn �2 þ � � �; for xj j < 1:

Hence, by the Multiple Rule, we have

1 � xð Þ�3 ¼ 1 þ 3x þ 6x 2 þ � � � þ n n� 1ð Þ2

x n� 2 þ � � �; for xj j < 1:


(c) Notice that, for the function f ( x ) ¼ tanh� 1 x, j x j< 1 , we have

f 0 xð Þ ¼ 1

1 � x 2 ¼ 1 þ x 2 þ x 4 þ � � � þ x 2n þ � � �:

Hence, by the Integration Rule for power series, the Taylor series for f at 0 is

tanh �1 x ¼ c þ x þ x 3

3þ x 5

5þ � � � þ x 2n þ1

2n þ 1 þ � � �; for xj j < 1;

since f (0) ¼ 0, it follows that c ¼ 0. Hence

tanh �1 x ¼ x þ x 3

3þ x 5

5þ � � � þ x 2n þ1

2n þ 1 þ � � �; for xj j < 1:


5. We know that

e x ¼ 1 þ x þ x 2

2!þ x 3

3!þ � � � þ x n

n !þ � � �; for x 2 R ;

and we know, from part (a ) of Problem 4, that

1 � xð Þ� 2¼ 1 þ 2x þ 3x 2 þ 4x 3 þ � � � þ n þ 1ð Þx n þ � � �; for xj j < 1:

Hence, by the Product Rule for power series, we have

e x 1 � xð Þ� 2 ¼ 1 þ x þ x 2

2þ x 3

6þ � � �

�

� 1 þ 2x þ 3x 2 þ 4x 3 þ � � ��

¼ 1 þ 2 þ 1ð Þx þ 3 þ 2 þ 1

2

�

x 2 þ � � �

¼ 1 þ 3x þ 11

2x 2 þ � � � ; for xj j < 1:


6. We are given that

f xð Þ ¼ x þ x3

1:3 þ x5

1: 3:5 þ � � � þ x 2n þ1

1: 3: ::: : 2n þ 1ð Þ þ � � �; for x 2 R :

(a) By the Differentiation Rule, we can differentiate the power series term-by-term;

this gives

f 0 xð Þ ¼ 1þ x2

1þ x4

1:3þ � � � þ x2n

1:3: ::: : 2n� 1ð Þ þ � � �; for x 2 R :

(b) It follows, from the definition of f, that

xf xð Þ ¼ x2 þ x4

1:3þ x6

1:3:5þ � � � þ x2nþ2

1:3: ::: : 2nþ 1ð Þ þ � � �; for x 2 R :

Hence, by the Combination Rules for power series,

f 0 xð Þ � xf xð Þ ¼ 1;

since all the other terms cancel out.

7. Since

ex ¼ 1þ xþ x2

2!þ x3

3!þ � � � þ xn

n!þ � � �; for x 2 R ;

we may deduce that

e�x2 ¼ 1� x2 þ x4

2!� x6

3!þ � � � þ �1ð Þn x2n

n!þ � � �; for x 2 R :

It follows, from the Integration Rule for power series, that

Z 1

0

e�x2

dx ¼ x� x3

3þ x5

5� 2!� x7

7� 3!þ � � � þ �1ð Þn x2nþ1

2nþ 1ð Þ � n!þ � � �

� �1

0

¼ 1� 1

3þ 1

10� 1

42þ � � � þ �1ð Þn 1

2nþ 1ð Þ � n!þ � � �:

8. By the General Binomial Theorem

1þ 6xð Þ14 ¼X

1

n¼0

14

n

�

xn; for 6xj j < 1;

where

14

n

�

¼14

� �

� 34

� �

� 74

� �

. . . 14� nþ 1

� �

n!:

It follows that

1þ 6xð Þ14 ¼ 1þ

14

� �

16xð Þ þ

14

� �

� 34

� �

26xð Þ2þ

14

� �

� 34

� �

� 74

� �

66xð Þ3þ � � �

¼ 1þ 3

2x� 27

8x2 þ 189

16x3 � � � �; for xj j < 1

6:


