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MA301 Notes:

Introduction to Proofs and Real Analysis

Richard C. Penney

Purdue University

Contents

Chapter 1. Numbers, proof and ‘all that jazz’. 5

Chapter 2. Inequalities 19

Chapter 3. Rates of Growth 35

Chapter 4. Limits of Sequences 53

Chapter 5. Limit Theorems 75

Chapter 6. Max, Min, Sup, Inf 83

Chapter 7. Positive Term Series 97

Chapter 8. Absolute convergence 123

Chapter 9. Irrational Numbers 133

Chapter 10. Limits of Functions 155

Chapter 11. Continuity 175

Chapter 12. Taylor (Maclaurin) Approximations 179

3

CHAPTER 1

Numbers, proof and ‘all that jazz’.

There is a fundamental difference between mathematics and othersciences. In most sciences, one does experiments to determine laws.A “law” will remain a law, only so long as it is not contradicted byexperimental evidence. Newtonian physics was accepted as valid untilit was contradicted by experiment, resulting in the discovery of thetheory of relativity.

Mathematics, on the other hand, is based on absolute certainty.A mathematician may feel that some mathematical law is true on thebasis of, say, a thousand experiments. He/she will not accept it as true,however, until it is absolutely certain that it can never fail. Achievingthis kind of certainty requires constructing a logical argument showingthe law’s validity–i.e. constructing a proof.

There is, however, a problem with the notion that everything shouldbe proved. If we insist on proving everything then we initially knownothing, and, if we know nothing, how can we prove anything? We haveno place to begin. Clearly, we must have some body of information thatwe know to be true on which to base our proofs.

So what can we assume known and what must be proved? Beforethe time of Euclid, the answer to this question was personal and sub-jective. You were allowed to assume anything that you could bully yourlistener into believing. If I could get you to agree that all integers areeven (which is false) I could use it to prove all sorts of other wonderful(and equally false) things. This often led to many mistakes, so muchso that it was very difficult to know what was true and what was not.

Euclid solved this problem for geometry by stating an explicit col-lection of “self evident” properties (called axioms) which were assumedwithout proof. Furthermore, the axioms are the only properties thatwere to be assumed without proof. All other properties must be provedusing either the axioms or their consequences.

Since the time of Euclid, lists of axioms for many fields of mathe-matics, such as set theory, logic, and numbers have been compiled. Inthese notes, we present one of the standard lists of axioms for the realnumbers, which are the numbers used in calculus. Thus, we are stating

5

6 1. NUMBERS, PROOF AND ‘ALL THAT JAZZ’.

“up front,” those properties that we are allowed to assume withoutproof. As will be seen, the list is rather long and will be covered overseveral sections. We begin with the field axioms, which describe thoseproperties of numbers that do not relate to inequalities.

In principle, every number fact we use should be proved using onlyour axioms. In fact, in these notes, we usually adopt a much looserstandard. As the reader will see, proving everything directly from theaxioms would take so long that we would never progress beyond thissection! It is, however, important that the reader prove a number ofbasic number facts using only the axioms in order to appreciate theirsignificance.

Before stating the number axioms, we state some properties ofequality. These are not number axioms since they apply to thingsother than numbers, e.g. triangles, circles, functions, chairs, etc. Theyare really axioms of logic. We take these properties as given and donot typically indicate their use in our proofs. This is not to say thatthey are unimportant. For example, it is (EQ2) that allows us to writethe distributive law (D1) as

ab+ ac = a(b+ c)

instead of

a(b+ c) = ab+ ac

which justifies factoring a out of ab+ ac.

Properties of Equality

EQ1: (Identity) For all a, a = a.EQ2: (Reflexive) If a = b, then b = a.EQ3: (Transitive) If a = b and b = c, then a = c.

The Field Axioms for the Real NumbersAxioms for Addition

A0: (Existence of Addition) Addition is a well defined processwhich takes pairs of real numbers a and b and produces fromthen one single real number a+ b.

A1: (Associativity) If a, b, and c are real numbers, then

a+ (b+ c) = (a+ b) + c.

A2: (Additive Identity) There is a real number 0 such that forall real numbers a

a+ 0 = a.

1. NUMBERS, PROOF AND ‘ALL THAT JAZZ’. 7

A3: (Additive Inverse) For every real number a there is a realnumber −a such that

a+ (−a) = 0.

A4: (Commutativity) If a and b are any real numbers, then

a+ b = b+ a.

Axioms for Multiplication

M0: (Existence of Multiplication) Multiplication is a well de-fined process which takes pairs of real numbers a and b andproduces from then one single real number ab.

M1: (Associativity) If a, b, and c are any real numbers, then

a(bc) = (ab)c.

M2: (Multiplicative Identity) There is a real number 1 such thatfor all real numbers a

a1 = a.

M3: (Multiplicative Inverse) For every real number a 6= 0, thereis a real number a−1 such that

aa−1 = 1.

M4: (Commutativity) If a and b are any real numbers, then

ab = ba.

Other Laws

D: (Distributive) For all real numbers a, b, and c,

a(b+ c) = ab+ ac.

Z: (Non-triviality) 0 6= 1

Notice that the axioms mention neither subtraction nor division.This is because they may be expressed using addition and multiplica-tion:

Definition 1. Let a and b be numbers. We define

a− b = a+ (−b)a

b= ab−1 (b 6= 0)


Example 1. Solve the following equality for x in a step-by-stepmanner, listing all of the axioms that you use.

3x+ 5 = 12

Solution: We begin by adding −5 to both sides of the equality. Axiom(A3) guarantees the existence of the negative of any number. Axiom(A0) guarantees that adding −5 to both sides preserves the equality.We proceed as follows.

3x+ 5 = 12

(3x+ 5) + (−5) = 12 + (−5) (A0), (A3), arithmetic

3x+ (5 + (−5)) = 7 (A1)

3x+ 0 = 7 (A3)

3x = 7 (A2)

Next we would like to divide both sides by 3. This is the sameas multiplying by 3−1, which exists due to (M3), and multiplicationpreserves the equality due to (M0). Hence:

3−1(3x) = 3−17 (M0), (M3)

(3−1 3)x = 3−17 (M1)

1x = 3−17 (M1), (M3), (M4)

x =7

3(M4), (M4), Definition 1, (M2)

Remark. We needed to use (M4) (commutativity of multiplica-tion) in the third step because (M3) tells us only that 33−1 = 1, andnot that 3−13 = 1. Similarly we needed (M4) in the last step because(M2) tells us only that 3 · 1 = 3. We also needed (M4) to say 3−17 = 7

3

because Definition 1 tells us only that 73−1 = 73.

We can check that our solution is correct. Suppose that x = 73,

then

3x+ 5 = 37

3+ 5

= 3(73−1) + 5

= 3(3−17) + 5

= (33−1)7 + 5

= 1 · 7 + 5

= 7 · 1 + 5

= 7 + 5 = 12 from arithmetic


showing that x = 73does indeed solve the equality. (We leave the

reasons for the reader.)

Remark: Checking that x = 7/3 is a solution was necessary. Ourfirst sequence of equalities proved only that if x is a solution, thenx = 7/3. It did not prove that 7/3 actually is a solution. Also, using“from arithmetic” as a justification for 7 + 5 = 12 is a “cheat”. Weshould prove this from the axioms, which would first require defining 5,7, and 12. This all can be done, but it is tedious. Some of the exercisesdiscuss this further. We will typically allow “from arithmetic” as ajustification for most simple numeric calculations. In fact, we wouldtypically justify 3(7/3)+5 = 12 by saying that it is “from arithmetic.”If we are simplifying an expression involving variables, we will typically(in this chapter) insist that all steps be put in. Thus we not allowedto justify 3−1(3x) = x by “from arithmetic”.

Many familiar number properties do not appear in our axiom list.This is because they can be proved from the axioms. The followinglist contains some of the more common ones. These properties arenot axioms: they are consequences of the axioms. We will leavemost of the proofs to the reader.

Theorem 1. Let a, b, and c be real numbers. Then

C1: (a+ b)c = ac+ bc.C2: 0a = 0C3: −a = (−1)a.C4: −(ab) = (−a)b = a(−b).C5: −(−a) = a.C6: If a 6= 0 6= b, then (ab)−1 = a−1b−1.C7: (a−1)−1 = a.C8: −(a+ b) = (−a) + (−b).

Example 2. Prove (C2).

Solution: We note that

(1)

a = a · 1 (M2)

= a(1 + 0) (A2)

= a · 1 + a · 0 (D)

= a+ a · 0 (M2)

Hencea = a+ 0 · a


We solve this equation for a · 0:

(2)

(−a) + (a+ 0 · a) = −a+ a = 0 (A0), (A3), (A4)

((−a) + a) + 0 · a = 0 (A3), (A1)

0 + 0 · a = 0 (A2), (A3), (A4)

0 · a = 0 (A2), (A4)

Example 3. Find all solutions to the following in a step-by-stepmanner, listing all of the properties that you use.

2x+ 1

x= 3

Solution: Suppose that x satisfies our equality. From the definitionof fractions, this is equivalent with

(2x+ 1)x−1 = 3

Our solution proceeds as follows:

((2x+ 1)x−1)x = 3x (M0)

(2x+ 1)(x−1x) = 3x (M1)

(2x+ 1)1 = 3x (M3), (M4)

2x+ 1 = 3x (M2)

Next, we would like to cancel 3x by adding −3x to both sides of theequation. If we do something to one side of an equality, we must doexactly the same thing to the other side. Thus, we add −3x to theleft of each side:

−(3x) + (2x+ 1) = −(3x) + 3x (A0), (A3)

= 0 (A4), (A3)


We finish the computation:

(−(3x) + 2x) + 1 = 0 (A1)

((−3)x+ 2x) + 1 = 0 (C4)

((−3) + 2)x+ 1 = 0 (C1)

(−1)x+ 1 = 0 from arithmetic

−x+ 1 = 0 (C3)

x+ (−x+ 1) = x+ 0 (A0)

(x+ (−x)) + 1 = x (A1), (A2)

0 + 1 = x (A3)

1 = x (A2), (A4)

Conversely, if x = 1,

2x+ 1

x=

2 · 1− 1

1= 1 from arithmetic

showing that x = 1 does indeed solve the equality.

Clearly, putting in every step can be quite tedious, even in a simplecalculation as in Example 3. Fortunately, it is not essential that youdevelop great skill at doing more complicated examples. The pointhere is only to stress that all of the computations done in elementaryalgebra can all be justified using only the axioms for the real numbers.

The next example demonstrates the necessity for checking the so-lution.

Example 4. Find all solutions to the following equality

x =√2− x

You need not indicate all of your steps.

Solution: We reason as follows. Suppose x satisfies the given inequal-ity. Then

x =√2− x

x2 = 2− x

x2 + x− 2 = 0

(x− 1)(x+ 2) = 0

Hence, our solution appears to be x = 1 and x = −2.If x = 1 √

2− x =√1 = 1 = x.


However, if x = −2√2− x =

√4 = 2 6= x.

Hence the only solution is x = 1.

Remark: The symbol√a, by definition, is the positive square root.

Hence√4 = 2, not ±2.

The axioms only discuss addition of pairs of numbers. We defineaddition of triples by

a+ b+ c = (a+ b) + c

which, from the associative law, is the same as a + (b + c). We definethe sum of four numbers by

a+ b+ c+ d = (a+ b+ c) + d

You will prove in the exercises that

(a+ b+ c) + d = (a+ b) + (c+ d) = a+ (b+ c+ d)

Hence, the sum is the same no matter how we group the terms.Similar comments apply to adding n numbers.

Definition 2. If a1, a2, . . . , an are n numbers, then we define

a1 + a2 + · · ·+ an = (a1 + a2 + · · ·+ an−1) + an.

This is a recursive definition, in that it defines the sum n numbers,assuming that we already know how to sum n− 1 numbers. Thus, forexample, it defines

a+ b+ c+ d+ e = (a+ b+ c+ d) + e

wherea+ b+ c+ d = (a+ b+ c) + d.

We define the product of n numbers similarly:

Definition 3. If a1, a2, . . . , an are n numbers, then we define

a1 a2 . . . an = (a1 a2 . . . an−1) an.

You will prove specific instances of the following theorem in theexercises. By “grouping” we refer to the manner in which parenthesesare put into the sum.

Theorem 2. The sum and product of n numbers are independentof both the order and the grouping of the terms.


Theorem 2 covers all applications of both the commutative andassociative laws in solving equalities.

Example 5. Solve the following equality listing all of your steps.

4x− 9 = x− 2

Solution: Assume that x satisfies the given equality. Then

4x− 9 + (−x) + 9 = x− 2 + (−x) + 9 (A0)

4x+ (−x) + 9 + (−9) = x+ (−x) + 9 + (−2) (Thm. 2)

4x+ (−1)x+ 0 = 0 + 7 (A3), (C3)

(4 + (−1))x = 7 (A2), (C1)

3x = 7

The solution now proceeds exactly as at the end of Example 1 on page 8.

Theorem 2 allows us to prove some number facts that the readerlearned in grade school. We define

2 = 1 + 1

3 = 2 + 1

4 = 3 + 1

5 = 4 + 1

. . .

The set N of natural numbers is, by definition, the numbers ob-tainable by adding 1 to itself any number of times. Thus,

N = {1, 2, 3, 4, . . . }

Example 6. Prove that

(a) 2 · 2 = 4

(b) − 3 + 2 = −1

Solution:


Part (a):

2 · 2 = (1 + 1)(1 + 1) Def. of 2

= 1(1 + 1) + 1(1 + 1) (C1)

= (1 + 1) + (1 + 1) (M2), (M4)

= 2 + (1 + 1) Def. of 2

= (2 + 1) + 1 (A1)

= 3 + 1 Def. of 3

= 4 Def. of 4

Part (b):

−3 + 2 = −(2 + 1) + 2 Def. of 3

= −(1 + 2) + 2 (A4)

= ((−1) + (−2)) + 2 (C8)

= −1 + ((−2) + 2) (A1)

= −1 + 0 (A4), (A3)

= −1 (A2)

Example 7. Prove that for real numbers a and b

(a+ b)2 = a2 + 2ab+ b2

Solution:

(a+ b)2 = (a+ b)(a+ b) = a(a+ b) + b(a+ b) (C1)

= a2 + ab+ ba+ b2 (D)

= a2 + ab+ ab+ b2 (M4)

= a2 + 1(ab) + 1(ab) + b2 (M2), (M4)

= a2 + (1 + 1)ab+ b2 (C1)

= a2 + 2ab+ b2 Def. of 2

Exercises:

(1) Solve the following equalities for x in a step-by-step manner,listing each property you use as in Examples 1 and 2. DO


NOT USE Theorem 2.

(a) 7x− 5 = 19

(b) 7x = 2x+ 3

(c)3x+ 2

x= −1

(2) Re-do Example 5 without using Theorem 2. You may stoponce you have reduced the equation to 3x = 7.

(3) The work below proves the following equality. Copy the proofonto your paper, giving reasons for each step. When using(A1), state what expression is being substituted for a, b, andc. DO NOT USE Theorem 2.

(x+ y)(z + w) = (xz + yz) + (xw + yw)

(x+ y)(z + w) = x(z + w) + y(z + w)

= (xz + xw) + (yz + yw)

= xz + (xw + (yz + yw))

= xz + ((xw + yz) + yw)

= xz + ((yz + xw) + yw)

= xz + (yz + (xw + yw))

= (xz + yz) + (xw + yw)

(4) Let x, y, z be real numbers. In the notes we defined

x+ y + z = (x+ y) + z

There are 6 different orders in which we could sum 3 numbers,all of which, of course, yield the same answer. As an illustra-tion of this, use the axioms and the above definition to provethe following equalities. DO NOT USE Theorem 2.(a) x+ y + z = z + x+ y.(b) x+ y + z = z + y + x

(5) Let x, y, z, and w be real numbers. Use (A1) and the defini-tions to prove the following equalities. In each case state whatexpression is being substituted for a, b, and c in (A1). DONOT USE Theorem 2.

x+ (y + z + w) = (x+ y) + (z + w)

= (x+ y + z) + w


Remark: This exercise proves that the sum of four numbersis independent of how they are grouped.

(6) Let x, y, z, w, and u be real numbers. Use (A1) and the resultof Exercise 5 to prove the following equalities. In each casestate what expression is being substituted for a, b, and c in(A1). DO NOT USE Theorem 2.

x+ (y + z + w + u) = (x+ y) + (z + w + u)

= (x+ y + z) + (w + u)

= (x+ y + z + w) + u

Remark: This exercise proves that the sum of five numbersis independent of how they are grouped.

(7) State and solve analogue of Exercise 5 for multiplication.(8) State and solve analogue of Exercise 6 for multiplication.(9) Why was (M4) needed in the first equation in the sequence of

equations (1)? How about in the fourth equation?(10) Why was (A4) needed in the first equation in the sequence of

equations (2)? How about in the fourth equation?(11) Reason as in Example 6 to prove the following number facts.

(a) 2 + 2 = 4(b) 2 · 3 = 6 Hint: From Example 6 on page 13, 2 · 2 = 4.(c) 3− 4 = −1(d) 3− 5 = −2

(12) (a) Prove property (C1).(b) Let c and d be real numbers. Suppose c+ d = 0. Use the

axioms to prove that c = −d. Hint: Solve c+ d = 0 for cin a step-by-step manner.

(c) Prove (C3). Hint: a+ (−1)a = (1)a+ (−1)a.(d) Prove (C5). Hint: (−a) + a =?.(e) Prove the first equality in (C4). Hint: ab+ (−a)b =?

(13) Let c and d be real numbers.(a) Suppose cd = 1. Use the axioms to prove that d = c−1.

Hint: First explain why c 6= 0. Then solve cd = 1 for d ina step-by-step manner.

(b) Prove (C6). Hint: Simplify (ab)(a−1b−1) and apply theresult proved in (a).

(c) Use the result proved in (a) along with the axioms toprove (C7). Hint: a−1a =?.

(14) If we wish to solve x2 + 3x + 2 = 0, we factor, finding that(x + 1)(x + 2) = 0; hence x = −1 or x = −2. This is based


on the property that if a and b are numbers and ab = 0, theneither a = 0 or b = 0 (or both). Prove this property usingthe axioms. Hint: If a = 0, there is nothing to prove. Henceassume a 6= 0 and solve for b.

(15) Let a, b, c and d be numbers with b and d non-zero. Use theaxioms and properties (C1)-(C8) to prove the following:(a) ab−1 = (ad)(bd)−1

(b) ab−1 + cd−1 = (ad+ bc)(bd)−1

(c) (ab−1)(cd−1) = (ac)(bd)−1

(d) (ab−1)(cd−1)−1 = (ad)(bc)−1

(16) Each of the equalities in the preceding problem expresses afamiliar law of fractions. Write each in fractional form.

CHAPTER 2

Inequalities

In this section we add the axioms describe the behavior of inequal-ities (the order axioms) to the list of axioms begun in Chapter 1.A thorough mastery of this section is essential as analysis is based oninequalities.

Before describing the additional axioms, however, let us first ask,“What, exactly, is an inequality?” Addition is a binary operation; ittakes two numbers a and b and produces a third, a + b. Less than isa binary relation: it takes two numbers a and b and produces eitherthe value ‘true’ or ‘false’. Mathematically, we would say that < is afunction whose domain is the set of all pairs of real numbers and whoserange is the set {true, false}. Thus 2 < 3 produces ‘true’ and 3 < 2produces ‘false’. If we write a < b without explanation, we are assertingthat a < b is true.

Order Axioms

I1: (Trichotomy) For real numbers a and b, one and only one, ofthe following statements must hold:(1) a < b(2) b < a(3) a = b.

I2: (Transitivity) If a < b and b < c, then a < c.I3: (Additivity) If a < b and c is any real number, then a+ c <b+ c.

I4: (Multiplicativity) If a < b and c > 0, then ac < bc.

Important! Throughout this text, in our proofs, we will typically onlygive reasons for material from the current chapter. Hence, in doingproofs with inequalities, we will typically not explicitly indicate the useof field axioms such as associativity, commutativity, etc. Similarly, inChapter 3, we will not typically indicate the use of the order axioms inour proofs.

19

20 2. INEQUALITIES

We define a > b to mean b < a. The statement a ≤ b is a compoundstatement. It is true if either a < b or if a = b. Thus 2 ≤ 3 and 3 ≤ 3are both true statements. The symbol ‘≥’ is defined similarly.

There are many rules for studying inequalities which are derivablefrom the axioms. The reader will be asked to prove many of them inthe exercises. These are not axioms.

Theorem 1. Let a, b, c, and d be real numbers. Then

E1: (Inequalities add) If a < b and c < d, then a+ c < b+ d.E2: (Positive inequalities multiply) If 0 < a < b and 0 < c < d,then 0 < ac < bd.

E3: (Multiplication by negatives reverses inequalities) If a < band c < 0, then ac > bc.

E4: (Inversion reverses inequalities) If 0 < a < b, then1a> 1

b> 0.

E5: (The product to two negatives is positive) If a < 0 and b < 0then ab > 0.

E6: If ab > 0 then either both a and b are positive or they areboth negative.

E7: For all a, a2 ≥ 0.E8: If a ∈ N, a > 0. (Recall that N is the set of natural num-bers.)

Remark: In the following example we use interval notation familiarfrom calculus. Thus, if a and b are real numbers with a < b, then(a, b) is the set of x such that a < x < b. Use of a bracket instead ofa parenthesis indicates that the corresponding end point is included.Hence, for example, [a, b) is the set of x such that a ≤ x < b. Use of∞ as a right end point, or −∞ as a left endpoint, indicates that theinterval has no endpoint on that side. Note, however, that ∞ is NOTA NUMBER! Thus, for example, there is no interval “(−1,∞].”

In general, a set is just a collection of objects. The objects in theset are the elements of the set. We write “x ∈ A” as shorthand for “xis an element of A.” Hence, x ∈ (2, 5] is equivalent with 2 < x ≤ 5.

Example 1. Find the solution set to the following inequality. Doyour work in a step-by-step manner so as to demonstrate the orderaxioms and properties used.

(1) 2x+ 1 < 3x+ 2

2. INEQUALITIES 21

Solution: Suppose that x satisfies the given inequality. Then

2x+ 1 < 3x+ 2

2x < 3x+ 1 (I3)

−x < 1 (I3)

x > −1 (E3)

Hence, if x satisfies inequality (1), then x ∈ (−1,∞).Conversely, suppose that x ∈ (−1,∞). Then

x > −1

−x < 1 (E3)

2x < 3x+ 1 (I3)

2x+ 1 < 3x+ 2 (I3)

Hence, x satisfies the given inequality, showing that the solution set is(−1,∞).

Note that our proof in the second part of the solution was just theproof from the first part “run in reverse.”

Remark: Notice that our solution required two proofs: we first provedthat if x solves (1), then x ∈ (−1,∞). We next proved that if x ∈(−1,∞), then it solves (1). Both parts are necessary because the solu-tion set is a SET. Two sets A and B are equal if they consist of exactlythe same elements–i.e. every element of A is an element of B ANDevery element of B is an element of A. Thus, in principal, proving thevalidity of a solution set to an inequality will always involve two proofs.We first show that if some number solves the inequality, then it belongsto a certain set. Next we show that every element of this set solvesthe inequality. This second proof is often just the first proof reversed.In fact, we often skip the second proof altogether since it is usuallyclear that our steps do reverse. However, in this section we will insistthat you do the reverse proof since we want to stress that both partsare necessary. Sometimes, in fact, the steps do not reverse, in whichcase your solution may require some modification, as in the followingexample.

Example 2. Solve the following inequality and prove your answer.

(2) x ≥√2− x

22 2. INEQUALITIES

Solution: We begin by noting that inequality (2) is meaningless ifx > 2, since then

√2− x in undefined. Thus, we assume that x ≤ 2.

We now reason as follows1:

x ≥√2− x

x2 ≥ 2− x (We squared both sides1)

x2 + x− 2 ≥ 0 (I3)

(x− 1)(x+ 2) ≥ 0

From (E6), this inequality holds if (x−1) and (x+2) both have thesame sign (or are both zero) which holds if either x ≥ 1 (both terms≥ 0) or x ≤ −2 (both terms ≤ 0). Since we have already assumedx ≤ 2, our solution appears to be (−∞,−2] ∪ [1, 2]. 2 This, however,is wrong. For example, if x = −2, inequality (2) says −2 ≥

√4 = 2.

To find our mistake, we attempt to reverse our sequence of inequal-ities:

Suppose that x ∈ (−∞,−2] ∪ [1, 2]. Then x − 1 and x + 2 bothhave the same sign (or are both zero). Hence

(3)

(x− 1)(x+ 2) ≥ 0

x2+x− 2 ≥ 0

x2 ≥ 2− x

We would like to take the square root of both sides of this inequality.We must, however, be careful. For negative x, it is not true that√x2 = x. For example

√

(−2)2 =√4 = 2 6= −2

Rather√x2 = |x|. Thus, “square-rooting” both sides of inequality (3)

produces|x| >

√2− x

which is not equivalent with equation (2).In fact, since square roots can never be negative, only non-negative

x can satisfy inequality (2). Thus, the interval (−∞,−2] cannot be partof our solution set. For x in [1, 2], the final inequality in formula (3)can be square-rooted, showing that our solution set is just [1, 2].

Our first step in solving Example 2 was to square both sides of theinequality (2). Is this allowed? More generally, if we do the same thing

1We discuss the validity of operations such as squaring and square rootinginequalities after the discussion of this example.

2In set theory, A ∪B is the set of elements which belong to A or B or both.

2. INEQUALITIES 23

0

0.2

0.4

0.6

0.8

1

y

-1 -0.5 0.5 1

x

Figure 1

to both sides of an inequality, is the inequality preserved? The answerto this last question is, “NO!” If we multiply both sides by −1, theinequality reverses: 2 < 3 but −2 > −3. If we take the inverse of bothsides the inequality can also reverse: 2 < 3 but 1

2> 1

3. On the other

hand, adding the same number to both sides preserves the inequality:2+1 < 3+1. So when does doing the same thing to both sides preservethe inequality and when does it reverse it?

The answer comes from calculus. Recall that a function y = f(x)is said to be increasing if y gets larger as x gets larger–i.e. x1 < x2

implies f(x1) < f(x2). This means that applying an increasing functionto both sides of an inequality preserves it. The function y = x2 isincreasing for x ≥ 0. (Figure 1) Hence squaring both sides of aninequality will be valid as long as both sides are non-negative. Sincesquare roots are non-negative, inequality (2) is only meaningful if bothsides are non-negative. Hence, squaring both sides was indeed valid.

Similarly, applying a decreasing function to both sides of an in-equality will reverse it. For example, the function y = x2 is decreasingfor x < 0. Hence, squaring inequalities involving negative numbers willreverse the inequality. For example −3 > −4 but 9 < 16.

Example 3. Find an interval I on which a < b implies

ae−a > be−b

Solution: Let f(x) = xe−x. The example asks for an interval I onwhich application of f(x) to both sides of a < b reverses the inequality.This will be true for any interval on which f is decreasing. Fromcalculus, f(x) will be decreasing on any interval where f ′(x) < 0. Wecompute

f ′(x) = e−x − xe−x = (1− x)e−x

24 2. INEQUALITIES

which is negative for x > 1. Hence we may take I = (1,∞). Toillustrate our answer, note that 2 and 3 belong to I and 2 < 3, but2e−2 = .271 which is larger than 3e−3 = .149.

Remark: In a formal proof, whenever you apply a functionto both sides of an inequality, you must justify your work interms of the increasing or decreasing nature of the functionin question.

When solving inequalities, one must be careful when multiplyingboth sides by a quantity which might potentially be negative.


(4)x

x+ 1≥ 1

Solution: If we are not careful, we will not find any such x. Specifi-cally, we might reason as follows:

x

x+ 1≥ 1

x ≥ x+ 1 (I4)

0 ≥ 1 (I3)

Since 0 < 1, there are no such x.But this is wrong. For x = −2,

−2

−2 + 1= 2 ≥ 1

Our mistake lay in the first step of our solution where we multipliedboth sides of the given inequality by x + 1 without reversing the in-equality. This is valid only if x+ 1 is positive.

If x+ 1 < 0, (i.e. x < −1)

x

x+ 1≥ 1

x ≤ x+ 1 (E3)

0 ≤ 1 (I3)

Since 1 = 12, (E7) implies that 0 ≤ 1; hence our inequality yields noadditional restriction on x. Thus, we guess that the inequality is validfor all x < −1. We can prove this by repeating the above inequalities

2. INEQUALITIES 25

in reverse order:

0 ≤ 1 (E7) and 1 = 12

x ≤ x+ 1 (I3)x

x+ 1≥ 1 (E3)

Thus, the solution to our inequality is x < −1. (Note that the inequal-ity is meaningless if x = −1 since division by 0 is not allowed.)

When multiplying (or dividing) an inequality by a quantity thatcan be either positive or negative, it is often necessary to the caseswhere the quantity may be positive separate from the cases where itmay be negative, as in the next example.


(5)x2

(x+ 1)< x+ 1

Solution: We begin by multiplying by x+ 1. Since this quantity maybe positive or negative, we split the argument into the correspondingcases:

Case 1: x > −1.Assume that x satisfies the given inequality and x > −1. Then

x2

(x+ 1)< x+ 1

x2 < (x+ 1)2 (I4)

x2 < x2 + 2x+ 1

0 < 2x+ 1 (I3)

−1 < 2x (I3)

−1

2< x (I4)

Conversely, if x > −12, then x > −1. Hence, we may reverse the above

sequence of inequalities to see that inequality (5) holds. We concludethat in Case 1, our inequality holds if and only if x ∈ (−1

2,∞).

Case 2: x < −1.

26 2. INEQUALITIES

Assume that x satisfies the given inequality and x < −1. Then

x2

(x+ 1)< x+ 1

x2 > (x+ 1)2 (E3)

x2 > x2 + 2x+ 1

0 > 2x+ 1 (I3)

−1 > 2x (I3)

−1

2> x (I4)

Since −12> −1, this is true for all x satisfying the assumptions of

Case 1. Conversely, if x < −1, then x < −12. Hence, we may reverse

the above sequence of inequalities to see that inequality (5) holds. Weconclude that our inequality for all x satisfying the hypotheses of Case2.

Conclusion: Putting Case 1 and 2 together, we see that the solutionset of the inequality is (−∞,−1) ∪ (−1

2,∞).

Many students learn to do proofs by starting with what they wishto prove and reasoning until they obtain a true statement. Withoutfurther qualification, this is not a valid proof technique. For example,if we square both sides of

−2 = 2

we obtain

4 = 4

which is true. This certainly does not prove that −2 = 2.To be valid, a proof must, in principle, begin with known

facts and end with what you want to prove. The bottom linemust be what you want to prove. A proof that begins with what youwant to prove and ends with a true statement is called backwards.

Often, as the following example illustrates, a backwards proof canbe reversed to produce a valid proof. In fact, it is a standard prooftechnique to begin (on a piece of scratch paper) with what you wantto prove, reason until you reach a known statement, and then producethe formal proof by reversing the argument.

In this example, as elsewhere in these notes, we write our“scratch work” in italics to distinguish it from the work wemight hand in if this were a homework assignment.

2. INEQUALITIES 27

Example 6. Let a and b be positive numbers. Prove that

a+ b ≤√2(a2 + b2)1/2.

Scratch Work: Squaring both sides and simplifying produces:

(a+ b)2 ≤ 2(a2 + b2)

a2 + 2ab+ b2 ≤ 2a2 + 2b2

0 ≤ a2 − 2ab+ b2 (Subtract a2 + 2ab+ b2 from both sides.)

0 ≤ (a− b)2

which is true from (E7). Our formal proof will be obtained by reversingthe above steps.

Proof:

0 ≤ (a− b)2 (E7)

0 ≤ a2 − 2ab+ b2

a2 + 2ab+ b2 ≤ 2a2 + 2b2 (I3): Add a2 + 2ab+ b2 to both sides

(a+ b)2 ≤ 2(a2 + b2)

a+ b ≤√2(a2 + b2)1/2 y =

√x is a increasing function

as desired.Note that in our scratch work we squared both sides of the inequal-

ity whereas in the actual proof we took the square root of both sides.

Remark: A backwards argument becomes a valid proof if we arecareful to mention (and check) the reversibility of each step in thebackwards argument. In this section we will insist that all “backwardsproofs” be reversed. Later we will allow you to simply note that thesteps do reverse.

We will often need to study inequalities involving absolute values.By definition,

|x| ={ x x ≥ 0

−x x < 0

Proving theorems about absolute values often requires the consid-eration of several cases which depend on the sign of the quantitiesinvolved.

28 2. INEQUALITIES

Example 7. Prove that for all x and y

|xy| = |x| |y|

Solution: There are 3 cases: (1) x > 0, (2) x = 0 and (3)x < 0. Wewill begin case (1) and leave the remaining cases as exercises.

Case (1) splits into three subcases: (2.1) y > 0, (2.2) y = 0, and(2.3) y < 0.

We will do only (2.3). In this case, |x| = x and |y| = −y so

|x| |y| = −xy

Also, multiplication of the inequality x > 0 by y reverses it, showingthat xy < 0. Hence

|xy| = −xy

Hence, in this case, |xy| = |x| |y|

The kind of absolute value inequalities we need are typically of theform

(6) |x| < a

which is equivalent with

−a < x < a

The inequality

a < |x|is equivalent with

x > a or − x > a

Example 8. Find all x such that

(7) |4− 3x| < .1

Solution: We reason

−.1 < 4− 3x < .1

−4.1 < −3x < −3.9

4.1

3> x >

3.9

3= 1.3

This shows that if x satisfies (7), then x ∈ (1.3, 4.13). We leave it as

an exercise to show that conversely, every x in this interval does satisfy(7).

2. INEQUALITIES 29

A very important absolute value inequality is the triangle inequalitywhich states that for all real numbers x and y

(8) |x+ y| ≤ |x|+ |y|

We leave the case-by-case proof as an exercise.Exercises

In doing inequality problems, you should quote the relevant inequal-ity axioms or properties in use, (I1-I4, E1-E8) but you need not quoteany of the axioms or properties from Chapter 1.

(1) In each part, solve the inequality and prove your answer. Jus-tify any application of functions (such as taking logs, exponen-tials, square roots, etc.) in terms of increasing and decreasing.(a) 2x+ 7 < 3x+ 4 ans. (3,∞)(b) |2y − 8| < .0002 ans. (3.9999, 4.0001)(c) ln(3 − 4t) < 7 ans. ((3 − e7)/4, 3/4). (Note: ln x is

defined only if x > 0.)(d) x

3x+1> 1

3ans. (−∞,−1

3)

(e) ((.5)x − 3)1/3 < 5 ans. ( ln(128)ln(.5)

,∞)

(f) (2x− 1)−1/3 > 2 ans. (12, 916)

(g) 0 ≤ arccos(3x + 1) < π3

ans. (−16, 0] Hint: Graph

y = cos x over [0, π3].

(h) 0 ≤ arcsin(3x+ 1) < π4

ans. [−13,√2−26

) Hint:Graphy = sin x over [0, π

4].

(i) 1arctanx

< .001 ans. (−∞, 0) ∪ (tan(1000),∞)

(j) (.4)2x−1 < 7x ans. ( ln .42 ln .4−ln 7

,∞)

(k) x2−3x

> x−2 ans. (−∞, 0)∪(32,∞). (See Example 5

on page 25.)

(l) x2

x+1< x+ 2. (See Example 5 on page 25.)

In the following 7 exercises, you are asked to proveproperties (E1)-(E8). In these exercises, you may onlyuse one of the properties (E1)-(E8) if it was proved in one ofthe preceding exercises. Otherwise, you should use onlythe axioms.

(2) Prove property (E1). Hint: Try adding c onto both sides ofa < b and b onto both sides of c < d.

(3) Prove property (E2). Show by example that (E2) can failif c < 0. Hint: Reason as in the preceding exercise, usingmultiplication instead of addition.

30 2. INEQUALITIES

(4) Can we subtract inequalities? I.e. if a < b and c < d does itfollow that a− c < b− d? If so, prove it. If not, find a counterexample.

(5) Can we divide positive inequalities? I.e. if 0 < a < b and0 < c < d does it follow that a/c < b/d? If so, prove it. If not,find a counter example.

(6) Suppose that a < b.(a) Prove that −b < −a. Hint: Begin by adding −b onto

both sides of a < b.(b) Prove (E3). Hint: From (a), −c > 0.(c) Prove (E5).(d) Prove (E7). Hint: Consider three cases: a > 0, a = 0,

and a < 0. Which axiom allows you to break this up intothese three cases?

(e) Use (E7) to prove that 1 > 0. Then prove that 2, 3, and4 are all positive.

(7) (a) Prove that if b > 0, then b−1 > 0. Hint: Either b−1 > 0or b−1 = 0 or b−1 < 0. If you can show that the latter twoconditions are impossible, then the first must hold.

(b) Prove (E4). Hint: Begin by multiplying both sides ofa < b by b−1.

In the remaining exercises you may use both theaxioms and properties (E1)-(E8).

(8) Prove that the average of two numbers lies between them I.e.if a < b then a < a+b

2< b. Hint: Do each inequality separately.

(9) Let a and b be positive numbers. Prove that

(i)√ab ≤ a+ b

2

(ii)√a2 + b2 ≤ a+ b

(iii)a

b+

b

a≥ 2

(iv) If 0 < a < b then a2 < ab < b2

(10) Do the remaining subcases from case (1) of Example 6 in thetext.

(11) Do Case (2) from Example 6.(12) Do Case (3) from Example 6.(13) Prove that for all x and y, y 6= 0,

∣

∣

x

y

∣

∣ =|x||y|

34 2. INEQUALITIES

Split your proof into three cases as in Example 6. Your in-structor may assing you to do only certain of these cases.

(14) The purpose of this exercise is to understand why the triangleinequality holds.(a) Find a non-zero value of y such that |2+ y| = 2+ |y| and

a value such that |2 + y| < 2 + |y|.(b) Describe (i) the set of all y such that |2+ y| = 2+ |y| and

(ii) the set of all y such that |2 + y| < 2 + |y|.(c) Suppose x > 0. For which y is |x+y| = |x|+|y|? |x+y| <

|x| + |y|? Note that from Axiom I1, this accounts for ally.

(d) Answer the questions from (c) under the assumption thatx = 0. Next answer them under the assumption thatx < 0.

Remark This all amounts, more or less, to a proof of inequal-ity (7). The reason that it is only ”more or less” is that youweren’t ask you to prove the answers to (c) and (d).

(15) Graph the set of points (x, y) defined by

(a) |x| = |y|(b) |x|+ |y| = 1

(c) |xy| = 2

(d) |x| − |y| = 2

(16) In parts (a)-(g), find all intervals I for which the stated in-equality holds for all elements x and y of I, x < y. Illustrateyour answer with a specific value of x and y. Prove your an-swer using calculus.(a) x3 < y3

(b) x−1/3 < y−1/3

(c) x4 < y4

(d) ln x < ln y

(e) ye−y2 < xe−x2

(f) y3 + 6y2 + 9y < x3 + 6x2 + 9x(g) x

x2+1< y

y2+1

CHAPTER 3

Rates of Growth

Inequalities are very useful in comparing rates of growth of func-tions.

Definition 1. If f and g are two functions, we say that f(x)dominates g(x) for large x if there is a value N such that

f(x) > g(x)

for all x > N .

f(x)

g(x)

N

Figure 1. f(x) dominates g(x) for large x

The function f(x) = Cxa where C and a are both positive, exhibitspower growth in that it grows like a power of x. The followingexample illustrates that the larger the power, the faster the growth,regardless of the value of C.

Example 1. Prove that (.01)x4 dominates x2 for large x.

Scratch work: According to Definition 1, we need to find a value Nsuch that

(.01)x4 > x2

for all x > N . Since we are only considering large x we assume x > 0.Division by (.01)x2 followed by taking the square root produces

x2 > 100

x > 10

35

36 3. RATES OF GROWTH

Thus, we can take N = 10.For our formal solution, we state the value of N and reverse the

above steps to prove that the stated value works.

Solution: LetN = 10. Assume x > 10. We may square this inequalitybecause y = x2 is an increasing function for x > 0, obtaining

x2 > 100

We may multiply by (.01)x2 since this term is positive, showing that

(.01)x4 > x2

for x > 10, fulfilling the requirements of Definition 1.

Example 2. For each pair of functions below (i) determine whichis the dominant function, (ii) prove your answer by finding a numberN fulfilling the requirements of Definition 1. Prove that your value ofn really works.

(1) 3x2 + 5x− 3, x3

(2) x5, x6 + 3x2 − 2x+ 1

Solution

Scratch work(1):For polynomials, the highest power of x determines the rate of

growth. Hence, in (1), we expect x3 to dominate. To prove our an-swer we must find an N such that

3x2 + 5x− 3 < x3

for all x > N .We see no way of solving this inequality. Instead we seek a simple

quantity “?” and a value N for which

3x2 + 5x− 3 <? < x3

for all x > N . In other words, we wish to replace 3x2 + 5x − 3 by alarger, simpler quantity, which is still dominated by x3. We begin bydropping negative terms:

3x2 + 5x− 3 < 3x2 + 5x.

Since1

2x3 +

1

2x3 = x3

3. RATES OF GROWTH 37

this latter quantity will be less than x3 provided

(1) 3x2 <1

2x3 and 5x <

1

2x3.

Solving these inequalities yields

(2) 6 < x and√10 < x.

Since√10 < 6, both of (1) hold for x > 6.

Proof (1): Let N = 6 and assume x > N . Then (2) both hold. Wemultiply the first inequality in (2) by x2/2 obtaining

3x2 <1

2x3.

Similarly squaring the second inequality in (2) and multiplying by x/2produces

10 < x2

5x <x3

2

Hence (1) both hold. It follows as in the scratch work that

3x2 + 5x− 3 < x3

for all x > 6, showing that x3 dominates 3x2 + 5x− 3.

Scratch Work (2):Now we expect x6 + 3x2 − 2x + 1 to dominate. To prove this we

must find an N such that for x > N ,

x5 < x6 + 3x2 − 2x+ 1.

Again we seek a quantity “?” satisfying

x5 <? < x6 + 3x2 − 2x+ 1,

meaning that we want to make x6 + 3x2 − 2x + 1 smaller. Hence, wedrop positive quantities producing

x5 < x6 − 2x < x6 + 3x2 − 2x+ 1.

We cannot drop the −2x term since this would make x6−2x larger.We can, however, replace 2x by a larger quantity, such as 1

2x6, since


subtracting a larger quantity produces a smaller result.1 We note that

(3) 2x <1

2x6

for 4 < x5–i.e. for 41/5 < x. For such x

x6 − 2x > x6 − 1

2x6 =

1

2x6.

Finally

(4)1

2x6 > x5

provided x > 2. Since 2 > 41/5 we choose N = 2.

Proof (2): Let N = 2 and assume that x > N . Then (3) and (4) bothhold. It follows as in the scratch work that

x5 < x6 + 3x2 − 2x+ 1

for all x > 2, showing that x6 + 3x2 − 2x+ 1 dominates x5.

In studying rates of growth, it is useful to note the following propo-sition which is proved in the exercises.

Proposition 1. Suppose that 0 < a < b. Then

xb > xa

for x > 1.

One of the goals in the study of rates of growth is to get informationhow fast a function grows by comparing it to another function whosegrowth we understand.

Definition 2. Let f and g be functions. We say that f grows likea multiple of g if there is are constants C > 0, D > 0 and N such thatfor all x > N ,

Cg(x) < f(x) < Dg(x).

(See Figure 2)

Example 3. Show that f(x) = x3 − 3x2 + 5x − 1 grows like amultiple of x3.

1We choose a multiple of x6 in order to obtain a simple expression. We used1

2x6 in order to obtain a positive result; no negative multiple of x6 would dominate

x5.


N

f(x)

Cg(x)

Dg(x)

Figure 2. f grows like a multiple of g

Solution: According to Definition 2, we need to find positive constantsC, D, and N such that

Cx3 < x3 − 3x2 + 5x− 1 < Dx3

for all x > N .Deleting negative terms makes a sum larger. Thus, for x > 1,

(5) x3 − 3x2 + 5x− 1 < x3 + 5x < x3 + 5x3 = 6x3

Hence, for the right hand inequality, we may take D = 6 and N = 1.Finding C is harder. Deleting positive terms makes a sum smaller.

Hence, for x > 0,

(6) x3 − 3x2 + 5x− 1 > x3 − 3x2 − 1

Subtracting more also makes a quantity smaller. Hence, for x > 1,

x3 − 3x2 − 1 > x3 − 3x3 − x3 = −3x3

Unfortunately, we cannot use −3 as C since a positive value is required.To avoid this problem, we replace 3x2 and 1 by sufficiently multiples ofx3.

We choose N so that both of the following hold for x > N :

(7)3x2 <

1

3x3

1 <1

3x3

The first inequality is valid for x > 9 and the second for x > 31/3 ≈ 1.44.For x > 9, both are valid and

(8) x3 − 3x2 − 1 > x3 − 1

3x3 − 1

3x3 =

1

3x3

Hence we may choose N = 9 and C = 1/3. This value of N also worksfor the D inequality with D = 6 because 9 > 1.


It is a general property of ratios of positive numbers that the largerthe denominator, the smaller the number. Thus, for example

3

9<

3

5

because 5 < 9. This principle is the basis for the next two examples.

Example 4. Find a value of m such that the following functiongrows like a multiple of xm. Prove your answer.

f(x) =3x5

x3 − 3x2 + 5x− 1

Solution: We need to find C > 0, D > 0, m, and N such that

Cxm <3x5

x3 − 3x2 + 5x− 1< Dxm.

for all x > N .Finding m is easy. The numerator is a multiple of x5 while, from

Example 2, the denominator grows like a multiple of x3. Hence thewhole fraction should grow like a multiple of x5/x3 = x2, implyingm = 2. More precisely, from the work done in Example 2, we knowthat for x > 9

(9)1

3x3 < x3 − 3x2 + 5x− 1 < 6x3

We may invert this inequality since, for x > 9, x3 > 0. We find

3

x3>

1

x3 − 3x2 + 5x− 1>

1

6x3

We may multiply by 3x5 since, again, for x > 9, x5 > 0. We find

9x2 >3x5

x3 − 3x2 + 5x− 1>

x2

2

Hence, we may choose m = 2, C = 1/2, D = 9 and N = 9.

Remark: The principle that increasing the size of the denominatormakes the fraction smaller is only valid if both the numerator anddenominator are positive. For example

5

−7<

5

3

despite the fact that −7 < 3. Thus, the positivity of the C and Dfound in Example 2 was crucial in Example 3.


Example 5. Find a value of m such that the following functiongrows like a multiple of xm. Prove your answer.

f(x) =3x2 + 1

x3 − 3x2 + 5x− 1

Solution: We need to find C > 0, D > 0, m, and N such that

Cxm <3x2 + 1

x3 − 3x2 + 5x− 1< Dxm.

for all x > N .

Solution: Since the denominator grows like a multiple of x3 and thenumerator like a multiple of x2, the whole fraction should grow likea multiple of x2/x3 suggesting that m = −1. To prove this we findthe C and D’s for the numerator and denominator. Specifically, frominequality (9), for x > 9,

1

3x3 < x3 − 3x2 + 5x− 1 < 6x3

which, upon inversion, becomes.

(10)3

x3>

1

x3 − 3x2 + 5x− 1>

1

6x3

For the numerator we find that for x > 1

(11) 4x2 > 3x2 + 1 > 3x2.

Multiplying inequalities (10) and (11) shows that, for x > 9,

12x2

x3>

3x2 + 1

x3 − 3x2 + 5x− 1>

3x2

6x3

(The multiplication is allowed since, for x > 9, both inequalities involveonly positive numbers.) Hence C = 1

2, D = 12, N = 9 works.

Logarithmic growth is particularly slow. Recall that

ln x =

∫ x

1

1

tdt

Thus, ln x is the area under the curve y = 1/x between 1 and x.Comparison of areas as in Figure 3 shows that for x > 1,

(12) ln x < x

The same inequality holds for 1 ≥ x > 0 since in this range, ln x ≤0. Thus ln x grows slower than x.

We summarize this discussion in the following important theorem:


��

��1

Area of big rectangle=x

Shaded area=ln x

x1

(1.1)

y=1/x

Figure 3. ln x < x

Theorem 1. For all x > 0, ln x ≤ x.

Actually, ax dominates ln x for any a > 0.

Proposition 2. Let 0 < a. Then

(13) ln x < ax

for x > 4/a2.

Proof From Theorem 1, for all b > 0,

ln(bx) ≤ bx

ln b+ ln x ≤ bx

ln x ≤ bx− ln b

= bx+ ln1

b

≤ bx+1

b

If x > 1b2, then bx > 1

bso

ln x ≤ bx+ bx = 2bx.

Our result follows by letting b = a2. �

It is a consequence of Proposition 2 that logarithmic growth is slowerthan power growth, regardless of the power, as long as the power ispositive, as the following example demonstrates.

Example 6. Find a value of N such that ln x < x1/2

3for all x > N .


Scratch Work: To make the powers of x on both sides of the inequalitythe same, we multiply by 1

2and use properties of the logarithm:

ln x <x1/2

31

2ln x <

x1/2

6

ln(

x1

2

)

<x1/2

6

According to inequality (13) on page 42

ln y <y

6for y >

4

(1/6)2= 144.

Hence, letting y = x1/2,

ln(

x1

2

)

<x1/2

6for x1/2 > 144

which is the same as x > (144)2. Hence, N = (144)2 works.

Proof: Let N = (144)2. Then, if x > N , x1/2 > 144 so according toProposition 2 on page 42,

ln(

x1

2

)

<x1/2

6

which, from the scratch work, is equivalent with the desired inequality.

Remark: The value N = (144)2 = 20736 found above is by no meansthe “best possible” answer. A graphing calculator indicates that in fact

ln x < x1/2

3is true for all x > 289. This shows that the value of N given

in Proposition 2 can be vastly larger than necessary. This, however,does not matter in determining which is the dominant function.

Exponential growth is faster than power growth, as the next exam-ple shows.

Example 7. Prove that 2x dominates x3.