9. (a) By the General Binomial Theorem

1� xð Þ�12 ¼X

1

n¼0

� 12

n

�

�xð Þn; for xj j < 1;

where

� 12

n

�

¼� 1

2

� �

� 32

� �

� 52

� �

. . . � 12� nþ 1

� �

n!:

It follows that

1� xð Þ�12¼ 1þ 1

2xþ 3

8x2 þ � � � þ �1ð Þn � 1

2

n

�

xn þ � � �; for xj j < 1:


(b) We know that

d

dxsin�1 x ¼ 1


1� x2p ; for x 2 �1; 1ð Þ:

Then, by the result of part (a), with x replaced by x2, we obtain


1� x2p ¼ 1þ 1

2x2 þ 3

8x4 þ � � � þ �1ð Þn � 1

2

n

�

x2n þ � � �; for xj j < 1:

Hence, by the Integration Rule for power series

sin�1 x ¼ cþ xþ 1

6x3 þ 3

40x5 þ � � �

þ �1ð Þn � 12

n

�

x2nþ1

2nþ 1þ � � �; for xj j < 1:

454 Appendix 4

Putting x¼ 0 into this equation, we see that c¼ 0. It follows that

sin�1 x ¼ xþ 1

6x3 þ 3

40x5 þ � � �

þ �1ð Þn � 12

n

�

x2nþ1

2nþ 1þ � � �; for xj j < 1:


Section 8.5

1. We use the Addition Formula

tan�1 xþ tan�1 y ¼ tan�1 xþ y

1� xy

�

; ( )

which holds provided that tan�1 xþ tan�1 y lies in the interval � p2; p

2

� �

.

First, we deduce, from the Addition Formula (), that

tan�1 1

3

�

þ tan�1 1

4

�

¼ tan�113þ 1

4

1� 13� 1

4

!

¼ tan�1 4þ 3

12� 1

�

¼ tan�1 7

11

�

:

This equation holds, since

tan�1 1

3

�

þ tan�1 1

4

�

’ 0:3218þ 0:2450 ¼ 0:5668;

and 0:5668 2 � p2; p

2

� �

’ �1:5708; 1:5708ð Þ:Next, we deduce, from the Addition Formula (), that

tan�1 1

3

�

þ tan�1 1

4

�

þ tan�1 2

9

�

¼ tan�1 7

11

�

þ tan�1 2

9

�

¼ tan�17

11þ 2

9

1� 711� 2

9

!

¼ tan�1 63þ 22

99� 14

�

¼ tan�1 1ð Þ ¼ p4:

This equation holds, since

tan�1 7

11

�

þ tan�1 2

9

�

’ 0:5667þ 0:2187 ¼ 0:7854;

and 0:7854 2 � p2; p

2

� �

’ �1:5708; 1:5708ð Þ:2. First

I0 ¼Z 1

�1

cos1

2px

�

dx

¼ 2

psin

1

2px

� � �1

�1

¼ 4

p:

It follows that p I0¼ 4.

Next, using integration by parts twice, we obtain

I1 ¼Z 1

�1

1� x2� �

cos1

2px

�

dx

¼ 1� x2� � 2

psin

1

2px

� � �1

�1

�Z 1

�1

�2xð Þ 2

psin

1

2px

�

dx

¼ 4

p

Z 1

�1

x sin1

2px

�

dx

¼ 4

px � 2

p

�

cos1

2px

� � �1

�1

� 4

p

Z 1

�1

� 2

p

�

cos1

2px

�

dx

¼ 8

p2

2

psin

1

2px

� � �1

�1

¼ 32

p3:

It follows that p 3I1 ¼ 32.