Scratch work:We must show that there is an N > 0 such that

(14) x3 < 2x


for all x > N . Since ln x is an increasing function, our inequality isequivalent with

ln x3 < ln 2x

3 ln x < x ln 2

ln x <ln 2

3x

which, according to Proposition 2, with a = ln 23, is true for

x > 36/(ln 2)2 ≈ 74.93.

We cannot, however, use 74.93 as the value of N since this valueis only an approximation. If the actual value of 36/(ln 2)2 is slightlygreater than 74.93, then it is conceivable that inequality (14) might failfor some x > 74.93. However, presuming that our calculator has atleast 2 decimal accuracy, we can be certain that, say, 80 > 36/(ln 2)2.Hence, we may use N = 80.

Solution: Assume that x > 80. Then

36/(ln 2)2 < x

Thus, from Proposition 2,

ln x <ln 2

3x

3 ln x < x ln 2

ln x3 < ln 2x

Since ex is an increasing function we may continue this sequenceof inequalities as follows:

elnx3

< eln 2x

x3 < 2x

as desired.

Remarks: Note that the formal solution required the fact that theexponential function is increasing; not that the logarithm function isincreasing, as our “scratch work” had suggested. Typically, if we ap-ply a particular function to both sides of an inequality in the “scratchwork,” then in the formal solution we will apply the corresponding in-verse function to both sides of an inequality. Fortunately, it is a generalprincipal that the inverse of an increasing function is increasing. (See


Exercise 13 below.) Similar comments apply for decreasing functions.(See Exercise 14 below.)

Example 7 also demonstrates that in rounding computed values ofN , we should never “round down.” If a particular inequality is knownto true for, say, all x > 4.01, then it might not be valid for all x > 4.It will, however, hold for all x > 5.

Another general principle is that if f(x) is dominated by g(x) whichis dominated by h(x), then f(x) is dominated by h(x). (See Exer-cise 11.) We apply this principle in the next example.

Example 8. Find a value of N such that 7x2 < 2x for all x > N .

Scratch work: If we attempt to solve this using the same idea as inExample 6, we take the log of the inequality obtaining

ln(7x2) < ln 2x

ln 7 + ln x2 < x ln 2

ln 7 + 2 ln x < x ln 2

Unfortunately, this cannot be transformed into something to which Propo-sition 2 applies.

Solution: We reason that 7x2 is dominated by x3 which is dominatedby 2x. Specifically, 7x2 < x3 for x > 7 and, from the solution toExample 7, x3 < 2x for x > 80. Hence, 7x2 < 2x for x > 80.

Example 9. Find an a > 0 such that the following function growslike a multiple of ax. Prove your answer.

2x + x ln x+ x3

4x + x33x + 1

Solution The fastest growing term in the numerator and denominatorare, respectively, 2x and 4x. Hence, we expect that the fraction shouldgrow like 2x/4x = (1/2)x, suggesting that we may use a = 1/2. Toprove our answer, we must find constants C > 0, D > 0 and N suchthat

(15) C2−x <2x + x ln x+ x3

4x + x33x + 1< D2−x

for all x > N .


We first consider the denominator. Since x33x and 1 should growmore slowly than 4x, there should exist an N such that for x > N ,

1 < 4x

x33x < 4x

The first inequality is true for all x > 0. Dividing by 3x and usingthe fact that ln x is an increasing function, we see that the secondinequality is equivalent with

x3 <

(

4

3

)x

3 ln x < x ln4

3

ln x < (1

3ln

4

3)x

which, from Proposition 2, holds for x > 4/((13ln 4

3)2 ≈ 434.9876.

Hence, for, say, x > 450

4x < 4x + x33x + 1 < 4x + 4x + 4x = 3(4x)

Inverting, we see

(16)1

3(4x)<

1

4x + x33x + 1<

1

4x

For the numerator, we note that for x > 1,

2x + x ln x+ x3 < 2x + x · x+ x3

= 2x + x2 + x3

< 2x + 2x3

where we used ln x < x in the first line.On the other hand, from Example 7, x3 < 2x for x > 80. Hence,

for such x,

(17) 2x + x ln x+ x3 < 3(2x).

Inequalities (16) and (17) will both hold for x > 450. Multiplyingthese inequalities shows that inequality (15) holds for all x > 450 withC = 1/3 and D = 3, finishing the proof.

The rate of growth of the sum of two functions is determined by thefastest growing term in the sum. The situation is more complicatedin the case of products. For example, x33x dominates either 3x or x3.However, since exponential growth is faster than power growth, x33x

grows slower than ax for any a > 3.


Example 10. Prove that x33x is dominated by (3.1)x.

Solution We must show that there is an N such that for all x > N

(18) x33x < (3.1)x

However, since ln x is an increasing function, this equation is equiv-alent with:

x3 <

(

3.1

3

)x

3 ln x < x(ln 3.1− ln 3)

which, from Proposition 2, holds for

x >36

(ln 3− ln 3.1)2≈ 33482.99.

Thus, we can be certain that the stated inequality holds for, say, x >40, 000.

The next example is based on this idea.

Example 11. Find positive numbers C, D, a, b, and N such that

Cax <2x + x ln x+ x3

x33x + 1< Dbx

for all x > N .

Solution From the solution to Example 10, equation (18) holds forx > 40, 000. Hence, for such x

3x < x33x + 1 < (3.1)x + (3.1)x = 2(3.1)x.

Inverting:1

2(3.1)x<

1

x33x + 1<

1

3x.

Multiplication by inequality (17) (which holds for x > 450) yields theestimate

2x

2(3.1)x<

2x + x ln x+ x3

x33x + 1< 3

2x

(3x)

1

2

(

2

3.1

)x

<2x + x ln x+ x3

x33x + 1< 3

(

2

3

)x

which is true for x > 40, 000. Thus, we may choose a = 2/3.1, b = 2/3,C = 1/2, D = 3, and N = 40, 000. In place of 3.1 we could, of courseuse any number strictly greater than 3, although different choices resultin different values of N . We cannot use 3.


Exercises

(1) Suppose that 0 < a < b.(a) Prove that if x > 1, then xa > 1. Hint: Use calculus to

prove that, y = xa is increasing on (0,∞).(b) Prove that if x > 1, then xb > xa. Hint: From (a),

xb−a > 1.(2) For each pair of functions below (i) determine which is the

dominant function, (ii) prove your answer by finding a numberN fulfilling the requirements of Definition 1 on page 35. Provethat your value of n really works.

(a) 3√x, x1/3

(b) 8x2, x3

(c) 8x2, x3 + 1(d) 8x2 − x, x3

(e) 8x2 + x, x3 Hint: x < x2 for x > 1(f) x5, x3 + 2x2 − x+ 1(g) x5 − 7x2 + 2x+ 1, x3 + x+ 1(h) x3 + 3x2 + 1, x3 + x+ 1(i) x3 + 3x2 + 1, x3 + 2x2 + x+ 7(j) 3

√x, x− x1/3

(k) x5 − 7x2 + 2x+ 1, x4

(l) x5 − 7x2 + 2x+ 1, (1.1)x5

(m) x5 − 7x2 + 2x+ 1, (.99)x5

(3) For each function f(x), find a value of m such that f(x) growslike a multiple of g(x) = xm. Prove your answer by findingconstants C, D and N fulfilling the requirements of Defini-tion 2 on page 38.(a) f(x) = x4 + 3x2 + 1(b) f(x) = 5x2 − 3x+ 7

(c) f(x) = 5x2−3x+7x4+3x2+1

(d) f(x) = x4+3x2+15x2−3x+7

(e) f(x) = x4 + 3x2 − 5x+ 1(f) f(x) = x2 − 3x−√

x+ 7

(g) f(x) = x2−3x−√x+7

x4+3x2−5x+1

(h) f(x) = x4+3x2−5x+1x2−3x−√

x+7

(i) f(x) = x5−4x2−5x5+14x−3

(j) f(x) = x5+14x−3x5−4x2−5


(4) In each part, find constants C > 0, D > 0 and N such thestated inequalities are valid for all x > N . State in words whateach inequality tells you about rates of growth.(a) C

(

23

)x< 2x+14x−3

3x−4x2−5< D

(

23

)x

(b) C(

1.53

)x< 2xx2+14x−3

3x−4x2−5< D

(

2.53

)x

(c) C(

54

)x< 5x−3x−√

x+74x+3x2−5x+1

< D(

54

)x

(d) C(

4.94

)x< 5xx3−3x−√

x+74x+3x2−5x+1

< D(

5.14

)x

(5) Find a value of N such that for x > N ,(a) x2 − 3x+ 27 < x2

(b) x2 − 3x+ 27 > .9x2

(c) Explain why there is no N such that x2 − 3x + 27 > x2

for all x > N .(6) Find a value of N such that for x > N ,

(a) x3 + 3x2 − 27x+ 1 < (1.01)x3

(b) x3 + 3x2 − 27x+ 1 > x3

(c) Explain why there is noN such that x3+3x2−27x+1 < x3

for all x > N .(7) For each pair of functions below (i) determine which is the

dominant function, (ii) prove your answer by finding a numberN fulfilling the requirements of Definition 1 on page 35. Provethat your value of n really works.(a) ln x, x1/4

(b) 12

√x, ln x

(c) x, (ln x)1

3

(d)√x, (ln x)5

(e) x4 ln x, x4.25

(f) 1, 3x3 lnxx3−3x2+5x−1

(g) 3x, x7

(h)√x, 1.5x

(i) (1.1)x, 5x2

(j) x1/5, (ln x)3

(k) 3x, 5x3 ln x(l) (ln x)10, x1/10

(8) (a) Prove that for n > 3, n! > 29(3n). Hint 3! = 3 · 2 = (33)2

9.

5! = 5 ·4 ·3! > 3 ·3 ·3! = (35)29. Repeat the same argument

in the general case.(b) Find an N such that 2

93n > 2n for n > N . How does it

follow that n! dominates 2n?


(c) Use the reasoning from (a) to find an N such that n! >3324n for n > N . Use this to find an N such that n! > 3n

for all n > N .(9) For each pair of functions below (i) determine which is the

dominant function, (ii) prove your answer by finding a numberN fulfilling the requirements of Definition 1 on page 35. Provethat your value of n really works.

(a) x2, ex

(b) x2 ln x, (1.5)x

(c) 3x, x32x

(d)x3

2x,

1

(1.5)x

(e)1

(1.5)x,

x2 ln x

2x

(f)3n

n!,

1

2n

(g)3n + n2 + 1

2n − 5,

3n

2n

(h)n!

100n, (1.1)n

(10) Are there positive constants C and N such that

Cx < ln x

for all x > N . If so, find them. If not, prove that suchconstants don’t exist.

(11) Suppose that f(x), g(x), and h(x) are functions such that f(x)dominates g(x) and g(x) dominates h(x). Use Definition 1 onpage 35 to prove that then f(x) dominates h(x).

(12) Let f be an invertible function and let g = f−1 be the inversefunction. (Hence, g(f(x)) = x for all x in the domain of f .)Prove that f(g(y)) = y for all y in the range of f . Hint: Sincey is in the range of f , y = f(x) for some x.

(13) Let f be an increasing, invertible function and let g = f−1.Prove that g is also increasing. Hint Suppose that there arenumbers a < b in the domain of g such that g(a) ≥ g(b). Whatdo you know about the effect of applying f to inequalities?

(14) State (carefully) a version of Exercise 13 that applies to de-creasing functions. Then solve your exercise.

(15) In the notes we used integrals to prove that for x > 0, ln x < x.


(a) Prove this by using calculus to find the minimum for thefunction f(x) = x − ln x. Use the second derivative testto prove that the value you find really is a minimum.

(b) Prove that ln x ≤ (1/e)x for all x > 0. Why is thisinequality false with “≤” replaced by “<”? Hint: Usethe same idea as you used in (a).

(c) Use the result from (b) to prove that ln x <√x for all

x > 0. Hint: Apply (b) with x replaced by x1/2.(d) Use the result from (c) to prove that Proposition (2) on

page 42 holds with 1/a2 in place of 4/a2.

CHAPTER 4

Limits of Sequences

In this section we study limits of sequences. As a preliminary def-inition, we might define a sequence to be a function whose domainconsists of the set N natural numbers. Thus, when we refer to “thesequence”

an = n/(n2 − 3)

we are implicitly stating that we will consider this expression only forn = 1, 2, 3, . . . . Hence, the preceding equality does define a sequence,despite the fact that the denominator is zero if n =

√3.

At times it is more convenient to consider the sequence as beginningwith values of n other than 1. For example, n = 0 is a common choice.Or if we were to study the following expression, we might want to beginwith n = 6 to avoid dividing by 0.

an =n

n− 5

At times, one even might want to begin with a negative number. Thus,we modify our original definition as follows. Recall that the set Z ofintegers is the set consisting of 0 together with ±n where n ∈ N. Hence

Z = {0,±1,±2,±3, . . . }.Definition 1. A sequence is a function whose domain consists of

all n ∈ Z, n ≥ a.

We will, however, adopt the convention that unless otherwise indi-cated, the domain of all sequences is considered to be the set of naturalnumbers.

We will discuss limits in terms of approximation theory. In science,when we make a statement such as “the value A is 4.7 ± .01,” wemean that although we do not know the value of A exactly, we arecertain that A is between 4.7 − .01 and 4.7 + .01. When discussingapproximations, we will always assume that the error term ispositive since negative values are accounted for by the “±.” Thus, wewill never say A = 4.7± (−.01) as this is the same as A = 4.7± .01

In general, for ǫ > 0, we define

A = B ± ǫ

53

54 4. LIMITS OF SEQUENCES

to mean1

B − ǫ < A < B + ǫ

which is the same as either of the following two equivalent statements

−ǫ < A−B < ǫ

|A−B| < ǫ.

To relate this to limits, consider the sequence

an =n

n+ 1

Below are a few values of an:

n 6 11 16 21 26 31 36 41 100 1000an .857 .917 .941 .955 .963 .969 .973 .976 .990 .9990

It appears that for large values of n, an is approximately equal to1. For example,

|a16 − 1| = |.917− 1| = .083 < .1

Hence

a16 = 1± .1

Similarly,|a100 − 1| = |.990− 1| = .01 < .02

|a1000 − 1| = |.9990− 1| = .001 < .002

so a100 = 1± .02 and a1000 = 1± .002.The next example illustrates that an is a very close approximation

to 1 for all sufficiently large values of n.

Example 1. Find a value of N such thatn

n+ 1= 1± 10−5

for n > N . Prove your answer.

Scratch work: We want

| n

n+ 1− 1| < 10−5

| 1

n+ 1| < 10−5

1Typically, A = B ± ǫ means B − ǫ ≤ A ≤ B + ǫ. Our convention excludes thepossibility of equality.

4. LIMITS OF SEQUENCES 55

Since the term inside the absolute value is positive, this is equivalentwith

1

n+ 1< 10−5

n+ 1 > 105

n > 105 − 1 = 99, 999

For our formal solution, we state a value of N and prove that itworks by reversing the above sequence of arguments.

Solution: Let N = 99, 999 and assume n > N . Then

n > 105 − 1

n+ 1 > 105

1

n+ 1< 10−5

| 1

n+ 1| < 10−5

| n

n+ 1− 1| < 10−5

showing thatn

n+ 1= 1± 10−5

as desired.

What if we wantn

n+ 1= 1± 10−10?

The argument from Example 1 shows that this will hold for all n >1010 − 1. More generally, for any ǫ > 0,

n

n+ 1= 1± ǫ

holds for n > 1e− 1.

This shows that n/(n+ 1) will approximate 1 to any desired degreeof accuracy for all sufficiently large n. How large n must be will, ofcourse, depend on the accuracy desired.

Example 1 illustrates a general principle: if an is a sequence suchthat limn→∞ an = L, then an will approximate L as closely as desiredfor all sufficiently large n. In fact, an informal definition of limit mightread, “ limn→∞ an = L provided an gets arbitrarily close to L as ngets large.” This means exactly that an approximates L as closely


as desired. Thus, we adopt the following statement as our “official”definition of limit for sequences:

Definition 2. Let an be some sequence of numbers and let L be anumber. We say that limn→∞ an = L provided that for every numberǫ > 0, there is a number N such that

(1) |an − L| < ǫ

for all n > N .

Remark Occasionally students feel that it is not really correct to saythat

limn→∞

n

n+ 1= 1

because n/(n + 1) never actually equals 1. They would prefer to saythat the limit is “very, very close to” 1. This is not correct. The limitrefers to the number that is being approximated; not the numbersdoing the approximation. There is one, and only one, number that theterms n/(n + 1) approximate better and better as n gets larger andlarger, namely 1. Actually, the fact that a convergent sequence canapproximate only one number is an important property of limits. Wewill leave the proof of this fact as an exercise (Exercise 31).

Proposition 1. A sequence can have only one limit: iflimn→∞ an = L and limn→∞ an = M , then L = M .

In a number of exercises, you will be asked to prove statements suchas limn→∞ an = L for some specific sequence an and some number L“using ǫ” i.e. directly from the definition. The formal solution to sucha problem will typically involve the following steps:

(1) Assume that a value of ǫ > 0 is given.(2) State an appropriate value of N . It will be given by some

expression involving ǫ.(3) Prove that the stated value of N really works i.e. assume that

n > N and prove that |an − L| < ǫ.

Example 2. Find

limn→∞

n4

n5 + 7

Find a number N such that n4

n5+7approximates the limit with error less

than .001 for all n > N . Prove, using ǫ, that your limit is correct.


Scratch work: Since n5+7 grows like n5 our fraction grows (actuallydecays) like n4/n5 = 1/n. Hence, we guess that the limit is 0.

To prove our guess we we must show that for any given ǫ > 0,n4

n5+7= 0± ǫ for all sufficiently large n. i.e.

(2)

∣

∣

n4

n5 + 7− 0

∣

∣ < ǫ

n4

n5 + 7< ǫ

Butn4

n5 + 7<

n4

n5=

1

n

Thus, formula (2) will be valid if 1/n < ǫ, i.e. n > 1/ǫ. In particular,we obtain ±.001 accuracy for n > 1000.

Formal Solution: For n ∈ N

n5 < n5 + 7

1

n5 + 7<

1

n5

n4

n5 + 7<

n4

n5=

1

n

Now, let ǫ > 0 be given and let N = 1/ǫ. Assume that n > N Then

n >1

ǫ1

n< ǫ

n4

n5 + 7<

1

n< ǫ

∣

∣

n4

n5 + 7− 0

∣

∣ < ǫ

This fulfills the requirements for the definition of the limit.The first part of the problem is solved by letting ǫ = .001, in which

case N = 1000.

Remark: In our solution to Example 2, we used the observation thatfor n > 1, 000

n4

n5 + 7<

1

n< .001


In general, if we are trying to show that some sequence an can be madeless than ǫ, we often try to find some relatively simple quantity “?”(which will depend on n) such that

an <? < ǫ

for all sufficiently large n.

Example 3. Find the following limit and prove your answer usingǫ.

limn→∞

2n3

n3 + 5n+ 1

Scratch Work: Considering only the fastest growing terms suggeststhat our fraction grows like 2n3/n3 = 2. Hence we guess the limit to be2.

For the proof, we must show that for any given ǫ there is an N suchthat for all n > N ,

(3)

∣

∣

2n3

n3 + 5n+ 1− 2

∣

∣ < ǫ

∣

∣

−10n− 2

n3 + 5n+ 1

∣

∣ < ǫ

10n+ 2

n3 + 5n+ 1< ǫ

(We omitted the absolute values in the last line since the quantitiesinvolved are clearly positive.)

We seek a simple quantity “?” such that, for large enough n,

10n+ 2

n3 + 5n+ 1<? < ǫ.

Making either the denominator smaller or the numerator larger in-creases the size of a fraction. Hence, for n > 2,

10n+ 2

n3 + 5n+ 1<

10n+ n

n3 + 5n+ 1<

11n

n3=

11

n2.

Hence, inequality (3) holds if n > 2 and

11

n2< ǫ

n2

11>

1

ǫ

n >√

11/ǫ


Formal Solution: Let ǫ > 0 be given and let N be the larger of√

11/ǫ and 2. Assume that n > N . Then from the “scratch work”,

| 2n3

n3 + 5n+ 1− 2

∣

∣ <11

n2< ǫ

Thus the requirements for the definition of the limit are fulfilled.

Remark: The proof in Example 3 is typical of many limit proofs.Often to prove that limn→∞ an = L we:

(1) Compute, and simplify |an−L| by, for example, putting termsover common denominator and/or factoring.

(2) Eliminate the absolute value by determining the sign of an−Lfor large n.

(3) Use rates of growth to estimate the growth of an − L so as toshow that it may be made less than ǫ

A common mistake is to forget to subtract the limit before estimat-ing the rate of growth. For example,

2n3

n3 + 5n+ 1<

2n3

n3= 2

All this tells us is that the limit, if it exists, is at most 2.

Often the techniques from Chapter 3 play a role, as in the nextexample.


limn→∞

2n3 − 6n2

n3 − 3n2 + 5n− 1

Scratch Work: Considering only the fastest growing terms suggeststhat our fraction grows like 2n3/n3 = 2. Hence we guess the limit to be2.

For the proof, we must show that we can approximate 2 to within±ǫ for any ǫ > 0. This means

(4)

∣

∣

2n3 − 6n2

n3 − 3n2 + 5n− 1− 2

∣

∣ < ǫ

∣

∣

2n3 − 6n2 − 2(n3 − 3n2 + 5n− 1)

n3 − 3n2 + 5n− 1

∣

∣ < ǫ

∣

∣

−10n+ 2

n3 − 3n2 + 5n− 1

∣

∣ < ǫ


We seek a simple quantity “?” such that, for large enough n,

∣

∣

−10n+ 2

n3 − 3n2 + 5n− 1

∣

∣ <? < ǫ.

We can make our fraction larger by making the denominator smaller.Specifically, we find positive numbers C and No such that

Cn3 < n3 − 3n2 + 5n− 1

for all n > No.We choose No so that both of the following hold for n > No:

3n2 <1

3n3

1 <1

3n3

The first inequality is valid for n > 9 and the second for n > 31/3 ≈1.44. For n > 9, both are valid and

n3 − 3n2 + 5n− 1 > n3 − 3n2 − 1

> n3 − 1

3n3 − 1

3n3 =

1

3n3

Hence we may choose No = 9 and C = 1/3.In particular the denominator is positive for such n. Furthermore,

for n ∈ N, −10n+ 2 < 0. Hence, for n > 6,

(5)

∣

∣

−10n+ 2

n3 − 3n2 + 5n− 1

∣

∣ =10n− 2

n3 − 3n2 + 5n− 1

<10n

n3/3

<30n

n3=

30

n2

Hence, inequality (4) holds if

30

n2< ǫ

n2

30>

1

ǫ

n >√

30/ǫ

Formal Solution: We should first prove inequality (5). However,since this is not the main point of the problem, we will allow ourselves(and the student) to use the “scratch work” as the proof.


Now let ǫ > 0 be given and let N be the larger of√

30/ǫ and 9.Assume that n > N . Then from the “scratch work”,

| 2n3 − 6n2

n3 − 3n2 + 5n− 1− 2

∣

∣ =10n− 2

n3 − 3n2 + 5n− 1<

10n

n3/3< ǫ

Thus the requirements for the definition of the limit are fulfilled.

What if we try to prove something which is false?

Example 5. Attempt an ǫ-proof of the following (incorrect) limitstatement. Describe carefully where your proof breaks down.

limn→∞

2n

n+ 1= 1

Solution: Let ǫ > 0 be given. If the limit statement is true there isan N such that the following holds for n > N :

∣

∣

2n

n+ 1− 1

∣

∣ < ǫ

∣

∣

n− 1

n+ 1

∣

∣ < ǫ

For n ∈ N the fraction is positive and the absolute value may bedropped:

n− 1

n+ 1< ǫ

n− 1 < ǫ(n+ 1)

n− 1 < ǫn+ ǫ

(1− ǫ)n < 1 + ǫ

If ǫ < 1, we may divide by 1− ǫ:

(6) n <1 + ǫ

1− ǫ.

But this inequality says that we achieve the desired accuracy only if n issufficiently small. Hence, for large n, 2n/(n+1) does not approximate1 to the desired accuracy for all large n, showing that the limit is not1.

If you see a difference of two square roots, either in a numerator orin a denominator, it is usually “wise to rationalize”-i.e. multiply bothnumerator and denominator by the sum of the square roots.



limn→∞

(√n−

√n+ 1).

Scratch Work: To find the limit, we write

√n−

√n+ 1 = (

√n−

√n+ 1)

√n+

√n+ 1√

n+√n+ 1

=(√n)2 − (

√n+ 1)2√

n+√n+ 1

=−1√

n+√n+ 1

As n → ∞ this will tend to zero. Furthermore,

∣

∣

√n−

√n+ 1− 0

∣

∣ =1√

n+√n+ 1

<1√n

This will be less than ǫ if√n > 1/ǫ i.e. n > 1/ǫ2.

Formal Solution: We claim that the limit is 0. To prove this, letǫ > 0 be given and set N = 1/ǫ2. Suppose that n > N . Then from the“scratch work”

∣

∣

√n−

√n+ 1− 0

∣

∣ <1√n< ǫ.

Thus, the conditions for the definition of limit are fulfilled.

Example 7. Guess the following limit and prove your answer usingǫ.

limn→∞

n2

2n2 + 3n+ n lnn.

Scratch Work: The denominator grows like 2n2 since both 3n andn lnn grow more slowly. Hence, the fraction behaves like n2/(2n2) =1/2. Hence, the limit should be 1/2.


For the proof, we compute

∣

∣

n2

2n2 + 3n+ n lnn− 1

2

∣

∣ =∣

∣

−3n− n lnn

2(2n2 + 3n+ n lnn)

∣

∣

=3n+ n lnn

2(2n2 + 3n+ n lnn)

<3n+ n lnn

4n2

Both 3n and n lnn grow more slowly than n3/2. Specifically

(7) n lnn < n3/2

is true if lnn < n1/2 which, as the reader may show, holds for n > (16)2.(See Example 5 of Chapter 3.) Also

(8) 3n < n3/2

holds for n > 9. Hence, for n > (16)2,

(9)

3n+ n lnn

4n2<

n3/2 + n3/2

4n2

=1

2n1/2

This will be less than ǫ if n1/2 > 1/(2ǫ) i.e. n > 1/(4ǫ2). Thus, wemay choose N to be any number larger than both 1/(4ǫ2) and (16)2 .

Formal Solution: From the argument in the “scratch work” (7), (8),and (9) all hold for n > (16)2.

Now, let ǫ > 0 be given and let N be the larger of (16)2 and 1/(4ǫ2)and let n > N . Then

n > 1/(4ǫ2)

n1/2 > 1/(2ǫ)

1

2n1/2< ǫ

Furthermore, from the scratch work

∣

∣

n2

2n2 + 3n+ n lnn− 1

2

∣

∣ =3n+ n lnn

2(2n2 + 3n+ n lnn)

<1

2n1/2< ǫ

Thus, the conditions for the definition of limit are fulfilled.


Not all sequences have a limit. A sequence which has a limit con-verges and one which does not diverges. The following exampleillustrates one particular type of divergence–going to infinity.

Example 8. I have a bank account that pays 5% interest per year,credited on Dec. 31. I deposit $1 on Jan. 1 of year 0 and make nofurther deposits. On Jan. 1 of year 1, I have 1 + (.05)1 = 1.05 dollarson deposit. On Jan. 1 of year 2, I have

P2 = 1.05 + (.05)(1.05) = (1.05)(1.05) = (1.05)2

dollars. After n years, I will have

(10) Pn = (1.05)n

dollars.

(1) How many years does it take for my balance to be over $100?$10,000? $100,000? $1,000,000?

(2) Prove that for any positive number M there is an n such thatPn > M . Thus, there is no limit to the amount of money I willeventually have in my account, as long as I leave it on depositlong enough.

Solution (1) To have more than $100, we need

(11)

(1.05)n > 100

ln ((1.05)n) > ln(100)

n ln(1.05) > ln(100)

n >ln(100)

ln(1.05)= 40.99

Hence, it would take 41 years. Similarly, to get:

$10, 000 we need n >ln(10, 000)

ln(1.05)≈ 81.98 years

$100, 000 we need n >ln(100, 000)

ln(1.05)≈ 102.47 years

$1, 000, 000 we need n >ln(1, 000, 000)

ln(1.05)≈ 122.97 years

Hence, after only 123 years our heirs will be millionaires!


Solution (2) For the proof we do the work from part (1) in the generalcase. Suppose we want $M . Let n be a natural number such that

n >lnM

ln(1.05).

Then, reversing the steps in (11):

n >lnM

ln(1.05)

n ln(1.05) > lnM

ln ((1.05)n) > lnM

(1.05)n > M

Example 8 demonstrates what we mean by limn→∞ an = ∞. Thebalance in our account grows without bound; for any value M we willeventually have more than $M in the account as long as we wait suf-ficiently long. Certainly, however, no matter how large, the balancealways much closer to 0 than to infinity! The “official” definition oftending to infinity is:

Definition 3. Let an be a sequence. We say that limn→∞ an = ∞if for all M > 0 there is an N such that

an > M

for all n > N .

We stress that infinity is not a limit: going to infinity is aspecial type of divergence.

Proving that limn→∞ an = ∞ involves assuming that a constantM > 0 is given and showing that for sufficiently large N , an > M .Often we find some simple quantity “?” such that

an >? > M

as in the following example.

Example 9. Prove, using M , that

limn→∞

n4

2n2 + 3n+ n lnn= ∞.

Scratch Work: Let M > 0 be given. We seek a quantity “?” suchthat

n4

2n2 + 3n+ n lnn>? > M


for all sufficiently large n. Making the denominator larger makes thefraction smaller. Since the fastest growing term in the denominator isn2, we seek constants C and No such that

2n2 + 3n+ n lnn < Cn2

for all n > No. Since, for n > 1, n < n2 and lnn < n, we see

(12) 2n2 + 3n+ n lnn < 2n2 + 3n2 + n2 = 6n2.

Hencen4

2n2 + 3n+ n lnn>

n4

6n2=

n2

6

This is > M provided n >√6M . We also need n > 1 for the validity

of inequality (12).

Formal Solution: Let M > 0 be given and let N be the larger of1 and

√6M . Let n ∈ N satisfy n >

√6M . Then, from the “scratch

work”n4

2n2 + 3n+ n lnn>

n4

6n2=

n2

6

>(√6M)2

6= M.

Thus, the conditions for the definition of limn→∞ an = ∞ are fulfilled.

A sequence can also diverge by tending to −∞. We could define thisnotion in a manner similar to our definition of tending to ∞. However,the simplest definition is just:

Definition 4. Let an be a sequence. We say that limn→∞ an = −∞if limn→∞ −an = ∞.

The following example demonstrates how to prove general theoremsusing the definition of limit.

Example 10. Suppose that limn→∞ an = 2. Prove, using ǫ, thatlimn→∞ a2n = 4.

Scratch Work: Let ǫ > 0 be given. We want to show that for allsufficiently large n,

|a2n − 4| < ǫ

|an − 2| |an + 2| < ǫ


Now an − 2 can be made as small as we wish while, for large n, an + 2should be near 4. More precisely, we may find an N so that an = 2± 1for n > N . Hence,

1 < an < 3

3 < an + 2 < 5

Thus, for n > N ,

|an − 2| |an + 2| < 5|an − 2|

This is less than ǫ if |an−2| < ǫ/5 which will be true for all sufficientlylarge N .

Formal Proof: Let ǫ > 0 be given. Since limn→∞ an = 2, there is anN such that for n > N

|an − 2| < ǫ/5

Furthermore there is an No such that for n > No

|an − 2| < 1

−1 < an − 2 < 1

1 < an + 2 < 5

|an + 2| < 5

Let N1 be the larger of N and No. Then for n > N1,

|a2n − 4| = |an − 2| |an + 2| < (ǫ/5)5 = ǫ

Hence the requirements for the definition of limit are fulfilled.

Exercises

(1) The figure below shows the first 20 terms of a sequence anwhich, apparently, is converging to 1. Assume that the termsnot shown are closer to one than any of the shown terms.What (approximately) is the first n where we become confidentthat all of the rest of the terms are within ±.2 of 1? ±.1?±.05? ±.025? (Note that the ‘tick marks’ on the vertical axisare .1 apart.) Explain in your own words, how this problemdemonstrates the general definition of limit of a sequence.


.

.

.

.

.

.

.

.

.

..

..

.. . . . . .

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20

.1

.

.

.

.

.

.

.

.

.

.

.

.

.

.

19

(2) For each sequence below, find the limit and prove your answerusing ǫ.(a) limn→∞(1− 1/n2)

(b) limn→∞3n2

n2+1

(c) limn→∞(√n2 + 1−

√n2 + 2) (See Example 5)

(d) limn→∞(√n2 + 1− n)

(e) limn→∞2n

n2+n−5.

(f) limn→∞2n lnnn2+n−5

. (See Example 6)

(g) limn→∞(

23

)nHint: Use logs.

(h) limn→∞2n

n!. Hint: See Exercise 8, Chapter 3.

(i) limn→∞1+(−1)n

n+ 7

(j) limn→∞n3

2n3−7

(k) limn→∞2n

2n+n−5.

(l) limn→∞2n

3(2n)+n−5.

(m) limn→∞sinnn2+1

(n) limn→∞(−1)n sinn

n2+1

(o) limn→∞n3

2n3+n lnn+1

(p) limn→∞n3

2n3−n2+2


(q) limn→∞n√n2+1

Hint: Put an − L over a common denomi-

nator and then rationalize the numerator.(r) limn→∞

√4n+1n

(s) limn→∞√n+1√n

(t) limn→∞5n3−10n3+n2−1

(3) In Example 7 on page 62, we used lnn <√n for n > (16)2.

Why would it not work to use lnn < n instead?(4) Prove, using ǫ, that for all k > 0, limn→∞

1nk = 0.

(5) Prove that the answer to part (b) of Exercise 2 is not 4. Reasonas in Example 4.

(6) The statements limx→∞ f(x) = L and limn→∞ an = L meanslightly different things. In the first statement, f(x) is a func-tion that is defined for all sufficiently large real numbers xand the limit considers all such x. In the second statement,an is a sequence, implying that n assumes only integral val-ues. As a demonstration of this, explain, using a graph off(x) = cos(2πx), why the first limit below exists and the sec-ond does not. What is the value of the first limit?

limn→∞

cos(2πn)

limx→∞

cos(2πx)

Remark: We define limx→∞ f(x) = L as follows

Definition 5. Let f(x) be a function that is defined forall real numbers x > 0 and let L be a number. We say thatlimx→∞ f(x) = L provided that for every number ǫ > 0, thereis a real number N such that

|f(x)− L| < ǫ

for all x > N .

(7) Use the preceding definition to prove the following limit state-ments.

(a) limx→∞

x

x+ 1= 1

(b) limx→∞

2x5

x5 − 7= 2

(8) Let f be a function that is defined for all positive real num-bers x. Suppose that limx→∞ f(x) = L. Prove (using ǫ) thatlimn→∞ f(n) = L where “f(n)” denotes the sequence an de-fined by an = f(n) for all n ∈ N.


(9) We say that an is bounded from above if there is a number Msuch that an < M for all n, in which case M is referred to asan upper bound for an. The following sequence of argumentsproves that if L = limn→∞ an exists, then an is bounded fromabove. Give a reason (or proof) for each statement.

There is a number N such that L− 1 < an < L+ 1 for alln > N .

There is a numberMo such that an < Mo for all 1 ≤ n ≤ N ,n ∈ N.

Let M be the larger of L+1 and Mo. Then M is an upperbound for an.

(10) Let an be a convergent sequence. Prove that an is boundedfrom below. Use a similar argument to that sketched in thepreceding exercise.

(11) Each of the following sequences an tends to ∞. Demonstratethis by finding a value of N such that for all n > N (i) an >100, (ii) an > 1000 and (iii) an > 1, 000, 000. Finally, proveusing M that limn→∞ an = ∞.(a) 2n.(b) 2n

lnn. Hint: For sufficiently large n, lnn < n < (1.5)n.

(c) lnn.

(d) n3

n−1.

(e) n5

n2−n−lnn.

(f) n3

n+1.

(g) n5

n2+n+1.

(12) Using logs, prove the following:(a) For a > 1, limn→∞ an = ∞.(b) For 0 < a < 1, limn→∞ an = 0.

(13) Let x be a positive number. We note that

(1 + x)2 = 1 + 2x+ x2 > 1 + 2x

Hence

(1 + x)3 = (1 + x)(1 + x)2

> (1 + x)(1 + 2x) = 1 + 3x+ 2x2 > 1 + 3x

Similarly

(1 + x)4 = (1 + x)(1 + x)3

> (1 + x)(1 + 3x) = 1 + 4x+ 3x2 > 1 + 4x

(a) Use the fact that (1 + x)4 > 1 + 4x to prove that(1 + x)5 > 1 + 5x.


(b) Use the fact that (1 + x)5 > 1 + 5x to prove that(1 + x)6 > 1 + 6x.

(c) Suppose that we have succeeded in proving that(1 + x)n > 1 + nx for some value of n. Use this to provethat (1 + x)n+1 > 1 + (n+ 1)x.Warning: You cannot simply replace n by n+1 since ourassumption is only that you have proved the inequalityfor some n, not all n. Instead, you should use reasoningsimilar to that used in (a).

(d) Explain how it follows from (b) that (1 + x)n > 1 + nxfor all n.

(e) Use (c) to prove that for all a > 1, limn→∞ an = ∞.

Remark: The proof sketched in (b) is a Mathematical Induc-tion argument. Mathematical Induction is a method of provingan infinite number of statements, one at a time. We first provethe first satement. We then use the first to prove the second,use the second to prove the third, use the third to prove thefourth, etc. To prove all of the statements we need to provethat this process can be continued indefinitely. In practice thismeans assuming that we have proved the theorem for some n(but not yet for n+ 1). We then use the nth case to prove the(n+ 1)st case.

(14) Suppose that limn→∞ an = 7. Use ǫ to prove that(a) limn→∞ a2n = 49(b) limn→∞

1an+1

= 18

(c) limn→∞√an + 2 = 3 (Hint: Rationalize)

(d) limn→∞an

an+7= 1

2

(15) Suppose that limn→∞ an = −∞. Prove that for each M < 0,there is an N such that an < M for all n > N . Hint: In thetext we defined limn→∞ an = −∞ by limn→∞ −an = ∞.

(16) Prove the converse of the preceding exercise: i.e. Suppose thatfor all M < 0, there is an N such that an < M for all n > N .Prove that limn→∞ an = −∞.

(17) Prove, using ǫ, that limn→∞ an = L if and only if limn→∞(an−L) = 0, i.e. first prove that if limn→∞(an − L) = 0 thenlimn→∞ an = L. Then prove that if limn→∞ an = L thenlimn→∞(an − L) = 0.

(18) Suppose that limn→∞ an = L. Prove (using ǫ) thatlimn→∞(−an) = −L.


(19) Suppose that limn→∞ |an| = 0. Does it follow that limn→∞ an =0? Prove your answer.

(20) Suppose that limn→∞ an = 0. Does it follow thatlimn→∞ |an| = 0? Prove your answer.

(21) Suppose that limn→∞ |an| = 1. Does it follow thatlimn→∞ an exists? Prove your answer.

(22) Suppose that limn→∞ an = L. Prove, using ǫ, that limn→∞ 5an =5L.

(23) If C is any constant, then limn→∞C = C–i.e. if an = C for alln, then limn→∞ an = C.

(24) Suppose that limn→∞ an = L. Prove (using ǫ) that for anyconstant C, limn→∞ Can = CL.

(25) Suppose that for all n, an ≤ bn ≤ cn and limn→∞ an =limn→∞ cn = L. Prove, using ǫ, that then limn→∞ bn = L.Hint: an − L ≤ bn − L ≤ cn − L.

(26) Suppose that limn→∞ an = L where L > 0. Prove that thereis an N such that an > 0 for all n > N . Hint: How close to Lmust an be to guarantee that an > 0. What tells you that canget this close?

(27) (a) The following statement is false. Find an example thatdemonstrates its falsehood. “If an < 0 for all n andlimn→∞ an exists, then limn→∞ an < 0.”

(b) Prove that if an < 0 for all n and limn→∞ an exists, thenlimn→∞ an ≤ 0. Hint: This is almost immediate from theresult of Exercise 26. Explain.

(28) Suppose that limn→∞ an = L where L < 0. Prove that thereis an N such that an < 0 for all n > N .

(29) Use Exercise 28 to prove that if an > 0 for all n and limn→∞ anexists, then limn→∞ an ≥ 0.

(30) Suppose that an > L for all n. Prove that limn→∞ an ≥ L.Hint: The result of Exercises 17 and 29 might help.

(31) Prove Proposition 1. Hint: For your proof, assume that L 6=M . Let ǫ = |L−M |.

(32) Suppose that an and bn are sequences where limn→∞ an = 0.Suppose also that and |bn| < 1 for all n. Prove, using ǫ, thatlimn→∞ anbn = 0.

(33) Let an be a sequence of nonnegative real numbers such thatlimn→∞ an = 0. Prove, using ǫ, that limn→∞

√an = 0.

(34) Let an be a convergent sequence of nonnegative real numbers.Prove that limn→∞

√an =

√limn→∞ an.


Hint: Let L = limn→∞ an. You may assume L > 0 as theL = 0 case was done in the preceding exercise. Rationalize|√an −

√L| and note that

√an +

√L ≥

√L.

(35) Suppose that limn→∞ an = 1. Prove, using ǫ, that limn→∞ 1/an =1. Hint: First prove that there is an N such that .5 ≤ an ≤ 1.5for all n > N . Then simplify |1/an − 1|.

(36) Let limn→∞ an = L . Prove that limn→∞ an+1 = L.(37) Let an be such that limn→∞ an = ∞. Prove, using ǫ, that

limn→∞1an

= 0.

(38) Is the converse of Exercise 37 true? That is, if limn→∞ 1/an =0, does it follow that limn→∞ an = ∞. If not, under whatcircumstances would it follow? Prove your answers.

(39) Most mathematical theorems can be expressed in the form “IfP, then Q” where P and Q are statements. (A statement is asentence that has the potential of being either true or false.)For example, the converse to the statement in Exercise 37 isthe statement in Exercise 38.

Below are some true statements. For each statement, for-mulate the converse statement and state whether or it is true.If false, give a counter example.(a) If limn→∞ an exists, then limn→∞ can exists for all con-

stants c.(b) If limn→∞ an exists and limn→∞ bn exists, then

limn→∞(an + bn) exists.(c) If limn→∞ an exists, then an is bounded.(d) If limn→∞ an = ∞, then an is unbounded.(e) If limn→∞ an = ∞, then an is not bounded from above.(f) Suppose that limn→∞ an = b. Then limn→∞(an − b) = 0.

CHAPTER 5

Limit Theorems

When doing calculations using approximations, the error can in-crease. For example, 1.99 approximates 2 with error .01 and 2.98 ap-proximates 3 with error .02. The sum 1.99+ 2.98 = 4.97 approximates2 + 3 = 5 with error .03. This demonstrates that when adding ap-proximations, the errors may add as well. It follows that if we wantto approximate a sum a + b with error less than ǫ, we will need toapproximate both a and b with error less than ǫ/2. This observation isthe basis for the proof of the following theorem.

Theorem 1 (Sum Theorem). Suppose that an and bn are conver-gent sequences. Then

limn→∞

(an + bn) = limn→∞

an + limn→∞

bn

Proof Let ǫ > 0 be given and set L = limn→∞ an and M = limn→∞ bn.Our theorem is equivalent with

limn→∞

(an + bn) = L+M

From our hypothesis, there are numbers N1 and N2 such that

|an − L| < ǫ

2for all n > N1

|bn −M | < ǫ

2for all n > N2

Let N be the maximum of N1 and N2. Then n > N implies

|an + bn − (L+M)| = |(an − L) + (bn −M)|≤ |an − L|+ |bn −M | < ǫ

2+

ǫ

2= ǫ

This fulfills the requirements for the definition of limit, proving ourtheorem.

Remark: The above theorem is not at all remarkable–we certainlyexpect that if an is close to L and bn is close to M , then an+ bn will beclose to L + M . What is remarkable is that we can prove this result

75

76 5. LIMIT THEOREMS

without resorting to statements such as “an gets real close to L and bngets real close to M so an + bn gets real close to L+M .”

In preparation for the proof of the product theorem, we prove thefollowing special case:

Lemma 1. Suppose that limn→∞ an = limn→∞ bn = 0. Then limn→∞ anbn =0.

Proof Let ǫ > 0 be given. We need to show that there is an N suchthat for all n > N ,

(1) |anbn − 0| = |anbn| = |an| |bn| < ǫ

for all n > N .This will be true if both of the following inequalities hold for all

n > N :

(2)|an| <

√ǫ

|bn| <√ǫ

However, since limn→∞ an = 0, we know that there is a number N1

such that the first inequality in (2) holds for all n ≥ N1. Similarly,there is a number N2 such that the second inequality in (2) holds forall n ≥ N2. Both inequalities hold for n > N = max{N1, N2}. Hence,inequality (2) holds for all n > N , proving our lemma.

We also note the following simple results, all of which were exercisesin Chapter 4. (Exercise 17 on page 71, Exercise 23 on page 72, andExercise 24 on page 72.)

Lemma 2. If C is any constant, then limn→∞C = C–i.e. if an = Cfor all n, then limn→∞ an = C.

Lemma 3. If C is any constant, then limn→∞ Can = C limn→∞ an.

Lemma 4. If an is a sequence, then limn→∞ an = L if and only iflimn→∞(an − L) = 0

We are now ready to prove the product theorem.

Theorem 2 (Product Theorem). Suppose that an and bn are con-vergent sequences. Then

limn→∞

(anbn) = ( limn→∞

an)( limn→∞

bn)

Proof Let L = limn→∞ an and M = limn→∞ bn. Our theorem is equiv-alent with the statement

limn→∞

(anbn) = LM

5. LIMIT THEOREMS 77

From Lemma 4, this in turn is equivalent with

limn→∞

(anbn − LM) = 0

Note that

anbn − LM = ((an − L) + L)((bn −M) +M)− LM

= (an − L)(bn −M) + L(bn −M) + (an − L)M

Hence

(3)limn→∞

(anbn − LM) = limn→∞

(an − L)(bn −M)

+ limn→∞

L(bn −M) + limn→∞

(an − L)M

From Lemma 3

limn→∞

(an − L) = limn→∞

(bn −M) = 0.

Hence Lemmas 1 and 4 show that each of the three terms on the rightin equation (3) equal 0, proving the Product Theorem.

The following result will be used to prove the Quotient Theorem.

Lemma 5. Suppose that limn→∞ an = 1. Then

limn→∞

1

an= 1

Proof We note that

(4)∣

∣

1

an− 1

∣

∣ =|1− an||an|

There is an N1 such that for n > N1,

|an − 1| < 1

2

−1

2< an − 1 <

1

2In particular,

(5) an >1

2for all n > N1.

It now follows from equation (4) and inequality (5) that for n ≥ N1,

(6)∣

∣

1

an− 1

∣

∣ <|1− an|1/2

= 2|an − 1|

Choose N2 such that

(7) |an − 1| < ǫ

2

78 5. LIMIT THEOREMS

for all n > N2. Then, if n > max{N1, N2}, inequalities (6) and (7)both hold, showing that for such n,

∣

∣

1

an− 1

∣

∣ < 2( ǫ

2

)

= ǫ

proving the lemma.

The following result follows almost immediately:

Theorem 3 (Inverse). Suppose that limn→∞ an = L where L 6= 0.Then

limn→∞

1

an=

1

L.

Proof From Lemma 4

limn→∞

anL

=1

Llimn→∞

an =1

LL = 1.

Hence, Lemma 5 shows that

1 = limn→∞

L

an= L lim

n→∞

1

an.

Our theorem follows by dividing both sides of the above equality by L.

Combining Theorem 2 with Theorem 3 yields the following result,which we leave as an exercise.

Theorem 4 (Quotient Theorem). Let an and bn be convergent se-quences where limn→∞ bn 6= 0. Then

limn→∞

anbn

=limn→∞ anlimn→∞ bn

Before demonstrating the use of the Sum, Quotient and ProductTheorems in solving limit problems. we note one further result whichwas an exercise in Chapter 4. (Exercise 4 on page 69.)

Proposition 1. For all k > 0

limn→∞

1

nk= 0.

Example 1. Find the following limit in a step-by-step fashion,stating the main limit theorem used in each step.

limn→∞

n5 + 1

3n5 − 4n+ 1


Solution:

limn→∞

n5 + 1

3n5 − 4n+ 1= lim

n→∞

n5(1 + 1n5 )

n5(3− 4n4 +

1n5 )

No limit theorems used

=limn→∞(1 + 1

n5 )

limn→∞(3− 4n4 +

1n5 )

Quotient Theorem

=limn→∞ 1 + limn→∞

1n5

limn→∞ 3− 4 limn→∞1n4 + limn→∞

1n5 )

Sum Theorem, Lemma 3

=1 + 0

3− 0 + 0=

1

3Prop. 1

Exercises

(1) Use limit theorems to evaluate the following. Justify your workas was done in Example 1. 1

(a) limn→∞

n3 − 3n2 + 1

2n3 − n− 5

(b) limn→∞

(

2n2

n2 + 4

)(

n− 3

n+√n2 + 1

)

(c) limn→∞

n

(n+ 1)(

2 + ln(1 + 1n2 )

)

(2) Use Theorems 3 and 2 to prove Theorem 4.(3) In Examples 1 and 3 of Chapter 4, it was shown that

(8)

n

n+ 1= 1± ǫ for n >

1

ǫ− 1

2n3 − 6n2

n3 − 3n2 + 5n− 1= 2± ǫ for n >

√

30/ǫ

Use these inequalities to answer the following questions.

1Part (b) requires the observation that limn→∞

√

1 + 1

n2 = 1, which is a con-

sequence of Exercise 34 of Chapter 4. Part (b) requires the observation thatlimn→∞ ln(1 + 1

n2 ) = 0, which is a consequence of the continuity of the log function.

(See Theorem 3 on page 180 in Chapter 11.) Both results may be assumed for thesake of this exercise.