456 Appendix 4

Index

ax, definition and continuity, 161

Abel’s limit theorem, 333

absolute convergence test, 104

absolute convergence theorem, 332, 337

absolute value, 12

absolutely convergent series, 104

addition formula for tan�1, 166

alternating test, 107

antipodal points theorem, 145

approximation by Taylor

polynomials, 317

Archimedean property of R , 7

arithmetic in R , 8, 30

arithmetic mean – geometric mean

inequality, 18, 21

asymptotic behaviour of functions, 176

basic

continuous functions, 142

differentiable functions, 224

null sequences, 48

power series, 325

series, 97

Bernoulli’s inequality, 20, 237

Bessel function, 329

Bijection, 355

binomial theorem, 358, 342

blancmange function, 244

Bolzano–Weierstrass theorem, 70

bound, greatest lower, 26

bound, least upper, 25

boundedness theorem, 62, 92, 149

Cauchy condensation test, 97

Cauchy’s mean value theorem, 239

Cauchy–Schwarz inequality, 20, 311

Chain Rule, 218

combination rules


convergent sequences, 55

convergent series, 89


functions which tend to1, 177

inequalities, 14

infimum, 275

integrable functions, 278

limits of functions, 172

null sequences, 46

power series, 338

primitives, 284

sequences which tend to1, 65

series, 89

supremum, 275

tilda, 305

common limit criterion, 268

common refinement, 262

comparison test, 94

composition rule

asymptotic behaviour of functions, 180




conditionally convergent series, 104

continuity

continuity, 132, 194


one-sided, 134, 194

uniform, 201

continuous functions

basic, 142

boundedness theorem, 149

combination rules, 136

composition rule, 136

extreme values theorem, 169

integrability, 277

intermediate value theorem, 143

inverse function rule, 153

squeeze rule, 137

convergence, interval of, 330

radius of, 330

convergent sequence, 53


quotient rule, 55

squeeze rule, 58

convergent series, 85


cos�1, 156

cosh, power series, 339

cosh�1, 158

cosine function

derived function, 214

power series, 326

Darboux’s theorem, 254

decimal representation of numbers, 3, 87

density property of R , 7

derivatives, 207

higher order, 215

one-sided, 210

standard, 359

difference quotient, 206

differentiability of

f(x)¼ ex, 214

f(x)¼ ax, 223

f(x)¼ x�, 223

f(x)¼ xx, 223

trigonometrical functions, 214

differentiable functions


composition rules, 218

inequalities involving, 237

inverse function rule, 221

differentiation, 210

rule for power series, 341

Dirichlet’s function, 197, 310

discontinuity, removable, 172

divergent sequence, 61

series, 85

domination hierarchy, 56

e, definition, 74

irrationality, 175

"� � game, 43, 187

Euclidean algorithm, 78

even subsequence, 66

ex, definition, 75, 122

fundamental property, 127

inverse property, 75

exponent laws, 163

exponential function

continuity, 141


differentiability, 223

inequalities, 141

power series, 326

extreme values theorem, 149

extremum, 228

field, 9

first subsequence rule, 67

function, nowhere differentiable, 244

fundamental inequality for integrals, 288

fundamental theorem of Algebra, 146

of Calculus, 283, 292

general binomial theorem, 342

geometric series, 87

457

Goethe, ix

greatest lower bound, 26, 272

property of R , 29

Gregory’s series for estimating p, 346

harmonic series, 93, 109

higher-order derivatives, 215

hyperbolic functions, differentiability, 224

hypergeometric series, 352

increasing–decreasing theorem, 234

inequalities, power rule, 10

rules for integrals, 289

infimum, 26, 272

infinite series, 85

inheritance property of subsequences, 66

Int I, 234

Integrability, 264



integral test, 297

modulus rule, 278

monotonic functions, 276

Riemann’s criterion, 267

integral, 264

additivity, 280

inequality rules, 289

lower, 264

upper, 264

integration by parts, 285

by substitution, 286

integration rule for power series, 341

integration, reduction of order

method, 293

intermediate value theorem, 143

interval image theorem, 149

interval of convergence, 230