(a) According to the proof of Theorem 1, for which value ofN does

n

n+ 1+

2n3 − 6n2

n3 − 3n2 + 5n− 1= 3± 10−3?

for all n > N? Explain.(b) According to the proof of Lemma 5, for which value of N

doesn+ 1

n= 1± 10−3?

for all n > N? Explain.(c) According to the proof of Lemma 5, for which value of N

does

2(n3 − 3n2 + 5n− 1)

2n3 − 6n2= 1± 10−3?

for all n > N? Explain. Hint: First divide both sides ofthe second expression in (8) by 2.

(4) Repeat Exercise 3 with 10−3 replaced by 10−5

(5) In Examples 1 and 6 of Chapter 4, it was shown that

n

n+ 1= 1± ǫ for n >

1

ǫ− 1

(√n−

√n+ 1) = 0± ǫ for n > 1/ǫ2

According to the proof of Theorem 1, for which value of N doesn

n+ 1+ (

√n−

√n+ 1) = 1± 10−3?

for all n > N? Explain.(6) In Example 6 of Chapter 4, it was shown that

(√n−

√n+ 1) = 0± ǫ for n > 1/ǫ2

It is clear that1

n= 0± ǫ

for n > 1/ǫ.According to the proof of Lemma 1, for which value of N

does √n−

√n+ 1

n= 0± 10−3?

for all n > N? Explain.

CHAPTER 6

Max, Min, Sup, Inf

We would like to begin by asking for the maximum of the functionf(x) = (sin x)/x. An approximate graph is indicated below. Lookingat the graph, it is clear that f(x) ≤ 1 for all x in the domain of f .Furthermore, 1 is the smallest number which is greater than all of f ’svalues.

o

y=(sin x)/x

1

Figure 1

Loosely speaking, one might say that 1 is the ‘maximum value’ off(x). The problem is that one is not a value of f(x) at all. There isno x in the domain of f such that f(x) = 1. In this situation, we usethe word ‘supremum’ instead of the word ‘maximum’. The distinctionbetween these two concepts is described in the following definition.

Definition 1. Let S be a set of real numbers. An upper bound forS is a number B such that x ≤ B for all x ∈ S. The supremum, if itexists, (“sup”, “LUB,” “least upper bound”) of S is the smallest upperbound for S. An upper bound which actually belongs to the set is calleda maximum.

83

84 6. MAX, MIN, SUP, INF

Proving that a certain number M is the LUB of a set S is oftendone in two steps:

(1) Prove that M is an upper bound for S–i.e. show that M ≥ sfor all s ∈ S.

(2) Prove that M is the least upper bound for S. Often this isdone by assuming that there is an ǫ > 0 such that M − ǫis also an upper bound for S. One then exhibits an elements ∈ S with s > M − ǫ, showing that M − ǫ is not an upperbound.

Example 1. Find the least upper bound for the following set andprove that your answer is correct.

S = {12,2

3,3

4, . . . ,

n

n+ 1. . . }

Solution: We note that every element of S is less than 1 sincen

n+ 1< 1

We claim that the least upper bound is 1. Assume that 1 is notthe least upper bound,. Then there is an ǫ > 0 such that 1− ǫ is alsoan upper bound. However, we claim that there is a natural number nsuch that

1− ǫ <n

n+ 1.

This inequality is equivalent with the following sequence of inequalities

1− n

n+ 1< ǫ

1

n+ 1< ǫ

1

ǫ< n+ 1

1

ǫ− 1 < n.

Reversing the above sequence of inequalities shows that if n > 1ǫ− 1,

then 1− ǫ < nn+1

showing that 1− ǫ is not an upper bound for S. Thisverifies our answer.

If a set has a maximum, then the maximum will also be a supremum:

Proposition 1. Suppose that B is an upper bound for a set S andthat B ∈ S. Then B = supS.

6. MAX, MIN, SUP, INF 85

Proof Let ǫ > 0 be given. Then B− ǫ cannot be an upper bound for Ssince B ∈ S and B > B − ǫ, showing that B is indeed the least upperbound.

Example 2. Find the least upper bound for the following set andprove that your answer is correct.

T = {1, 12,2

3,3

4, . . . ,

n

n+ 1. . . }.

Solution: From the work done in Example 1, 1 is an upper bound forS. Since 1 ∈ S, 1 = supS.

Example 3. Find the max, min, sup, and inf of the following setand prove your answer.

S = {2n+ 1

n+ 1| n ∈ N}.

Solution: We write the first few terms of S:

S ={3

2,5

3,7

4,9

5,11

6, . . .

}

.

The smallest term seems to be 32and there seems to be no largest term,

although all of the terms seem to be less than 2. Since limn→∞2n+1n+1

= 2we conjecture that:

(a) There is no maximum, (b) supS = 2, and (c) minS = inf S = 32.

Proof

SupWe must first show that 2 is an upper bound–i.e.

2n+ 1

n+ 1< 2

2n+ 1 < 2n+ 2

1 < 2

which is always true. Reversing the above argument shows that 2 is anupper bound.

Next we show that 2 is the least upper bound. If 2 is not the LUB,there is an ǫ > 0 such 2− ǫ is an upper bound. However, we claim thatthere are n ∈ N such that

2− ǫ <2n+ 1

n+ 1


showing that 2− ǫ is not an upper bound.To prove our claim, note that the above inequality is equivalent

with

−ǫ <2n+ 1

n+ 1− 2

ǫ >1

n+ 1

n >1

ǫ− 1

Since there exist n ∈ N satisfying the above inequality, our claim isproved. Hence 2 is the LUB.

Next we show that 2 is not a maximum. This means showing thatthere is no n such that

2 =2n+ 1

n+ 1.

This however is equivalent with

2(n+ 1) = 2n+ 1

2n+ 2 = 2n+ 1

2 = 1

which is certainly false.Next we prove that minS = 3

2. We first note that 3

2∈ S since

3

2=

2 · 1 + 1

1 + 1.

Hence, it suffices to show that 32is a lower bound which we do as

follows:2n+ 1

n+ 1≥ 3

22(2n+ 1) ≥ 3(n+ 1)

n ≥ 1

which is true for all n ∈ N. Reversing the above argument shows that32is a lower bound. �

The central question in this section is “Does every non-empty setof numbers have a sup?” The simple answer is no–the set N of nat-ural numbers does not have a sup because it is not bounded fromabove. O.K.–we change the question:“Does every set of numbers whichis bounded from above have a sup?” The answer, it turns out, dependsupon what we mean by the word “number”. If we mean “rationalnumber” then our answer is NO!.


Recall that the set of integers is the set of positive and naturalnumbers, together with 0. I.e.

Z = {0,±1,±2, . . . ,±n, · · · | n ∈ N}.The set of rational numbers is the set

Q = {pq| p, q ∈ Z, q 6= 0}.

Thus, for example, 23and −9

7are elements of Q. In Chapter 9 (Theo-

rem 3) we prove that√2 is not rational.

Now, let S be the set of all positive rational numbers r such thatr2 < 2. Since the square root function is increasing on the set ofpositive real numbers,

S = {0 < r <√2 | r ∈ Q}.

Clearly,√2 is an upper bound for S. It is also a limit of values

from S. In fact, we know that√2 = 1.414213562 + .

Each of the numbers 1.4, 1.41, 1.414, 1.4142, etc. is rational and hassquare less than 2. Their limit is

√2. Thus, supS =

√2. (See Exer-

cise 6 below.) Since√2 is irrational, S is then an example of a set of

rational numbers whose sup is irrational.Suppose, however, that we (like the early Greek mathematicians)

only knew about rational numbers. We would be forced to say that Shas no sup. The fact that S does not have a sup in Q can be thoughtof as saying that the rational numbers do not completely fill up thenumber line; there is a missing number “directly to the right” of S.The fact that the set R of all real numbers does fill up the line is sucha fundamentally important property that we take it as an axiom: thecompleteness axiom. (The reader may recall that in Chapter I, wementioned that we would eventually need to add an axiom to our list.This is it.) We shall also refer to this axiom as the Least Upper BoundAxiom. (LUB Axiom for short.)

Least Upper Bound Axiom: Every non-empty set of real numberswhich is bounded from above has a supremum.

The observation that the least upper bound axiom is false for Q

tells us something important: it is not possible to prove the least upperbound axiom using only the axioms stated in Chapters 1 and 2. This isbecause the set of rational numbers satisfy all the axioms from Chapters


1 and 2. Thus, if the least upper bound axiom were provable from theseaxioms, it hold for the rational numbers.

Of course, similar comments apply to minimums:

Definition: Let S be a set of real numbers. A lower bound for S is anumber B such that B ≤ x for all x ∈ S. The infinum (“inf”, “GLB,”“greatest lower bound”) of S, if it exists, is the largest lower bound forS. A lower bound which actually belongs to the set is called a minimum.

Fortunately, once we have the LUB Axiom, we do not need anotheraxiom to guarantee the existence of inf’s. The existence of inf’s is atheorem which we will leave as an exercise. (Of course, we could havelet the existence of inf’s be our completeness axiom, in which case theexistence of sup’s would be a theorem.)

Greatest Lower Bound Property: Every non-empty set of realnumbers which is bounded from below has a infimum.

Proving that a certain number M is the GLB of a set S is similarto a LUB proof. It requires:

(1) Proving thatM is a lower bound for S–i.e. proving thatM ≤ sfor all s ∈ S.

(2) Proving that M is the greatest lower bound for S. Often thisis done by assuming that there is an ǫ > 0 such that M + ǫis a lower bound for S. One then exhibits an element s of Ssatisfying s < M + ǫ, showing that M + ǫ is not a lower boundfor S.

Example 4. Prove that the inf of S = (1, 5] is 1.

Solution: By definition S is the set of x satisfying 1 < x ≤ 5. Hence 1is a lower bound for S. Suppose that 1 is not the GLB of S. Then thereis an ǫ > 0 such that 1 + ǫ is also a lower bound for S. To contradictthis, we exhibit x ∈ S such that 1 < x < 1 + ǫ. Since

0 <ǫ

2< ǫ

we see that

x = 1 +ǫ

2satisfies

1 < x < 1 + ǫ.


Since 1+ǫ is (by assumption) a lower bound for S and 5 ∈ S, 1+ǫ ≤ 5,showing that x ∈ (1, 5]. Thus, 1 + ǫ is not a lower bound, proving that1 is the greatest lower bound.

Example 5. Find upper and lower bounds for y = f(x) for x ∈[−1, 1.5] where

f(x) = −x4 + 2x2 + x

Use a graphing calculator to estimate the least upper bound and thegreatest lower bound for f(x).

Solution: From the triangle inequality

|f(x)| = | − x4 + 2x2 + x| ≤ |x4|+ |2x2|+ |x|= |x|4 + 2|x|2 + |x|

The last quantity is largest when |x| is largest, which occurs when|x| = 1.5. Hence

|f(x)| ≤ 1.54 + 2(1.5)3 + 1.5 = 13.3125

Hence, M = 14 is an upper bound and M = −14 is a lower bound.As a check, we graph y = −x4 + 2x2 + x with xmin= −.5, xmax= 1.5,ymin= −15 and ymax= 15, as well as the lines y = 14 and y = −14.Since the graph lies between the lines, the value of M is acceptable,although considerably larger than necessary. To estimate the leastbound, we trace the curve using the trace feature of the calculator,finding that the maximum and minimum y-values are approximately2.0559 and −.130 respectively. These values are (approximately) theleast upper bound and the greatest lower bound respectively.

Remark: It is important to note that in the preceding example, thevalues of the function at the end points of the interval are not bounds forthe function because f(x) is not monotonic (i.e. it is neither increasingnor decreasing) over the stated interval. On the other hand, the largestvalue of |x|4+2|x|2+ |x| is at the largest endpoint because this functionincreases as |x| increases.

The next example uses both the triangle inequality and the obser-vation that making the denominator of a fraction smaller increases itsvalue.

Example 6. Find a bound for y = f(x) for x ∈ [−2, 2] where

f(x) =x3 − 3x+ 1

1 + x2



∣

∣

x3 − 3x+ 1

1 + x2

∣

∣ =|x3 − 3x+ 1|

1 + x2

Since x2 > 0, we see

x2 + 1 > 1

1

x2 + 1<

1

1= 1

Hence

|x3 − 3x+ 1|1 + x2

≤ |x3 − 3x+ 1| ≤ |x3|+ |3x|+ 1 ≤ 23 + 3 · 2 + 1 = 15

Thus, our bound is M = 15.

In all of the examples considered above, the least upper bound forf(x) is the maximum of f(x). This is always the case if f(x) has amaximum. Similarly, the greatest lower bound is the minimum of f(x)if f(x) has a minimum.

In Chapter 4, we studied sequences which diverge because they tendto infinity. Another class of sequences that have no limit are ones whichmight be dubbed “wishy-washy.” These are sequences that can’t makeup their minds what their limit is, tending simultaneously to severalnumbers. An example is

an =n+ (−1)nn

n+ 1

The first 10 values are listed below.

n 1 2 3 4 5 6 7 8 9 10an 0 1.33 0 1.60 0 1.71 0 1.78 0 1.82

It appears that some values approach (in fact equal) 0 while othersapproach 2. Indeed, for odd n

an =n− n

n+ 1= 0

which tells us that if the limit exists, it must be 0. For even n

an =n+ n

n+ 1

2n

n+ 1

which tends to 2. From Proposition 1, a sequence can have only onelimit. Hence, there is no limit.


A related type divergence is what might be referred to as “wan-dering,” where values wander in a seemingly random manner, nevergetting close to particular number. An example is

an = cosn.

The first 10 values are listed below.

n 1 2 3 4 5 6 7 8 9 10cosn .540 −.416 −.990 −.654 .284 .960 .754 −.146 −.911 −.839

There certainly seems to be no tendency toward any one number.

A sequence an is non-decreasing if

a1 ≤ a2 ≤ a3 ≤ · · · ≤ an < . . .

The sequence an is increasing if each of the above inequalities is strict.For example, an = n2 is increasing :

1 < 22 < 32 < 42 < . . .

The sequence from Example 1 is also increasing:

1

2<

2

3<

3

4<

4

5< . . .

The two preceding sequences demonstrate that an increasing se-quence can either go to infinity or can converge. However, an increasingsequence cannot wander; once it has exceeded a certain value, it cannever return to that value. Hence we arrive at the following theorem:

Theorem 1 (Bounded Increasing Theorem). For a non-decreasingsequence an, either limn→∞ an exists or limn→∞ an = ∞.

Proof limn→∞ an = ∞ implies that for all M > 0 there is an N suchthat an ≥ M for all n ≥ N . For a non-decreasing sequence, this isequivalent with the statement that for all M there is an n such thatan ≥ M . Hence if limn→∞ an 6= ∞, there is an M such that an < Mfor all n. Thus

S = {a1, a2, . . . , an, . . . }.is bounded from above. Let a = supS. Then

an ≤ a

for all n. Furthermore, since a is the smallest upper bound, a− ǫ is notan upper bound for any ǫ > 0. Hence, for all ǫ > 0, there is at leastone aN such that

a− ǫ < aN ≤ a.


However, since an is increasing and a is an upper bound for S, we seethat for all n ≥ N ,

aN ≤ an ≤ a.

It follows that for all n ≥ N ,

a− ǫ ≤ an ≤ a+ ǫ

proving that a is the limit of an, as desired.

The bounded increasing theorem is one of the most important waysof proving that sequences converge. In particular, we will make gooduse of in Chapters 7 and 8.

There is, of course, nothing special about increasing as opposed todecreasing. It is an immediate consequence of the Bounded-IncreasingTheorem that a decreasing sequence which is bounded from below alsohas a limit. (See Exercise 10 below.) Exercises

(1) Compute the sup, inf, max and min (whenever these exist) forthe following sets. 1

(a) {1 + 1/n | n ∈ N}(b) [0, 2)

(c) {n2+15n+1

| n ∈ N}(d) {x | x ∈ Q and x2 < 2}(e) {y | y = x2 − x+ 1 and x ∈ R}(f) {x | x2 − 3x+ 2 < 0 and x ∈ R}(g) { 1

n− 1

m|n,m ∈ N}

(h) {1 + 1+(−1)n

n|n ∈ N}

(i) {12, 13, 23, 14, 34, 15, 25, 35, 45, . . . }. (This is a list of the fractions

in the interval (0, 1). The pattern is that we list fractionsby increasing value of the denominator. For a given valueof denominator, we go from smallest to largest, omittingfractions which are not in reduced form.)

(j) {n/(1 + n2) | n ∈ N}(k) {3n2/(1 + 2n2) | n ∈ N}(l) {n/(1− n2) | n ∈ N, n > 1}

(m) {(1− 2n2)/3n2 | n ∈ N}(n) {2n2/(3n2 − 1) | n ∈ N}(o) {3n/

√1 + n2 | n ∈ N}

(p) {3/√1 + 2n2 | n ∈ N}

1In set theory, the symbol ‘|’ is read “where.” Thus, the set in part (a) is theset of numbers of the form 1 + 1/n where n is a natural number.”


(q) { n−1n2+5

| n ∈ N}Hint: In the proof of the upper bound (which is requestedin Exercise 2), you will discover that you need to proven2 − 7n + 12 ≥ 0. Factor this polynomial and determinewhen it can be negative.

(r) {3n+1n+1

| n ∈ N}(2) Prove your answers in Exercise 1.(3) In Example 5, use a graphing calculator to estimate the sup

and the inf for f(x).(4) For the following functions f(x), (i) find a number M such

that |f(x)| ≤ M for all x in the stated interval. (ii) Use agraph to estimate the sup s and inf t for f(x). Sketch yourgraph on a piece of paper. (iii) As evidence for the statementthat s is the LUB, find an x such that f(x) > s − ǫ for thestated value of ǫ. (iv) As evidence for the statement that t isthe GLB, find an x such that f(x) < t+ ǫ for the stated valueof ǫ.

(a) f(x) = x5 − x4 + x3 − x2 + x− 1 x ∈ [−1, 1], ǫ = .1

(b) f(x) =x3 − 3x2

1 + x+ x2x ∈ [0, 2], ǫ = .01

(c) f(x) = x3 − 3x2 + 2x− 1 + 5 sin x x ∈ [−1, 2], ǫ = .05

(d) f(x) =x− 4 cos x

x2 + 5x ∈ [2, 4], ǫ = .02

(e) f(x) = x2 + 5 sin x x ∈ [−4, 4], ǫ = .2

(f) f(x) =x4 + 1

x2 + 2 + cosxx ∈ [0, π], ǫ = .03

(5) The figure below shows a rectangle inside the circle x2+y2 = 1.One side of the rectangle is formed by the x axis while twovertices lie on the circle. Find the sup and inf of the set ofpossible areas for the rectangle. Is the sup also a max? Is theinf also a min?


(x,y)(-x,y)

2 21=x +y

(6) Suppose that M is an upper bound for the set S. Supposealso that there is a sequence sn ∈ S such that M = limn→∞ sn.Prove that M = supS.

(7) Prove the converse of Exercise 6–i.e. suppose that M = supS.Prove that there is a sequence an ∈ S such thatM = limn→∞ an.Hint: For all n ∈ N , M − 1/n is not an upper bound.

(8) Suppose that M is a lower bound for the set S. Suppose alsothat there is a sequence sn ∈ S such that M = limn→∞ sn.Prove that M = inf S.

(9) Prove the converse of Exercise 8–i.e. suppose that M = inf S.Prove that there is a sequence an ∈ S such thatM = limn→∞ an.Hint: Modify the argument from Exercise 7.

(10) A sequence is said to be non-increasing if for all n, an+1 ≤ an.Prove that a non-increasing sequence either tends to −∞ orconverges. This is the Bounded Decreasing Theorem. Hint:Consider −an.

(11) If S is any set of numbers, we define

−S = {−x|x ∈ S}

Compute the sup and inf of −S for each set S in Exercise 9.How do the sup and inf of S relate to the sup and inf of −S?

(12) This exercise uses the notation from Exercise 11.(a) Let S be a set of numbers that is bounded from below by

a number B. Show that −S is bounded from above by−B.

(b) Let S be as in part (a). It follows from (a) that −S isbounded from above. The Least Upper Bound Theoremimplies that −S has a least upper bound L. Prove that−L is the GLB for S. For this you must explain why (i)−L is a lower bound for S and (ii) why there is no greaterlower bound.


Remark It follows from (b) that S has a GLB. Hence theabove sequence of arguments proves the GLB property.

(13) Let a1 = 1. Let a sequence an be defined by

(1) an+1 =√2an.

Thus, for example,

a2 =√2 · 1 = 1.414

a3 =√2 · a2 = 1.6818

a4 =√2 · a3 = 1.8340

(a) Compute a5, a6, and a7 (as decimals.) You should findthat each is less than 2.

(b) Prove that if an < 2 then an+1 < 2 as well. How does itfollow that an < 2 for all n?

(c) Prove that for all n, an+1 > an.(d) Explain how it follows from the Bounded Increasing The-

orem that limn→∞ an = L exists.(e) Prove that L =

√2L. What then is the value of L? Hint:

Take the limit of both sides of (1).(14) Let a1 = 1. Let a sequence an be defined by

(2) an+1 =√2 + an.

(a) Compute a2, a3, and a4 (as decimals.) You should findthat each is less than 2.

(b) Prove that if an < 2 then an+1 < 2 as well. How does itfollow that an < 2 for all n?

(c) Prove that for all n, an+1 > an.(d) Explain how it follows from the Bounded Increasing The-

orem that limn→∞ an = L exists.(e) Prove that L =

√2 + L. What then is the value of a?

Hint: Take the limit of both sides of (2).(15) The following exercise develops the “divide and average” method

of approximating r =√2. From r2 = 2, we see that r = 2/r.

If r1 is some approximation to r, then 2/r1 will be another.The average

r2 =1

2

(

r1 +2

r1

)

will (hopefully) be a better approximation. We repeat thisprocess with r2 in place of r1 to produce r3. In general, we set

(3) rn+1 =1

2

(

rn +2

rn

)

.


For example, if r1 = 2, then

r2 =1

2(2 + 2/2) = 1.5

r3 =1

2(r2 + 2/r2) = 1.416666667

r4 =1

2(r3 + 2/r3) = 1.414215686

.(a) Compute r5, r6 and r7.(b) Prove that r2n > 2. Hint: Write this as r2n+1 > 2 and use

formula (3) above together with some algebra.(c) Prove that for all n, 0 < rn+1 ≤ rn. How does it follow

from the Bounded Decreasing Property that limn→∞ rnexists.

(d) Show that r = limn→∞ rn satisfies the equation

r =1

2

(

r +2

r

)

.

Use this to prove that r2 = 2. Hint: Take the limit ofboth sides of formula (3).

Remark: This exercise proves the existence of√2.

CHAPTER 7

Positive Term Series

One very important goal in mathematics is computation. The num-ber π, for example, has been computed to thousands of decimals. Howis this done? It turns out that there are some remarkable formulasthat can be used to approximate π. For example, the following for-mula, which comes from the theory of Fourier series, can be used toapproximate π2/6 and, thus, π.

(1)π2

6= 1 +

1

22+

1

32+ · · ·+ 1

n2+ . . .

This formula means that we may approximate π2/6 as accuratelyas desired by summing sufficiently many of the terms on the right ofthe equality. Let sn be the sum of the first n terms on the right. Thus,for example, using 4 terms:

(2)π2

6≈ s4 = 1 +

1

22+

1

32+

1

42= 1.423

Using 6 terms produces

π2

6≈ s6 = 1 +

1

22+

1

32+

1

42+

1

52+

1

62= 1.4911

Ten terms produces

π2

6≈ s10 = 1 +

1

22+

1

32+ · · ·+ 1

102= 1.5497

Using 100 terms (and a computer) we find that

(3)π2

6≈ s100 = 1 +

1

22+

1

32+ · · ·+ 1

1002= 1.6349

The approximations we just produced are essentially useless unlesswe can determine their accuracy. We could, of course, simply com-pare them with the value of π2/6 computed using, say, a calculator.However, our goal is to understand how calculators and computers cancompute numbers such as π2/6. Hence we must assume that we haveno means of computing the answer other than using the series; we mustdetermine the accuracy without first knowing the value of π2/6.

97

98 7. POSITIVE TERM SERIES

For this we use geometry. The term 1/n2 is the length of the linesegment drawn vertically from n on the x-axis to the curve y = 1/x2

as indicated in Figure 1.

1/361/251/161/91/4

1

2y=1/x

2 4 5 6 7 8 91 3

Figure 1

2 3 4 5 6 7 8 91

1

1/4

1/91/16 1/25 1/36

Figure 2

We use each of these line segments as the right edge of a rectangleof width 1 as in Figure 2.

According to formula (1), the sum of the areas of all of the rectanglesis π2/6. The approximation (2) is the sum of the areas of the first 4rectangles. Thus, π2/6 − s4 is the sum of the areas of the rectanglesover the interval [4,∞). (See Figure 3.)

7. POSITIVE TERM SERIES 99

��

2 3 4 5 6 7 8 91

1

Graph not drawn to scale!

area=s area=S-s

area=S

4 4

Figure 3

Since y = 1/x2 decreases on x > 0, each rectangle lies entirelyunder the graph. Hence

(4)

π2

6− s4 <

∫ ∞

4

1

x2dx

= −1

x

∣

∣

∞4

=1

4= .25

Thus, we can guarantee that our approximation (2) is accuratewithin within ±.25

For n terms,

(5)0 <

π2

6− sn <

∫ ∞

n

1

x2dx

= −1

x

∣

∣

∞n

=1

n

Thus, for n terms, the error is at most 1/n. Thus the approximation(3) is accurate to within ±1/100. Indeed, our calculator tells us that

π2/6 ≈ 1.644934068

which is within ±10−2 of (3).To guarantee 10−8 accuracy, we need 108 terms. Even for a com-

puter, computing this many terms is out of the question. This tells


us that this series is not practical for computing π to high degrees ofaccuracy.

As mentioned previously, there are other formulas for π. One thatyou will analyze in the exercises is

(6)π4

90= 1 +

1

24+

1

34+ · · ·+ 1

n4+ . . .

This series converges considerably more rapidly than the precedingseries because 1/n4 tends to zero much faster than 1/n2.

In general, if an is any sequence, we define the sequence of partialsums sn by

sn = a1 + · · ·+ an =n

∑

1

ak

We then define the infinite sum of the an by

(7)∞∑

1

ak = limn→∞

sn = s

provided this limit exists. When analyzing a summation∑

ak, whenwe refer to “sn” and “s” we mean the above defined expressions.

More generally, in the summation

∞∑

no

ak

we define

sn =n

∑

no

ak

∞∑

no

ak = limn→∞

sn.

Thus the “n” in sn denotes the final index of summation, not the num-ber of terms being summed. Hence, for example, in the summation

∞∑

n=0

2n

we have

s3 = 20 + 21 + 22 + 23.


Remark: When we refer to “∑∞

1 an”, we are referring to sn, not an.Thus, for example the sequence

n+ 1

n

converges since its limit is 1. However, the series

∞∑

1

n+ 1

n

diverges since

sn =1 + 1

1+

2 + 1

2+

3 + 1

3+ · · ·+ n+ 1

n> 1 + 1 + 1 + · · ·+ 1 = n

which goes to ∞. This example demonstrates a general principle:

Proposition 1. If limn→∞ an 6= 0, then∑∞

1 an cannot converge.

Proof We note that

sn+1 − sn = an+1.

If follows from Exercise 36 in Chapter 4 that

limn→∞

an = limn→∞

an+1

= limn→∞

sn+1 − limn→∞

sn

= s− s = 0

proving our proposition.

The method we used to analyze the series (1) was based on thefollowing facts:

(1) an ≥ 0(2) There is a decreasing function f(x) such that an = f(n).(3) s =

∑∞1 an exists.

In this case, the same geometrical argument shows:

Theorem 1. Suppose f , an and s are as described in (a)-(c) above.Then

s− sn ≤∫ ∞

n

f(x) dx

It is important that in Theorem 1, f be decreasing to guaranteethat the rectangles all lie below the graph of f .


In the preceding example, the convergence of series (1) was given.We can prove the convergence using the Bounded Increasing Property.

Example 1. Prove the convergence of the series (1).


sn+1 = 1 +1

22+

1

32+ · · ·+ 1

n2+

1

(n+ 1)2

= sn +1

(n+ 1)2

It follows that sn+1 > sn so the sn form an increasing sequence. Fromthe Bounded Increasing Property, either limn→∞ sn = ∞ or limn→∞ snexists. We will prove that the limit is not ∞ by showing that sn < 2for all n.

First Attempt: From Figure 3,

sn <

∫ n

0

1

x2dx

= −1

x

∣

∣

n

0= − 1

n+

1

0

But 1/0 is nonsense. The problem is that in Figure 3, the area ofthe region above the first rectangle and below the graph is infinite.

Second Attempt: To avoid the problem with the first rectangle, webegin our sum with n = 2, Then, from Figure 3,

1

22+

1

32+ · · ·+ 1

n2<

∫ n

1

1

x2dx

<

∫ ∞

1

1

x2dx = −1

x

∣

∣

∞1

= 1

Hence

sn = 1 + (1

22+

1

32+ · · ·+ 1

n2) < 1 + 1 = 2

as claimed, proving convergence.

In this section we only consider “positive term series”, meaning thatan > 0 for all n. Such series are special because then sn is an increasingsequence. Specifically

sn+1 = sn + an+1 > sn


since an+1 > 0. It follows that if a positive term series does not sum to∞, it must converge. The student should be aware that non-positiveterm series have many other ways of diverging. (See Exercise 1 below.)

The following theorem generalizes the computation done in Exam-ple 1. The proof is just a repeat of the explanation of Example 1 andwill be omitted.

Theorem 2. Suppose an > 0 for all n and f(x) is an integrable,decreasing function on [0,∞) such that an = f(n) for all n ∈ N. Thens =

∑∞1 an exists if there is a k such that

∫ ∞

k

f(x) dx < ∞

An important consequence is the following theorem, which is leftas an exercise.

Theorem 3. The following series converges for all p > 1.

(8)∞∑

1

1

np

Here is another example of using integrals to prove convergence andapproximate the sum.

Example 2. Prove that the following series converges. Compute sto ±.001

(9) s =∞∑

1

1

n2 + 1

Solution: We interpret s as the sum of the areas of rectangles underthe graph of f(x) = 1/(x2 + 1) as shown in Figure 4.

To apply Theorem 2, we need to know that f(x) is a decreasingfunction. In this case this is true since y = 1 + x2 gets larger as xgrows, which implies that 1/(1 + x2) gets smaller as x grows.

Comparing areas, we see that

sn ≤∫ ∞

0

1

x2 + 1dx = arctan x

∣

∣

∞0

=π

2

since limx→∞ arctan x = π2. It follows that limn→∞ sn 6= ∞, proving

that the limit exists.


2 3 4 5 6 7 8 91

1/21/5

1/10 1/17 1/26

y=1/(x +1)2

Figure 4

To determine how many terms of the series are required to approx-imate s to within ±.001 we use Theorem 1:

s− sn ≤∫ ∞

n

1

x2 + 1dx = arctan x

∣

∣

∞n

=π

2− arctann

Hence, we wantπ

2− arctann < .001

π

2− .001 < arctann

tan(π

2− .001) < n

999.999 < n

(Note that applying the tangent function to the inequality is justifiedsince both it and the arctangent function are increasing.) Hence 1000terms suffice. We compute (using a computer) that

s ≈ 1

12 + 1+

1

22 + 1+ · · ·+ 1

(1000)2 + 1= 1.074

Hence, s = 1.074± .001.

In general, the rate at which the terms tend to zero determines howfast the series converges. In fact, it is possible for the sum to divergeif the terms go to 0 too slowly, as the next example shows.

Example 3. Prove that

(10) 1 +1

2+

1

3+

1

4+ · · ·+ 1

n+ · · · = ∞


Solution: Let us compute a few of the partial sums

s1 = 1

s2 = 1 +1

2= 1.5

s3 = 1 +1

2+

1

3= 1.888 . . .

s4 = 1 +1

2+

1

3+

1

4= 2.083333333

Thus, if the limit of sn exists, it is greater than 2.In fact, experimental evidence suggests that we can make the sum

as large as desired by summing sufficiently many terms. For example

s11 = 1 +1

2+

1

3+ · · ·+ 1

11≈ 3.019877345 > 3

s35 = 1 +1

2+

1

3+ · · ·+ 1

35≈ 4.146781419 > 4

Thus, we guess that the sum in (10) just keeps getting larger andlarger. To prove this, we interpret 1/n as the length of a line segmentdrawn vertically from n on the x-axis to the curve y = 1/x, producing afigure similar to Figure 1. Now, however, we interpret this line segmentas the left edge of rectangle of width 1. These rectangles now lie overthe curve as in Figure 5. Then sn is a sum of areas of the first nrectangles. Comparison of areas shows that

(11) 1 +1

2+ · · ·+ 1

n>

∫ n+1

1

1

xdx = ln(n+ 1)


2 3 4 5 6 7 8 91

1

Graph not drawn to scale!

1

1/41/3

1/2

R is the region formed by all of the rectangles

The n th rectangle has height 1/n

y=1/x

area=S

Figure 5

It follows that sn may be made larger than any given number M .Specifically, sn > M will hold if ln(n+1) > M which is equivalent with

n+ 1 > eM

We see, therefore, that the sum (10) tends to ∞, despite the factthat the terms being summed tend to zero. Intuitively, this says thata lot of very little things can still total to something big.

The reader should note that in Example 1, the number of termsnecessary for sn to rise above 1000 is astronomically large. This seriestends to ∞ at a very slow rate.

The method we used to analyze Example 1 also applies generalseries:

Theorem 4. Suppose f , an and s are as described in (a)-(c) aboveTheorem 1. Then

sn ≥∫ n+1

1

f(x) dx

Using the proceeding theorem, we can prove that the series inTheorem 3 diverges if p ≤ 1. Hence, for this type series, p = 1 isthe line between divergence and convergence.


Theorems 1, 2, and 4 are, together, referred to as the ”integral testfor convergence.” In practice, use of the integral test is complicatedby the facts that (i) proving that the function f(x) is decreasing canbe difficult and (ii) explicitly evaluation the integral of f(x) might beimpossible. Fortunately, there are other techniques.

Example 4. Determine whether or not the following sum convergesand prove your answer.

(12) s =∞∑

1

n

n3 + n lnn+ 5

Solution: The fastest growing term in the denominator is n3. Hence,the sum should behave like

∑∞1 1/n2 which converges from Theorem 3.

In fact, from the discussion at the beginning of this section, it convergesto π2/6.

To prove our answer, we note that

n

n3 + n lnn+ 5<

n

n3=

1

n2

Hence1

13 + 1 ln 1 + 5+

2

23 + 2 ln 2 + 5+ · · ·+ n

n3 + n lnn+ 5

<1

12+

1

22+ · · ·+ 1

n2<

π2

6

Since sn < π2/6, limn→∞ sn 6= ∞, proving convergence.

Remark: In the above example, we did not really need to know theactual value of

∑∞1 1/n2. All we required was that it not be infinity.

The work done in Example 4 illustrates the comparison test forconvergence:

Theorem 5. Suppose that 0 ≤ an ≤ bn for all n. Then∑∞

1 an willconverge if

∑∞1 bn converges.

As stressed above, it is important to know how many terms wemust sum to obtain a given accuracy of approximation. When usingthe comparison test, the rule for determining this information is simple:

Theorem 6. Suppose that in Theorem 5, the sum of the first N bnapproximates

∑∞1 bn to within ±ǫ. Then the same will be true for an:

i.e. the sum of the first N an will approximate∑∞

1 an to within ±ǫ.


Thus, in Example 4 we proved convergence by comparison with∑∞

1 1/n2. We saw using formula (5) that summing 100 terms of thisseries approximates π2/6 to within ±.001. Hence, the sum of the first100 terms of the series in Example 4 will approximate s to within±.001.Thus

s =1

13 + 1 ln 1 + 5+

2

23 + 2 ln 2 + 5+ . . .

+100

1003 + 100 ln(100) + 5= .647± .001

Proof (of Theorem 6)Looking, for example, at Figure 2, we see that the error in approx-

imating s by sk is

s− sk = ak+1 + ak+2 + . . .

Similarly, letting the nth partial sum of the bn be tn and t =∑∞

1 bn,we see that

t− tk = bk+1 + bk+2 + . . .

From an < bn we see that

s− sk < t− tk

Hence, t− tk < ǫ implies that s− sk < ǫ which proves Theorem 6.

It is also possible to use the comparison test to prove that a seriesdiverges.

Example 5. Does the following sum converge?

s =n

∑

1

1/√n

Solution: We note that for all natural numbers n, 1/√n ≥ 1/n. Hence

1 +1√2+ · · ·+ 1√

n> 1 +

1

2+ · · ·+ 1

n

From the work done in Example 2, this latter quantity tend to ∞ as ntends to infinity. Thus, the sum in this example diverges.

Thus, in Theorem 5, if∑

an diverges, then∑

bn will also diverge.Occasionally, one can find a formula for sn. In the exercises, you

will prove the following:

(13) (1− x)(1 + x+ x2 + · · ·+ xn) = 1− xn+1


which, for x 6= 1, yields

(14) 1 + x+ x2 + · · ·+ xn =1− xn+1

1− x

Example 6. The following sum diverges to ∞. What is the firstvalue of n such that sn > 1010?

s = 1 + 2 + 22 + · · ·+ 2n + . . .

Solution: From formula (14), with x = 2, we see

(15) 1 + 2 + 4 + · · ·+ 2n = 2n+1 − 1

Hence, we can find the desired n by the following sequence of in-equalities:

2n+1 − 1 > 1010

(n+ 1) ln 2 > ln(1010 + 1)

n+ 1 >ln(1010 − 1)

ln 2≈ 33.22

Hence the first n such that sn > 1010 is n = 33.

The next example uses formula (14) in a convergent series.

Example 7. Find the value of the following sum and prove (usingǫ) that your answer is correct.

s = 1 +1

3+

1

32+ · · ·+ 1

3n+ . . .

Solution: From formula (14) with x = 13we see that

sn = 1 +1

3+

(

1

3

)2

+ · · ·+(

1

3

)n

=1−

(

13

)n+1

1− 1/3

=3

2− 3

2

(

1

3n+1

)

Thus the limit is 3/2 = 1.5.


If we want to approximate the limit to within ±ǫ, we need

|[

3

2− 3

2

(

1

3n+1

)]

− 3

2| < ǫ

3

2

(

1

3n+1

)

< ǫ

which will be true if

n > − ln ǫ

ln 3− 1

Since an appropriate N exists for all ǫ > 0, the limit is proved.

Examples 6 and 7 demonstrate a general theorem.

Theorem 7. Let x be a real number. Then the series on the rightside of the following equality converges if, and only if, |x| < 1. Fur-thermore, when it converges, it converges to the stated value.

(16)1

1− x= 1 + x+ x2 + · · ·+ xn + . . .

Theorem 7 gives us another class of series to compare with: thoseof exponential growth or decay.

Example 8. Determine the convergence or divergence of the fol-lowing series and prove your answer.

s =∞∑

1

n2

4n

Solution: We think that 4n grows so much faster than n2 that theseries should converge. Specifically, there is an N such that

(17) n2 < 3n for all n > N .

In fact, the preceding inequality is equivalent with

2 lnn < n ln 3

which, from Proposition 2 in Chapter 3, with a = (ln 3)/2, is valid forn > 4/a2 = 13.2. Hence, for n > 13,

n2

4n<

3n

4n=

(

3

4

)n


Hence, for n > 13, using Theorem 7,

sn = s13 +142

414+

152

415+ · · ·+ n2

4n

< s13 +

(

3

4

)14

+

(

3

4

)15

+ · · ·+(

3

4

)n

< s13 +∞∑

0

(

3

4

)n

= s13 +1

1− 34

= s13 + 4

It follows that sn is an increasing sequence which cannot tend to infin-ity; hence converges.

Remark: We didn’t really need to find an N for which (17) holds; allwe needed was the existence of such an N . Our argument would havebeen the same, except that we would have used “N” in every placethat “13” occured. In fact, for any N ∈ N and any sequence an, wecan write

∞∑

1

an =N∑

1

an +∞∑

N

an

The issue of convergence doesn’t arrise for the middle sum since it isfinite. Hence the sum on the left converges if and only if thaton the right does. We use these ideas in the next few examples.

Example 9. Determine the convergence or divergence of the fol-lowing series.

s =∞∑

1

1

(lnn)2

Solution: Since lnn grows more slowly than any power of n there isa N such that lnn <

√n for n > N . Hence

∞∑

1

1

(lnn)2= sN +

∞∑

N+1

1

(lnn)2

> sN +∞∑

N+1

1

n

This latter sum tends to ∞ from Example 3 so our original sum di-verges.



∞∑

1

lnn

n3 − n+ 2

Solution: Our thinking is that the fastest growing term in the de-nominator is n3 while the numerator grows more slowly than n. Sincen/n3 = 1/n2 which has a finite sum, the series should converge. Forthe proof, we note that there are positive constants C and N such that

Cn3 < n3 − n+ 2 for n > N

Hencelnn

n3 − n+ 2<

lnn

Cn3

<n

Cn3=

1

Cn2

Hence,∞∑

N+1

lnn

n3 − n+ 2<

∞∑

N+1

1

Cn2=

1

C

∞∑

N+1

1

n2

which converges from Theorem 3.


∞∑

1

5n

4n + n2 + 1

Solution: Our thinking is that the fastest growing term in the denom-inator is 4n. Hence, the terms being summed grow like

5n

4n=

(

5

4

)n

which tends to ∞. Thus, the terms being summed don’t even tend tozero, making convergence impossible. For the proof, since exponentialgrowth is faster than power growth, there is an N such that for n > N

n2 + 1 < 4n

Hence for n > N ,

5n

4n + n2 + 1>

5n

4n + 4n=

1

2

(

5

4

)n

which tends to ∞, proving divergence.


Exercises

(1) For the following series compute sn for n = 1, 2, . . . , 5. Do theseries seem to be converging? What theorem from this sectionof the notes does this exercise illustrate?

(a)1

2+

2

3+

3

4· · ·+ n

n+ 1+ . . .

(b) − 1 + 1 + (−1) + · · ·+ (−1)n + . . .

(c)1

2+

−2

3+

3

4· · ·+ (−1)n+1n

n+ 1+ . . .

(2) How many terms of the series (6) on page 100 does it take toapproximate π4/90 to ±10−2? Compute this approximationand compare with the answer your calculator gives for π4/90.Hint: Repeat the argument done for π2/6 on page 97.

(3) Determine the convergence or divergence of the following se-ries. Prove your answers.(a)

∑∞1

1√1+n

(b)∑∞

11√

1+n2

(c)∑∞

11√

1+n3

(d)∑∞

1n√1+n3

(e)∑∞

1n2√1+n2

(f)∑∞

12√n

3n2+1

(g)∑∞

1n11+3n5+7√

n25+1

(h)∑∞

11

n√3n−1

(i)∑∞

1lnnn

(j)∑∞

1n2+n+5

n4−lnn+1

(k)∑∞

1n2n

(l)∑∞

1 ne−n

(m)∑∞

1 ne−n2

(n)∑∞

1lnnn2n

(o)∑∞

11+cosn

n2

(p)∑∞

132n

n!

(4) For each of the infinite sums∑

an from the preceding exercise,find all x > 0 for which

∑

anxn converges. We solve (a) as an

example:


Solution: (To (a))∞∑

1

nxn

2n=

∞∑

1

n(x

2

)n

If x2> 1 then

(

x2

)nexhibits exponential growth in n. Thus,

the sum cannot converge since the terms don’t tend to 0.If 0 < x

2< 1, then

(

2x

)nexhibits exponential growth in n.

Hence, there is an N such that for all n > N(

2

x

)n

> n3

Thus, for n > N ,

n(x

2

)n

<n

n3=

1

n2

Hence, the infinite sum converges. The sum clearly divergesfor x = 2, so the set of positive x for which the sum convergesis exactly (0, 2).

(5) Assume that it is given that y = f(x) is decreasing on [3,∞).(See Figure 5 below for a possible graph of f .). Let an = f(n)and s =

∑∞1 an.

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

4

Figure 5

(a) Find a specific value of n and m such that the followinginequality is guaranteed to hold. Choose n as small aspossible and m as large as possible, consistent with theinformation provided. Justify your answer with a dia-gram.

s− sn <

∫ ∞

m

f(x) dx


(b) Find a specific value of n and m such that the followinginequality is guaranteed to hold. Choose m as small aspossible and n as large as possible, consistent with the in-formation provided. Justify your answer with a diagram.

∫ ∞

m

f(x) dx < an + an+1 + . . .

(c) Find a specific value of n and m such that the followinginequality is guaranteed to hold. Choose n as small aspossible and m as large as possible, consistent with theinformation provided. Justify your answer with a dia-gram.

∫ ∞

m

f(x) dx < s− sn

(6) Assume that it is given that y = f(x) is is increasing on [0, 2.5]decreasing on [2.5,∞). (See Figure 5 above for a possiblegraph of f .) Let an = f(n) and s =

∑∞1 an.

(a) Find a specific value of n, a and b such that the followinginequality is guaranteed to hold. Choose both n and bas large as possible and a as small as possible, consistentwith the information provided. Justify your answer witha diagram.

a1 + a2 + · · ·+ an >

∫ b

a

f(x) dx

(b) Find a specific value of n, a and b such that the followinginequality is guaranteed to hold. Choose both n and aas large as possible and b as small as possible, consistentwith the information provided. Justify your answer witha diagram.

a1 + a2 + · · ·+ an <

∫ b

a

f(x) dx

(7) Assume that it is given that y = f(x) is increasing on [0, 5]and decreasing on [5,∞). (See Figure 6 below for a possiblegraph of f .) Let an = f(n) and s =

∑∞1 an.

(a) Find a specific value of n, a and b such that the followinginequality is guaranteed to hold. Choose both n and aas large as possible and b as small as possible, consistentwith the information provided. Justify your answer with


a diagram.

a1 + a2 + · · ·+ an <

∫ b

a

f(x) dx


.

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

Figure 6

(b) Find a specific value of n and a such that the following in-equality is guaranteed to hold. Choose a and n as small aspossible, consistent with the information provided. Jus-tify your answer with a diagram.

s− sn <

∫ ∞

a

f(x) dx

(8) Let

s =∞∑

1

3n2

2n + 1

(a) Find values of k and m such that the following inequalityis satisfied. Justify with a diagram. Choose k and m assmall as possible, consistent with your diagram.

∫ ∞

k

3x2

2x + 1dx > am + am+1 + . . .

(b) Let k and m be as in part (a) and let n ≥ k. Find p(depending on n) such that the following inequality holds.Choose p as large as is consistent with the diagram in (a).Justify your answer in terms of your diagram.

∫ p

k

3x2

2x + 1dx > am + am+1 + · · ·+ an

(9) Which of the following series converge and which diverge? Ifconvergent, approximate the sum within ±10−3. If divergent,determine a value of N so that sn > 100 for all n ≥ N .


(a)∑∞

23n2

(n3+1)4Reason as in Example (2) on page 103. The-

orem 1 on page 101 is since in this exercise the summanddecreases for n ≥ 2.

(b)∑∞

13n2 cos2 n(n3+1)4

(c)∑∞

11√

2n+5Reason as in Example 3

(10) In Exercise 9, part (c), prove using M that limn→∞ sn = ∞.(11) Let

s =∞∑

1

n2 +√n+ 1

n4 + 3n+ 7

(a) Prove that this series converges.(b) Write a sum which computes s to within ±10−3.

(12) Let

s =∞∑

1

n

en2/50

(a) Find a specific value of n and m such that the following

inequality is guaranteed to hold where f(x) = xe−x2/50.Justify your answer with a diagram. Hint: Use a graphingcalculator to graph f(x). You may assume that f(x) isdecreasing where the calculator so indicates.

s− sn <

∫ ∞

m

f(x) dx

(b) Find a specific value of n and m such that the followinginequality is guaranteed to hold. Justify your answer witha diagram.

∫ ∞

m

f(x) dx < an + an+1 + . . .

(c) Does this series converge? If so prove it and approximateits limit to within ±10−3. If not find a value of n for whichsn > 1000 for all n ≥ N .

(d) Prove convergence of the following series and determinehow many terms are necessary to approximate its sum towithin ±10−3. Hint: Use Theorem 7.

s =∞∑

1

n

4n2/50 + n+ 1


(13) For which p does

∞∑

2

1

np lnn

converge? Hint: For p 6= 1, consider rates of growth. Forp = 1, use Theorem 4. The integral may be evaluated withthe substitution u = ln x.

(14) Does the following sum converge? If so, compute its value towithin ±10−2. If not, find a value of n such that sn > 1000.

1 +1

23+

1

33+ · · ·+ 1

n3+ . . .

(15) Prove the convergence of the following sum and compute itsvalue to within ±10−2.

∞∑

1

1

n3 + 3n+ 1

(16) Diverge or converge? Prove your answer.

(a)∞∑

1

n3 − n2 + lnn+ 1

n6 + n5 + lnn+√n

(b)∞∑

1

(n lnn)(15)n

n!

(17) Prove the convergence of the following sum and compute itsvalue to within ±10−3.

s =∞∑

1

sin2 n

2n

(18) Use Theorem 4 to determine how many terms of the seriesfrom Example 5 are required for the sum to exceed 100.

(19) Find all values of x for which the following series converges.Prove your answer using rates of growth.

∞∑

1

nxn

3n lnn

(20) Let p > 1. Use Theorem 2 to prove Theorem 3. Supposethat p = 1.001. How many terms will approximate the sum towithin ±10−1?


(21) Let p < 1. Use Theorem 4 to prove divergence of the series (8).Suppose that p = .99. How many terms will make the sumlarger than 10?

(22) Let p > 1. Use Theorem 2 to prove convergence of the fol-lowing series. Suppose that p = 1.001. How many terms willapproximate the sum to within ±10−1?

∞∑

2

1

n(lnn)p

(23) Use Theorem 4 to prove that the series in Exercise 22 divergesfor p < 1. How many terms will make the sum larger than 10?

(24) The following exercise outlines a different proof of the diver-gence of

∞∑

1

1

n

For the proof, let sn = 1 + 12+ 1

3+ · · ·+ 1

n.

(a) Explain why s4 > 1 + 12+ 1

4+ 1

4= 1 + 1

2+ 2

4.