inverse function rule for continuous

functions, 153


inverse hyperbolic functions, 158

inverse trigonometric functions, 156

irrational numbers, 5

K" lemma, 49, 193

least upper bound, 25, 262

property of R , 29

Leibniz notation, 207

Leibniz test, 107

Leibniz’s series for estimating p, 346

l‘Hopital’s rule, 241

limit comparison test, 95

limit inequality rule for limits, 175

sequences, 60

limit of a function, 169, 177, 185

as x!1, 177

one-sided, 175

limit of a sequence, 53, 182

limits of functions and continuity, 171

limits of functions, combination

rules, 172

composition rule, 173

one-sided, 175

squeeze rule, 174

local extremum theorem, 229

loge 1þ xð Þ, power series for, 325

lower integral, 264

lower Riemann sum, 259

Maclaurin integral test, 297

mathematical induction, 357

maximum, 23

mean value theorem, 233

minimum, 24

modulus function, continuity, 135

modulus rule for integrable

functions, 278

modulus, 12

monotone convergence theorem, 68

monotonic function, 153

integrability, 276

monotonic sequence theorem, 69

multiplication of series, 114

n!, Stirling’s formula for, 306

neighbourhood, 169

non-null test, 91

nth partial sum of series, 85

nth root function, continuity, 155

nth root of a positive real number,

32, 155

null partition criterion, 269

null sequence, 43

odd subsequence, 66

one-one function, 152, 355

one-sided derivative, 210

one-sided limit, 175

onto function, 355

order properties of R , 7

p, 76

irrationality of, 348

numerical estimates, 346

partial sum of series, 85

partition of [a, b], 258

standard, 258

power series

absolute convergence, 332

basic, 325


differentiation rule, 341

integration rule, 341

uniqueness theorem, 342

primitives, 282


scaling rule, 278

standard, 360

uniqueness theorem, 284

radius of convergence theorem, 329

ratio test, 96

for radius of convergence, 330

rational function, continuity, 136


rational power, 34

rearrangement of series, 109

reciprocal rule for

functions which tend to1, 177


tilda, 305

recursion formula, 71

reduction of order method, 293

refinement, 262

remainder estimate, 322

removable discontinuity, 172

Riemann’s rearrangement theorem, 112

Riemann’s function, 198

Riemann’s-criterion for integrability, 267

Rolle’s theorem, 231

scaling rule for primitives, 284

second derivative test, 236

second subsequence rule, 67

sequences, 38

basic null, 48


convergent, 53

divergent, 61

limit inequality rule, 60

monotonic, 40

null, 43

unbounded, 62

which tend to1, 63

which tend to �1, 65

squeeze rule, 58

series, 85

absolutely convergent, 104

basic, 97

convergent, 85

divergent, 85

geometric, 87

harmonic, 93, 109

hypergeometric, 352

integral test, 297

multiplication, 114

product rule, 114

Taylor, 325

telescoping, 88

sine function


power series, 326

sine inequality, 140

sin�1, 156

sinh, 158

power series, 339

square root function, continuity, 134

squeeze rule

as x!1, 178


458 Index

convergent sequences, 58


null sequences, 47


standard derivatives, 359

standard partition, 258

standard primitives, 360

Stirling’s formula, 306

strategy for testing for convergence, 116

sub-interval theorem, 280

subsequence, 66

sum function, 325

supremum, 35

tan�1, 157

power series, 341

tanh�1, 159

power series, 344

tangent approximation, 314

Taylor polynomial, 316

Taylor series, 325

Taylor’s theorem, 320

tilda notation, 300, 304


transitive property of R , 7

transitive rule for inequalities, 291

triangle inequality, 15, 16

backwards form, 15

infinite form, 105

integrals, 291

trichotomy property of R , 7

trigonometric functions

continuity, 139


unbounded sequence, 62

uniform continuity, 201

uniqueness theorem, power series, 342

primitives, 284

upper integral, 264

upper Riemann sum, 259

Wallis’s formula, 293

Weierstrass, K., x

zero derivative theorem, 235

zero of polynomial, 295

zeros localisation theorem, 147

Index 459

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