(b) Explain why s8 > 1 + 12+ 2

4+ 4

8.

(c) From similar lines of reasoning, how big will s16 be?(d) Find a value of n such that sn > 1000.(e) Prove that limn→∞ sn = ∞.

(25) For the following series(a) Compute s1, s2, s3 and s20. Hint: Use formula (14).(b) Find a value of n such that 3− sn < 10−100.(c) Prove, using ǫ, that limn→∞ sn = 3.

s =∞∑

0

(

2

3

)n

.

(26) For the following series(a) Compute s1, s2, s3 and s20. Hint: Use formula (14).(b) Find a value of n such that sn > 10100.(c) Prove, using M , that limn→∞ sn = ∞.

s =∞∑

0

(

3

2

)n

.

(27) Here is a proof of formula (14) for the n = 2 case:

(1− x)(1 + x+ x2) = (1− x)(1 + x) + (1− x)x2

= (1− x2) + (x2 − x3) = 1− x3


Next we use the n = 2 case to do n = 3:

(1− x)(1 + x+ x2 + x3) = (1− x)(1 + x+ x2) + (1− x)x3

= (1− x3) + (x3 − x4) = 1− x4

Next we use the n = 3 case to do n = 4:

(1− x)(1 + x+ x2 + x3 + x4) = (1− x)(1 + x+ x2 + x3) + (1− x)x4

= (1− x4) + (x4 − x5) = 1− x5

(a) Use a similar argument to do the n = 5 and n = 6 cases.(b) Use mathematical induction to prove the formula in gen-

eral.(28) Let

sn =1

1 · 3 +1

3 · 5 +1

5 · 7 + · · ·+ 1

(2n− 1) · (2n+ 1).

(a) Use mathematical induction to prove that sn = n2n+1

.(b) Compute

∞∑

1

1

(2n− 1)(2n+ 1)

(29) Let

an = 1 +1

2+

1

3+ · · ·+ 1

n− lnn

(a) Show that a1 = 1, a2 ≈ .8069, and a3 ≈ .7347. Computea4, a5 and a6. You should find that the values appear tosteadily decrease.

(b) Prove that

1

n+ 1< ln(n+ 1)− lnn

Hint: Use a single rectangle below the curve to bound∫ n+1

n

1

xdx

(c) Show that that for all n,

an+1 − an =1

n+ 1− (ln(n+ 1)− lnn)

Note that it now follows from (b) that an+1 − an < 0;hence the an decrease.

(d) Use formula (11) to prove that an > 0.


It now follows from the Bounded Decreasing Property thatγ = limn→∞ an exists and is less than .69. Hence, for large n

1 +1

2+

1

3+ · · ·+ 1

n≈ lnn+ γ

CHAPTER 8

Absolute convergence

The reader should note that all of the techniques demonstrated sofar are based on the Bounded Increasing Theorem, which requires thatthe an be non-negative.

Any series with positive and negative terms can be written as thedifference of two positive term series. For example,

(1)

1

12− 1

22+

1

32− 1

42+ . . .

=1

12+

1

32+

1

52+ . . .

−(

1

22+

1

42+

1

62+ . . .

)

Both of the positive series in the last equality converge since each isless than

π2

6=

1

12+

1

22+

1

32+ · · ·+ 1

n2+ . . .

The convergence of (1) follows.Notice that the preceding series is just the series on the left of the

equality in (1) with all minus signs changed to pluses. The argumentjust done generalizes to prove the following theorem.

Theorem 1. Let an be a sequence of real numbers. Then∑∞

1 anwill converge if

∑∞1 |an| converges.

Proof Let tn be the sum of the non-negative ai for i ranging from 1to n and let un be the sum of the negative ai for i in the same range.Then

sn = tn + un

Note that tn is increasing since it is a sum of non-negative terms. Forsimilar reasons, −un is increasing. (Note that if ai < 0, −ai > 0.)Furthermore, both tn and −un are less than

|a1|+ |a2|+ · · ·+ |an|+ . . .

123

124 8. ABSOLUTE CONVERGENCE

which is (by hypothesis) finite. It follows from the bounded increas-ing theorem that both tn and −un converge. Thus sn = tn − (−un)converges, proving the theorem.

Theorem 1 allows us to apply the notion of rates of convergence tonon-positive series.

Example 1. Does the following series converge?.

1

13 − 3−

√2

23 − 3+

√3

33 − 3+ · · ·+ (−1)n+1

√n

n3 − 3+ . . .

Solution: Replacing each term by its absolute value yields the series

1

2+

√2

23 − 3+

√3

33 − 3+ · · ·+

√n

n3 − 3+ . . .

Our thinking is that for large values of n,

√n

n3 − 3≈

√n

n3=

1

n2.5

indicating convergence.More precisely, there is are positive values of C and N such that

Cn3 < n3 − 3

for all n > N . Hence, for such n

√n

n3 − 3<

√n

Cn3=

1

Cn2.5

whose sum converges from Theorem 3 in Chapter 6, proving the con-vergence.

The rate at which the an tends to 0 is not, however, the whole story.Consider, for example,

(2) s =∞∑

1

(−1)n+1

n

On the basis of rates of growth one might expect this series to divergesince the an grow like 1/n which has infinite sum. Remarkably, how-ever, the sum actually converges. To see why, we compute a few values

8. ABSOLUTE CONVERGENCE 125

of sns1 = 1

s2 = 1− 1

2= .5

s3 = 1− 1

2+

1

3= .833

s4 = 1− 1

2+

1

3− 1

4= .583

s5 = 1− 1

2+

1

3− 1

4+

1

5= .783

s6 = 1− 1

2+

1

3− 1

4+

1

5− 1

6= .617

We make the following observations:

(1) The values follow a high-lower-higher pattern. Specifically,s1 > s2 < s3, s3 > s4 < s5 i.e. each odd numbered sumis greater than the next even numbered sum and eacheven numbered sum is less than the next odd num-bered sum.

(2) s1 > s3 > s5 while s2 < s4 < s6 i.e. the odd numberedsums decrease and the even numbered sums increase.

In Exercise 7, you will prove that these observations are indeed truefor all n. Granted them, we may prove the convergence as follows:

Since the even sums increase, either they have a limit or they tendto ∞. They cannot tend to ∞ since for n even

sn < sn−1 < s1

since the odd numbered sums decrease. Hence they converge.Similarly, the odd term must either have a limit or must tend to

−∞. They cannot tend to −∞ since, for n odd,

sn > sn+1 > s2

LetE = lim

n→∞s2n

O = limn→∞

s2n+1

be the limits of the even and odd terms.Since

s2n+1 = s2n +1

2n+ 1we see

O − E = limn→∞

(s2n+1 − s2n) = limn→∞

1

2n+ 1= 0

showing that E = O.


It now follows from Exercise 9 that s = limn→∞ sn exists and equalsboth E and O, showing convergence.

Remark: Since the odds decrease and the evens increase, we see thatfor n odd

sn ≥ s ≥ sn+1

Letting n = 5 we find, for example,.617 < s < .783. In general, since slies between sn and sn+1,

|s− sn| < |sn+1 − sn| =1

n+ 1

Thus, for sn to approximate s to within ±10−2, we require 1/(n +1) < 10−2 which corresponds to 100 terms. Approximating s to within±10−8 would require 100,000,000 terms! It turns out that the exactvalue of the sum is ln 2.

The argument used to analyze the series (2) generalizes to provethe following theorem, which is left as an exercise.

Theorem 2. Suppose that an is a positive, decreasing sequencewhere limn→∞ an = 0. Then

s =∞∑

1

(−1)nan

converges. Furthermore

|s− sn| < an+1

The series in Example 1 converges for relatively simple reasons: antends to 0 fast enough for

∞∑

1

|an|

to be finite. Such series are said to be absolutely convergent.The series (2) is not absolutely convergent since

∞∑

1

|an| =∞∑

1

1

n= ∞

Its convergence is due to the cancellation of the pluses and minuses inthe summation. A convergent series which is not absolutely convergentis said to be conditionally convergent. Conditional convergenceis a particularly unpleasant form of convergence. For example, the


commutative law fails for conditionally convergent series asthe next example shows.

Example 2. We commented that

1− 1

2+

1

3− 1

4+ · · ·+ (−1)n+1

n+ . . .

sums to a number less than 1. (ln 2 to be precise.) Show how torearrange the terms so as to sum to exactly 2.

Solution: Note that neither the odd nor the even numbered an havefinite sum:

1 +1

3+

1

7+ · · ·+ 1

2n+ 1+ . . . =

∞∑

1

1

2n+ 1= ∞

−1

2− 1

4− 1

6− · · · − −1

2n− . . . = −

∞∑

1

1

2n= −∞

We begin by summing just enough odd numbered an to make thesum ≥ 2. We find

1 +1

3+

1

7+ · · ·+ 1

17= 2.02 ≥ 2

Next, we subtract the first even numbered an

1 +1

3+

1

7+ · · ·+ 1

17− 1

2= 1.52

which brings us below 2. We knew that this would happen because ourodd sum can be at most 1/17 units above 2 and a2 = 1/2 > 1/17.

Next, we add on just enough additional odd terms to make the sum≥ 2:

1 +1

3+

1

7+ · · ·+ 1

41− 1

2= 2.004

We then subtract a4 which, as before, brings the sum below 2:

1 +1

3+

1

7+ · · ·+ 1

41

−1

2− 1

4= 1.754

We keep going:

1 +1

3+

1

7+ · · ·+ 1

69

−1

2− 1

4= 2.009


and

1 +1

3+

1

7+ · · ·+ 1

69

−1

2− 1

4− 1

6= 1.842

1 +1

3+

1

7+ · · ·+ 1

95

−1

2− 1

4− 1

6= 2.0007

and

1 +1

3+

1

7+ · · ·+ 1

95

−1

2− 1

4− 1

6− 1

8= 1.8757

In each step of this process, we use many of the odd numbered anand one of the even. Thus, we eventually use all of the an. At the nthstep the difference between the sum and 2 is at most the value of theeven numbered term that was subtracted. Since these terms tend tozero, this difference can be made less than any given ǫ, showing thatthe limit is indeed 2.

There is nothing special about 2. By rearranging the terms of (2)we can make the sum converge to and value we wish! There is alsonothing special about series (2) other than that it is only conditionallyconvergent: any conditionally convergent series can be arranged toconverge to any desired value!

The alternating series test is often used to answer questions suchas the following:

Example 3. Find all x for which the following series converges.For which x is the convergence absolute?

∞∑

0

xn

2n√n

Solution:

|an| =1√n

( |x|2

)n

For |x| > 2,(

|x|2

)n

exhibits exponential growth which is so much faster

than the decay of 1/√n that this should tend to∞ making convergence


impossible. In fact, there is a N such that for n > N ,

( |x|2

)n

> n

showing that for such n, |an| >√n, making convergence impossible.

For |x| < 2,

1√n

( |x|2

)n

<

( |x|2

)n

These terms have a finite sum, showing that our original series con-verges absolutely for |x| < 2.

If |x| = 2 then x = ±2. For x = 2 our series becomes

∞∑

0

1√n

which diverges. For x = −2 our series becomes

∞∑

0

(−1)n√n

which converges by the alternating series test. Hence, the series con-verges for x ∈ [−2, 2).

Exercises

(1) Decide which of the following series (a) converge absolutely(b) converge conditionally or (c) diverge. This exercise is tobe done by “inspection.” No proofs or reasons required (for


the moment).

(a)∞∑

1

(−1)n lnn

n

(b)∞∑

2

(−1)nn

lnn

(c)∞∑

1

(−1)nn

n+ 1

(d)∞∑

1

(−1)nn

n2 + lnn

(e)∞∑

1

(−1)nn

n3 + 1

(f)∞∑

2

cosn

n2 lnn

(g)∞∑

1

(−1)n3n

2n − n5 + 1

(h)∞∑

1

(−1)n2n

2n − n5 + 1

(i)∞∑

1

(−1)n(1.5)n

2n − n5 + 1

(j)∞∑

1

(−1)n(n3 −√n)

n3 + lnn

(k)∞∑

1

(−1)n(n3 −√n)

n3.001 + lnn

(2) Prove the answers given in Exercise 1.(3) For each of the series

∑

an in Exercise 1, find all x such thatthe series

∑

anxn converges. For which x is the convergence

absolute? Prove your answers.(4) For each of the series

∑∞1 an in Exercise 3 of Chapter 7, find

all x such that the series∑∞

1 anxn converges. For which x is

the convergence absolute? Prove your answers.


(5) Create 5 interesting examples (readers choice) of series whichconverge only conditionally. Do not use any of the series fromExercise 1 or from the text.

(6) Create 5 interesting examples (readers choice) of alternatingseries which diverge. Do not use any of the series from exercise1 or from the text.

(7) For the series (2):(a) Prove that for n odd, sn < sn+1 Hint: sn+1 = sn+?.(b) Prove that for n even, sn > sn+1 Hint: sn+1 = sn+?.(c) Prove that for n odd, sn > sn+2 Hint: sn+2 = sn+?.(d) Prove that for n even, sn < sn+2 Hint: sn+2 = sn+?.

(8) Suppose that for all n an > 0 and an > an+1. Show that(a)-(d) from Exercise 7 hold for

∑∞1 (−1)n+1an.

(9) Let bn be a sequence such that limn→∞ b2n and limn→∞ b2n+1

both exist and are equal. Prove, using ǫ, that limn→∞ bn exists.(10) Repeat the discussion from Example 3 to show how to rear-

range the series (2) to sum to 1. Do the first three iterations.(11) Modify the discussion from Example 3 to show how to rear-

range the series (2) to sum to -2. Do the first three iterations.Hint: A programmable calculator might help in computing thesums.

(12) Suppose that∑∞

0 anxn converges absolutely. Prove that

∑∞0 any

n

converges absolutely for |y| < |x|.(13) Suppose that

∑∞0 an converges, but not necessarily absolutely.

Prove that∑∞

0 anxn converges absolutely for |x| < 1. Hint:

From ? (you fill in ?) in Chapter 7, limn→∞ an = 0. Hence,there is an N such that for n > N , |an| < 1. (Why?)

(14) Suppose that∑∞

0 anxn converges, but not necessarily abso-

lutely, where x 6= 0. Prove that∑∞

0 anyn converges abso-

lutely for |y| < |x|. Hint: Apply the preceding exercise withan replaced by anx

n and x replaced by y/x.

Remark: It follows from this exercise that if∑∞

0 anxn doesn’t

converge for all x, then there is a number r (the radius ofconvergence) such that the series converges absolutely for|x| < r and diverges for |x| > r.

(15) Suppose that an is a positive, decreasing sequence such that∑∞

0 (−1)nan converges conditionally. For which x does∑

anxn

converge? For which x does the sum diverge? Explain.(16) (a) Suppose that there is an N such that for all n > N ,

|an|1/n < 12. Prove that

∑∞1 an converges absolutely.


(b) Suppose that limn→∞ |an|1/n = 1/4. How does it followfrom (a) that

∑∞1 an converges absolutely.

(c) Suppose that limn→∞ |an|1/n = c < 1. Prove that∑∞

1 anconverges absolutely.

(d) Use part (c) to prove that∞∑

1

(

n3 + 3

3n3 − 7

)n

converges.(e) Suppose that limn→∞ |an|1/n = c > 1. Prove that limn→∞ |an| =

∞, showing that∑∞

1 an cannot converge.

Remark: The theorem proved in (c) and (e) is called the roottest.

(17) (a) Suppose that for all n ≥ 0, |an+1| ≤ 12|an|. Prove that

∑∞0 an converges absolutely. Hint: Show that for all n ≥

0, |an| ≤(

12

)n |a0|.(b) Suppose that limn→∞

|an+1||an| = 1

4. Prove that

∑∞0 an con-

verges absolutely.

(c) Suppose that limn→∞|an+1||an| = c < 1. Prove that

∑∞0 an

converges absolutely.(d) Suppose that limn→∞ |an|1/n = c > 1. Prove that limn→∞ |an| =

∞, showing that the series cannot converge.

Remark: The theorem proved in (c) and (d) is called theratio test.

CHAPTER 9

Irrational Numbers

We are familiar with the equality

1

3= .3333333 . . .

This is really a statement about an infinite series. Explicitly, it states

1

3=

3

10+

3

102+ · · ·+ 3

10n+ . . .

=3

10

(

1 +1

10+

1

102+ · · ·+ 1

10n−1+ . . .

)

=3

10

(

1

1− 1/10

)

=1

3

where we used the formula for the sum of the geometric series in thelast equality.

In general, if an is a sequence of digits (i.e. an is a sequence ofintegers between 0 and 9) then we define

.a1a2a3 · · · =∞∑

1

an10n

This series converges. In fact, for each n

sn = .a1a2 . . . an < 1

Hence, limn→∞ an exists from the Bounded-Increasing Theorem. Henceevery infinite decimal you can write represents some specificnumber. Thus, for example,

a = .101001000100001000001000000100000001000000001000 . . .

represents a real number, where each 1 is followed by one more 0 thanthe previous 1.

Conversely, every real number is representable as an infinite deci-mal:

Theorem 1. Let x be a positive real number. Then there is anatural number n and sequence of digits ak such that

x = n.a1a2 . . . ak . . .

133

134 9. IRRATIONAL NUMBERS

Proof There is a non-negative integer n such that

n ≤ x < n+ 1.

This n is the integer part of the decimal expansion of x. Next, we dividethe interval [n, n + 1] into 10 subintervals, each of length 1/10. Sincex belongs to one of these subintervals, there is an integer 0 ≤ a1 ≤ 9such that

n.a1 ≤ x < n.a1 +1

10

Next, by dividing the interval [n.a1, n.a1 +110] into 10 subintervals,

each of length 1/100, we find that there is an integer a2 between 0 and9 such that

n.a1a2 ≤ x < n.a1a2 +1

100

Continuing, we produce a sequence of digits an such that

n.a1a2 . . . an ≤ x < n.a1a2 . . . an +1

10n

To prove that this sequence converges to x, let ǫ > 0 be given. Then

0 ≤ x− n.a1a2 . . . an <1

10n

|x− n.a1a2 . . . an| <1

10n

This will be less than ǫ provided 10−n < ǫ which is true if n > N whereN = − ln ǫ/ ln 10, proving our theorem.

A constant theme in our studies has been that, as much as possible,we should only use the axioms or their consequences in our proofs. Thepreceding proof used the fact that for every positive real number x,there is a natural number n such that x < n + 1. This is a versionof what is usually referred to as the Archimedean Property. Since itis not one of our axioms, we should either prove it or assume it asanother axiom. Remarkably, it is a consequence of the Least UpperBound Axiom. We have, in fact, already used this property severaltimes without comment. For example, in our solution to Example 1on page 84, we stated at one point “if n > 1

ǫ− 1, then 1 − ǫ < n

n+1

showing that 1−ǫ is not an upper bound for S.” It is the ArchimedeanProperty that guarantees the existence of such n. We leave it as anexercise (Exercise 16 on page 149 below) to explain how this propertyfollows from the GLB axiom.

9. IRRATIONAL NUMBERS 135

Theorem 2. For all numbers M there is a non-negative integer nsuch that

n ≤ M < n+ 1.

A decimal expansion is finite if eventually all of the an are zero.Thus, 1 = 1.0 and 2.74 are finite. Every non-zero number that hasa finite expansion is also has an infinite expansion. We claim, forexample, that

1.0 = .99999 . . .

In fact,

.9999 . . . =9

10+

9

102+ · · ·+ 9

10n+ . . .

=9

10

(

1 +1

10+

1

102+ · · ·+ 1

10n−1+ . . .

)

=9

10

(

1

1− 1/10

)

= 1

where we used the formula for the geometric series in the last equality.

Remark: Many students find the equality

(1) 1 = .99999 . . .

confusing. They argue that no matter how many 9’s we write after thedecimal, we will never reach 1. Hence, the best we can say is that

1 ≈ .9999 . . .

This is akin to saying that we should write

1 ≈ limn→∞

n

n+ 1

instead of1 = lim

n→∞

n

n+ 1since the values of n/(n + 1) never actually equal 1. The explanationis that the limit refers to the number being approximated and not tothe numbers that are doing the approximating. The preceding equal-ity means that 1 is the unique number that the fractions n/(n + 1)approximate as n gets large. Similarly, the equality (1) means that 1is the unique number that we approximate as we write more and more9’s after the decimal.

Dividing both sides of equation (1) by powers of 10 shows that:

.1 = .09999 . . .

.01 = .009999 . . .


and so forth. Hence

2.74 = 2.73 + .01

= 2.73 + .0099999 . . . = 2.7399999 . . .

This example illustrates the following general principle: every numberwhich has a finite decimal representation will also have an infinite dec-imal representation. It is a theorem (which we will not prove) that thenumbers that have finite expansions are the only ones whichcan have two expansions. For all other numbers the numbers anare unique.

In general, when we write a bar over a sequence of digits, we meanthat the sequence repeats forever. For example

2.736 = 2.73636363636 . . . .

We say that such an expansion is a repeating expansion.

Example 1. Express 2.736 as a fraction.


2.73636363636 . . . = 2.7 + .03636363636 . . .

= 2.7 +1

10.3636363636 . . .

Furthermore,

.36 =36

100+

36

1002+ · · ·+ 36

100n+ . . .

=36

100

(

1 +1

100+

1

1002+ · · ·+ 1

100n−1+ . . .

)

=36

100

1

(99/100)=

4

11

Hence, our answer is27

10+

1

10

4

11=

301

110

Recall that a number x is said to be rational if x = p/q where pand q are integers with q 6= 0. For example 3/7, 4 = 4/1, and −5/3are all rational. A number that is not rational is irrational. Examplesinclude π, e and

√2.

The general repeating decimal may be converted into a fractionusing the same technique as in the preceding example. The result will


be rational due to the observation that if x is rational, then

1 + x+ x2 + · · ·+ xn + · · · = 1

1− x

is also rational. Hence every repeating expansion represents a ra-tional number. Conversely, the expansion of a rational numbermust repeat. (The exercises explore why.)

Remark: Students often describe irrational numbers as “those num-bers whose decimal expansions go on forever.” This is incorrect. Theexpansion 2.736363636 . . . “goes on forever” and yet represents the ra-tional number: 4

11. The proper statement is that an irrational number

is a number whose decimal expansion does not repeat. It should benoted that numbers with finite expansions do repeat. For example2.7 = 2.70.

One very important irrational number is√2. Here is a proof of its

irrationality.

Theorem 3. The number√2 is irrational.

Proof If√2 were rational, then there would be integers p and q such

that√2 = p/q. We assume that the fraction p/q is in lowest terms so

that p and q have no common factors.Note that

2 = (√2)2 =

p2

q2

Hence

p2 = 2q2.

This makes p2 even. Hence p must be even, since the square of an oddnumber would be odd. Thus p = 2k where k is an integer. Substituting:

4k2 = 2q2

2k2 = q2.

Thus, q must be even also. But this contradicts the assumption that pand q have no common factors, proving our theorem.

Remark: The irrationality of√2 was discovered by the early Greek

mathematicians who proved that the length of the hypotenuse of anisosoles right triangle was an irrational multiple of the side lengths.This actually precipitated a crisis in Greek mathematics. Much of


Greek mathematics, including the theorems about parallel lines cut-ting transversals, was based on an assumption (called commensurabil-ity) which was equivalent to the statement that only rational numbersexisted. When this was found to be wrong, all of mathematics seemedto be coming apart at the seams. For many Greek mathematicians, thiswas more than a loss of a career. It was also a failure of their religion,since a number of them felt that mathematics was the medium throughwhich God spoke to them. It is reported that some even committedsuicide! Eventually, the crisis was resolved by Eudoxus who said, inessence, that between any two real numbers there is a rational num-ber. With this additional axiom, they were able to correct the proofsof their theorems.

Once we know one irrational number, we can produce as many aswe want.

Example 2. Prove that x = 23+ 4

5

√2 is irrational.

Solution: We work by contradiction, showing that assuming that x isrational leads to nonsense. Specifically, suppose that there are integersp and q such that

p

q=

2

3+

4

5

√2

We solve the above equality for√2:

p

q− 2

3=

4

5

√2

5

4

3p− 2q

3q=

√2

15p− 10q

12q=

√2

Since p and q are integers, 15p− 10q and 12q are both integers. Thus,√2 is a ratio of integers, which is nonsense since

√2 is irrational.

The argument just described generalizes to the following resultwhich you will prove in the exercises:

Proposition 1. Let Z be an irrational number and let x and y berational numbers with x 6= 0. Then xZ + y is irrational.

There are so many irrational numbers that there is an irrationalnumber between every pair of rational numbers.


Theorem 4. Let x and y be rational numbers with x < y. Thenthere is an infinite number of irrational numbers z satisfying x < z < y.

Proof Since limn→∞√2/n = 0, there is an N such that

√2

n< y − x

for all n > N . For such n

x <

√2

n+ x < y

From Theorem 1,√2

n+ x is irrational. This proves the theorem.

It is also true that between any two irrational numbers there arean infinite number of rational numbers.

Theorem 5. Let x and y be irrational numbers with x < y. Thenthere is an infinite number of rational numbers z satisfying x < z < y.

Proof We may write

y = M + .a1a2a3 . . .

where M is an integer and, of course, .a1a2a3 . . . is a decimal. Then

y = limn→∞

(M + .a1a2 . . . an).

Since y > x, there is an N such that for all n > N

M + .a1a2 . . . an > x

each of the numbers on the left side of the inequality is rational andlies between x and y, proving our theorem.

There are an infinite number of both irrational and of rational num-bers. However, there is a very real sense in which the set of irrationalsis vastly larger than the set of rationals. Below, we have begun a listwhich will eventually include all all rational numbers in the interval(0, 1):

1

2

1

3

2

3

1

4

3

4

1

5

2

5

3

5

4

5

1

6

5

6. . .

The pattern here is that we list fractions by increasing value of thedenominator. For a given value of denominator, we go from smallest tolargest, omitting fractions which are not in reduced form. Let us think


of the elements of this sequence as being expressed as decimals wherewe use the finite expansion whenever we have a choice:

(2)

a1 =.33333333 . . .

a2 =.66666666 . . .

a3 =.50000000 . . .

a4 =.75000000 . . .

a5 =.20000000 . . .

. . .

On the other hand, listing the irrational numbers in (0, 1) is, weclaim, quite impossible. To see this, let us imagine that we have some-how managed to list all irrationals in this interval. Our list might looksomething like:

b1 =.31415027 . . .

b2 =.14936815 . . .

b3 =.22719664 . . .

b4 =.97652234 . . .

b5 =.62718891 . . .

...

We imagine the decimal expansions extending out to infinity and thelist extending down the page to infinity. We claim that no matter whatthe specific numbers in the list, there will always be some irrationalnumber r which is not in the list. To see this, look at the first digit ofthe first number in the list. In our case, it is a three. We choose somenumber between 0 and 9, other than 3, and make it be the first digit ofr. Lets choose 4, so r = .4+. This insures that r 6= b1. Next, we lookat the second digit of the second number on the list: 4. We change it,declaring, say, r = .47+. This guarantees that r is also not equal tob2. We continue this way, at each step choosing the nth digit of r to besome some number between 0 and 9 which differs from the nth digit ofbn. For the list above, r might look like r = .47647+. It is clear thatin this manner we produce a number r which appears nowhere on ourlist.

We also must be careful that the number r we produce is not ratio-nal, since we claimed that there is an irrational number not on the list.For this, all we need do is to guarantee that r does not appear on thelist of rationals on page 140. This is easily accomplished; we simplyselect the nth digit of r so that it also differs from the nth digit of an.We also want to avoid selecting a 9. This is to avoid the problem of


non-uniqueness of representation of rational numbers. Specifically, thenumber .27999 . . . appears on the list of rationals as 2.28. If we were tochoose r = .279999 . . . , then r appears on the list of rational numbers,even though its expansion is different from all of the listed expansions.Once this is done, r appears on neither list; hence our list has excludedat least one irrational. In fact, by making different selections of thedigits, we prove that our list excludes an infinite number of irrationals.

Sets of numbers which may be listed are called countable. Thosewhich cannot are called uncountable. There is a very close relation-ship between these notions and the notion of counting. Consider, forexample, what happens if you ask a kindergartner how many objectsthere are in the set S = {+, ∗,×}. She might first hold up one finger,then another and finally a third. She would then say ‘T’ree’. What shehas done is set up a correspondence between the elements of the set Sand the first three fingers on her hand:

First Finger → +

Second Finger → ∗Third Finger → ×

Her correspondence is one-to-one; each finger corresponds to a differentsymbol. Her correspondence is ‘onto’; she hasn’t left out any symbols.For her, a set has three objects if she can find correspondence betweenthe first three fingers on her hand and the objects of the set which isboth one-to-one and onto.


When we listed the rational numbers in the interval (0, 1), we setup a correspondence between the natural numbers and the rationals:

1 → 1

2

2 → 1

3

3 → 2

3

4 → 1

4

5 → 3

4

6 → 1

5

7 → 2

5

8 → 3

5. . .

It is as if we had a hand with an infinite number of fingers (one foreach natural number) and we were using our infinite number of fin-gers to count the rationals in (0, 1). Notice that each of our ‘fingers’corresponds to a different rational number. (This is why we omittednon-reduced fractions from our list.) Thus our correspondence is one-to-one. Each rational in the interval eventually gets counted. Thus ourcorrespondence is onto. In the case of the real numbers, our infinitenumber of fingers was not enough to get the job done. No matter howwe try, we always have uncounted real numbers. It is as if we askedour kindergartner to count a set with six objects. She might reply “Ican’t do it. Don’t got enough fingers on my hand.”

Let us raise the level of the discussion a little. What, mathemati-cally, do we mean by a “correspondence?” We know that a sequence isjust a function whose domain is the set of natural numbers. Thus, ourlisting of the rational numbers is just a function whose domain is thenatural numbers and whose range is contained in the set of rationalnumbers.

In general, a function f from a set A to a set B is a correspon-dence between points of A and points of B with the property that toeach point a of A there corresponds a well defined point f(a) of B.In elementary calculus, the set B is almost always a set of numbers,although, in general, it can be any set. The set A is called the domain


of the function. The range of f is the set of points f(a). It is oftendenoted by the symbol f(A). The function is said to map A onto B ifB = f(A). In counting, this means that each point of B gets counted.The function is said to be one-to-one if for all y in the range of f , thereis only one x in the domain such that f(x) = y–i.e. if f(a) = f(b), thena = b. In terms of counting, this means that each point of B is countedonly once. The function f is said to be a one-to-one correspondenceif it is both one-to-one and onto.

A Ba

b

Range

f(a)=f(b)

Not one-to-one, not onto.

Thus, we arrive at the following definition:

Definition 1. An infinite set B is countable if there is a one-to-one function f whose domain is the set N of natural numbers andwhose range is B. All finite sets are also countable.

More generally, if A and B are two sets, then we say that A hasthe same cardinality as B if there is a one-to-one function f whosedomain is A and whose range is B.

Example 3. Prove that the intervals A = (−1, 1) and B = (1, 5)have the same cardinality by finding an explicit one-to-one correspon-dence. Prove your answer.

Solution: We hope for a correspondence of the form y = ax + b. Weassume that the points correspond somewhat as in the diagram below.

Then f(−1) = 1 and f(1) = 5. Hence

1 = a(−1) + b

5 = a+ b

Solving this pair of equations yields a = 2 and b = 3; hence the corre-spondence is

(3) y = 2x+ 3


−1

5y

x 1

1

Figure 1. Corresponding intervals

Proof:

Into: We first show that f(x) maps A into B. Assume x ∈ A. Then

−1 < x < 1

−2 < 2x < 2

1 < 2x+ 3 < 5

1 < y < 5

Hence f maps (−1, 1) into (−1, 5).

Onto:Reversing the preceding argument shows that if −1 < y < 5 then

x ∈ (−2, 1). Hence f maps (−2, 1) onto (−1, 5). 3 pt.

One-to-one: Suppose that f(x1) = f(x2). Then

2x1 + 3 = 2x1 + 3

2x1 = 2x2

x1 = x2

showing one-to-one.In solving the preceding example we could have use B as the do-

main of our correspondence and A as the range. Hence, instead ofcorresponding x to y in Figure 1, we correspond y to x. We obtaina formula defining such a correspondence by solving equation (3) onpage 143 for x:

(4) x =1

2(y − 3).

In general, if f : A 7→ B is one-to-one and onto, then for everyy ∈ B, there is one, and only one, x ∈ A such that f(y) = x. We writex = f−1(y). The function f−1 : B 7→ A is the inverse of f(x). Thus,


formula (4) says that the function g(y) = 12(y− 3) is the inverse of the

function f(x) = 2x+ 3.

Remark. Two functions f and g are said to be equal if they havethe same domain, the same range, and f(x) = g(x) for all x in thedomain. The particular symbols we use to denote the elements ofthe domain and range have no intrinsic meaning. Thus, the followingformulas both define the inverse to the function defined by formula (3)on page 143, provided that it is understood that the domainand rangeare, respectively, (1, 5) and (−1, 1):

g(y) =1

2(y − 3)

g(x) =1

2(x− 3).

The following proposition tells us that if A has the same cardinalityas B, then B has the same cardinality as A. Its proof is an exercise.(Exercise 24 on page 153.)

Proposition 2. Suppose that f : A 7→ B is one-to-one and onto.Then f−1 : B 7→ A is one-to-one and onto.

Remarkably, when we get to infinite sets, this notion of size becomesvery unintuitive. For example, it is easy to define a one-to-one corre-spondence between the natural numbers and the even natural numbers.We simply define f(n) = 2n. Hence,

1 → 2

2 → 4

3 → 6

4 → 8

. . .

Thus, the set of all natural numbers has the same cardinality as theset of all even natural numbers, despite the fact that the even numbersare a proper subset of the set of all natural numbers!

This causes many people immense difficulty. They absolutely refuseto accept that the set of even natural numbers could be the samesize as the set of all natural numbers. They are not questioning ourmathematics; the fact that we have a one-to-one function whose domainis N and whose range is the even natural numbers is clear. What theyare objecting to is our interpretation of our mathematics. They aretelling us that they cannot accept any notion of size which allows a set


S to have the same size as one of its proper subsets. This is O.K. Ifthe term ‘size’ bothers you, call it ‘one-to-one correspondence’.

Remark: The notion of cardinality was introduced by the mathe-matician Georg Cantor in 1874. At the time, it was quite controversial.Eventually, however, the notion was found to be very useful and gainedwide acceptance.

Exercises

(1) Use an infinite series to express the following as fractions:

(a) .68

(b) 2.7468

(c) .7324

(2) Let x = .68. Use the decimal expansion to explain why 100x =68 + x. Use this information to express x as a fraction.

(3) Express 2.7468 as a fraction using the technique of Exercise 2.(4) Express 47.56789 as a fraction using the technique of Exer-

cise 2.(5) Use the technique of Exercise 2 to prove that .9 = 1.(6) If we compute 4/13 on a hand held calculator with an 8 digit

display, we get

4

13≈ .30769231

which certainly does not seem to repeat. Our calculator is notcapable of displaying enough digits for us to see the repetition.There is, however, a cleaver ‘trick’ for getting our calculatorto produce as many digits as we want. We first multiply theabove equation by 10000000 getting

(5)40000000

13≈ 3076923.1.

The decimals for this number are the same as those for 4/13,just shifted over seven places.

Formula (5) tells us that 13 goes into 40000000, 3076923times with some remainder. To find the remainder, we com-pute (using our calculator!)

40000000− 13 · 3076923 = 1.

Hence

40000000/13 = 3076923 + 1/13.


Now, comes the main point: 1/13 represents the part of for-mula (5) which is after the decimal point. The decimal expan-sion of 1/13 will yield the 8 digits of the expansion of 4/13after .3076923. Using our calculator we find

(6)1

13= .07692308.

Hence

4/13 ≈ .307692307692308.

It seems clear that what we are obtaining is .307692. Justto be sure, though, let’s compute another batch of digits. Wemultiply equation (6) by 10000000 getting

10000000

13= 769230.8.

The remainder is

10000000− 13 · 769230 = 10

Hence100000000

13= 76923076 +

10

13.

The decimal expansion for 10/13 is .76923077 which yields thenext 8 digits of 1/13; hence the next 8 digits for 4/13. Thus

4/13 ≈ .3076923076923076923077.

This certainly seems to confirm the repetition. Of course, ifwe wish to be absolutely certain, we could use the technique ofExercise 2 to express this repeating decimal as a fraction andsee if we really get 4/13.

Finally, we are ready for the problem!(a) Use the calculator technique described above to compute

the decimal expansion of 7/17 until it starts repeating.(b) Using long division, compute the decimal expansion of

3/7 until it repeats. Explain how you know that it willkeep repeating. Explain why the repetition cycle in p/7is at most 7 for any natural number p.

(c) More generally, explain why the repetition cycle in a/n isat most n where a and n are natural numbers.

(7) Prove that (a) the square of an even number is even and (b)the square of an odd number is odd. Hint: An integer n iseven if n = 2k for some integer k. It is odd if n = 2k + 1 forsome integer k.


(8) The concepts of even and odd group all integers into twoclasses: those of the form 2k and those of the form 2k + 1.We can also divide the set of integers into 3 classes: those ofthe form 3k (class 0), those of the form 3k + 1 (class 1) andthose of the form 3k + 2 (class 2).(a) Determine the class of each of the following:

0,1,2,3,4,5,6,7,8,9,10,11.(b) Prove that the square of a class 0 number is class 0.(c) Prove that the square of a class 1 number is class 1.(d) Prove that the square of a class 2 number is class 1.

Note that there are no numbers whose square is class 2.Hence a class 2 number can never be a perfect square.

(9) This exercise in a continuation of Exercise 8.(a) Prove that if p in an integer and p2 is class 0, then p is

class 0. (Hint: Explain why p cannot be class 1 or 2.) Itfollows that p = 3k for some integer k.

(b) Prove that√3 is irrational. Hint: Repeat the proof of

Theorem 3, using class instead of even and odd.(10) Let x and y be rational numbers. Prove that the following

numbers are all rational. You may assume that the productand sum of two integers is an integer. (In part (a) we alsoassume y 6= 0 to avoid division by 0.) We prove (a) as anexample.

(a) x/y

(b) x+ y

(c) x− y

(d) xy

Proof of (a) Since x and y are rational, there are integers a,b, c, and d such that x = a/b and y = c/d. Then

x/y =a/b

c/d=

a

b

d

c=

ad

bc

This represents a rational number since both ad and bc areintegers.

(11) Using the technique of Example 2 on page 138, prove that thefollowing numbers are irrational. You may not use Propo-sition 1. You may, however, assume that π and

√3 are ir-

rational. You may also assume that the sum, product, andquotient of any two rational numbers is rational.(a) 4

5π


(b) 49− 17

11

√3

(c) 13−5

√2

(d) y + x√2 where x and y are rational numbers with x 6= 0.

(e) y+ xZ where x and y are rational numbers, x 6= 0 and Zis an irrational number.

(f)√2 +

√3 Hint Let x =

√2 +

√3 and assume that x is

rational. Ust the rationality of x2 to prove that√6 is ra-

tional. Hence both x and x√6 = 2

√3+3

√2 are rational.

So?(12) Only one of the following “theorems” is true. Prove the true

theorem and find an example showing the falsehood of each ofthe others.(a) The sum of two irrational numbers is irrational.(b) The product of two irrational numbers is irrational.(c) If x+y is irrational, then either x or y must be irrational.

(13) (a) Find a rational number between√101 and

√102. Ex-

press your answer as a fraction. (Hint: Use the decimalexpansions.)

(b) Find a rational number between ln 101 and ln 102. Ex-press your answer as a fraction.

(14) Find an explicit irrational number between 517578

and 519577

. Hint:Look at the proof of Theorem 4.

(15) Find an explicit irrational number between 31664799

and 31654797

. Hint:Look at the proof of Theorem 4.

(16) (a) Prove that for each M ∈ R, M ≥ 0, there is an m ∈N such that m ≥ M . Hint: Assume that this is false.Explain how it follows that the set of natural numbersis bounded from above. Let s = supN. Then there is anatural number n satisfying s − .5 < n ≤ s. (Explain!)What does this say about n+ 1?

(b) Let x = inf{n ∈ N | n ≥ M}. Then there is an n ∈ N

such that x ≤ n < x+ .5. (Why?) Hence n− 1 < x. Howdoes it follow that n − 1 < M . How does Theorem 2 onpage 135 follow?

(17) Find a one-to-one correspondence between the set of naturalnumbers and the set of multiples of 3.

(18) Describe a listing of the integers. This shows that the set ofintegers has the same cardinality as the set of natural numbers.

(19) The picture below describes a method of listing all positiverational numbers. We simply follow the indicated path, listingeach rational number in the order we hit it. Let us call the


resulting sequence an. Thus, a1 = 1, a2 = 2, a3 = 1/2, a4 =1/3, etc.

1 2 3 4 6

1/2 2/2 3/2 4/2 5/2 6/2

1/3 2/3 3/3 4/3 5/3 6/3

1/4 2/4 3/4 4/4 5/4 6/4

1/5 2/5 3/5 4/5 5/5 6/5

5

Counting the rationals

(a) What is the first n such that an = 5/3?(b) This is not a one-to-one correspondence. Demonstrate

this by finding values of n and m with n 6= m such thatan = am = 2/3.

(c) We may obtain a one-to-one correspondence by omittingall non-reduced fractions. What is the 14th fraction inthis reduced list?

(d) Describe a listing of the set of all rational numbers, in-cluding negatives and 0.

(20) The Hotel Infinity has an infinite number of rooms numbered1,2,3,... All rooms are occupied. A guest comes in and asksfor a room. You respond, “No problem. We’ll just move eachguest over one room, so that the guest in room 1, moves toroom 2, the one in room 2 moves to room 3, etc., leaving room1 free for you.”(a) Next a bus with 10 people shows up. How do you accom-

modate them?(b) A bus with an infinite (but countable) number of guests

shows up. How do you accommodate them?(c) Now (HELP!) an infinite (but countable) number of buses,

each with an infinite number (but countable) of guestsshows up. You can still accommodate all of them. How?

(21) You might think that large intervals contain more points thansmall intervals. Not so. Find an explicit one-to-one correspon-dence between the given intervals.(a) (1, 2) and (3, 7). (Hint: Try a function of the form f(x) =

ax+ b.)(b) (0, 1) and (1,∞)(c) (0, 1) and (0,∞)(d) (−∞,∞) and (−1, 1).


(22) Find a one-to-one correspondence between [0, 1] and (0, 1).Hint: Using lists similar to that on p. 139, describe a one-to-one correspondence between the rational numbers in [0, 1]and those in (0, 1). What is left?

(23) Well surely there are more points in a square than in an in-terval. NOT SO! Here is a function f whose domain is theinterval [0, 1] and whose range is the square S = {(x, y) |0 ≤x ≤ 1 and 0 ≤ y ≤ 1}.

0 1 0 1

1

f

a

f maps the interval onto the square!

f(a)=(x,y)

Let a ∈ [0, 1]. Write a as a decimal as a = .a1a2a3 . . . .We shall stipulate that, for the purposes of computing ourfunction, if a has two expansions, then we shall always choosethe expansion with an infinite number of 9’s. Thus, if a = .27,we choose the expansion a = .269.

We define

f(a) = (x, y)

where

x = .a1a3a5 . . . a2n+1 . . .

and

y = .a2a4a6 . . . a2n . . .

(Here, (x, y) is a point in R2, not an open interval in R!) Thus,for example

f(7/55) = f(.1272727 . . . )

= (.1777 . . . , .222 . . . )

= (8/45, 2/9)

f(.27) =f(.26999 . . . )

= (.2999, .6999)

= (.3, .7) = (3/10, 7/10)

(a) Compute f(1/n) for 1 ≤ n ≤ 6. Express the answer infractional form. Remember to use the expansion with aninfinite number of 9’s if you have a choice!


(b) Find a values of a and b such that f(a) = (1/3, 1/4) andf(b) = (1, 1/3). Remember to use the expansion with aninfinite number of 9’s if you have a choice!

(c) Prove that the range of f is the whole square.(24) Prove Proposition 2 on page 145.

CHAPTER 10

Limits of Functions

In Chapter 4 we described the limit of a sequence in terms of ap-proximations. We may describe limits of functions similarly.

Example 1. The Acme Geometry Company, manufacturers of pre-cision geometric figures, has received an order for a batch of squares,each with area 9 square inches, ±10−6. How accurately must the sidesof the squares be cut?

Solution: Let s be the side length of our square. To be within toler-ance, we require

9− 10−6 < s2 < 9 + 10−6

Taking square roots:

(1) 2.999999833 · · · < s < 3.000000167 . . .

This will certainly be true if

2.9999999 < s < 3.0000001

Thus, it suffices to make our cut accurate to within ±10−7.

It is rare that we can explicitly solve the inequalities as was donein Example 1. The following example shows how to deal with morecomplicated cases.

Example 2. We wish to make a rectangular box with volume 6±.1cubic units by cutting a square of side length 2 from each corner of a7×5 piece of paper and bending up the sides. How accurately must thesides of the squares be cut? (Assume that all 4 of the cut out squaresare identical.)

Solution: According to Figure 1, the volume V of the box is

V (x) = (5− 2x)(7− 2x)x = 4x3 − 24x2 + 35x

155

156 10. LIMITS OF FUNCTIONS

5-2x xx

xx 5-2x

7-2x

7-2x

xxx

x

7-2x

5-2x x

Figure 1

where x is the side length of each square. We need to find a numberδ > 0 such that

(2) |V (x)− 6| < .1

whenever x = 2± δ. We note that

V (x)− 6 = 4x3 − 24x2 + 35x− 6

= (x− 2)(4x2 − 16x+ 3)

We refer to the (x − 2) factor as the “gold” since it is small for xnear 2. We refer to the (4x2 − 16x + 3) factor as the “trash” since itdoes not help in making |V (x)− 6| small. Our first step will be to finda specific number B such that for all x sufficiently close to 2,

|4x2 − 16x+ 3| ≤ B.

(We refer to this as “bounding the trash.”)Specifically, suppose that x = 2± 1–i.e. 1 < x < 3. Then

|4x2 − 16x+ 3| ≤ 4|x|2 + 16|x|+ 3

< 4 · 9 + 16 · 3 + 3 = 87

Hence, for x = 2± 1,

|V (x)− 6| = |x− 2| |4x2 − 16x+ 3| < 87|x− 2|Hence, inequality (2) holds if

87|x− 2| < 10−1

|x− 2| < 10−1

87

Hence we can set δ to be the smaller of 1 (to guarantee that x = 2± 1)

and 10−1

87–i.e. δ = 10−1

87.

10. LIMITS OF FUNCTIONS 157

Remark: Examples 1 and 2 demonstrate a type of approximationproblem in which we are given (1) a function y = f(x), (2) a number L(the target value), (3) a number ǫ > 0 (the tolerance) and (4) a valuea on the x-axis. Our goal is to find a number δ > 0 (the confidencevalue)such that

f(x) = L± ǫ

whenever x = a±δ. In Example 1 our target value was 9, our tolerancewas ǫ = 10−6 and our confidence value was δ = 10−7. In Example 2 ourtarget value was 6, our tolerance was ǫ = .1 and our confidence valuewas δ = .1/87.

The solution to Example 2 demonstrates one technique for solvingapproximation problems:

(1) Choose an initial value δo for δ and assume initially that x =a± δo–i.e. a− δo < x < a+ δo.

(2) By factoring, write

f(x)− L = (x− a)g(x)

where g(x) is some function. We refer to the (x − a) term asthe “gold” because it is small when x is near a. The g(x) termis referred to as “trash.”

(3) Bound the trash over the interval (a− δo, a+ δo). : i.e. find anumber B such that

|g(x)| ≤ B

for all x ∈ (a− δo, a+ δo).(4) It then follows that

|f(x)− L| = |x− a| |g(x)| ≤ |x− a|Bfor all x ∈ (a− δo, a + δo). This will be < ǫ if |x− a| < ǫ/M .Thus, we can choose δ to be the smaller of δo and ǫ/M .

There is also a geometric way of solving Example 2:

Graphical Solution of Example 2: Using, a graphing calculator,we plot V using the viewing window defined by [5.9, 6.1] on the y-axisand some small interval on the x-axis containing x = 2, say [1.8, 2.2].(Figure 2.)

Any point on the visible portion of the graph has its V coordinatewithin tolerance. Using the trace feature on the calculator, we findthat the x-values for this piece of the graph range from x = 1.993 to


1 2 3

(2,6)

Graph of V(x):Viewing window greatly magnified

Viewing window

y=5.9

y=6.1

(1.993,6.096)

(2.005,5.932)

Figure 2. Graphical Solution of Example 2

x = 2.005. Any x within ±.005 of 2 will lie within this range. Thus, ifour graph is correct, we can choose δ = .005.

As a check, we compute that V (1.995) = 6.065 and V (2.005) =5.94, both of which are within ±.1 of 6. Our graph indicates that V isdecreasing over [1.995, 2.005]. Granted this, it follows that for all x inthis interval

6.065 ≥ V (x) ≥ 5.94

proving that V (x) is within ±.1 of 6. Hence our value of δ is correct.

Remark: The analytical solution suggests that we need to cut oursides to a tolerance of .1/87 ≈ .001 units. However, the graphicalsolution shows that in fact we only need ±.005 accuracy. Thus, thegraphical solution is “better.” However, the analytical solution has theadvantage that it yields a formula for δ as a function of ǫ. Specifically,the work done in Example 2 showed that for 0 < x < 3,

|V (x)− 6| < 87|x− 2|.Hence, |V (x) − 6| < ǫ will be true if |x − 2| < ǫ/87 showing that wecan set δ to be the smaller of 1 and ǫ/87.

To relate Examples 1 and 2 to the concept of limit, consider whatwe mean by the statement

limx→a

f(x) = L.

One way of describing this might be “f(x) becomes close to L as xbecomes close to, but not equal to, a.” (We exclude x = a sincewe are describing what happens as x “approaches” a.) In terms of


approximations we could say that f(x) can be made to approximateL as closely as desired by making x sufficiently close to a, withoutnecessarily equaling a. In more precise terms, we adopt the followingdefinition. The condition |x− a| > 0 excludes x = a.

Definition 1. We say that

limx→a

f(x) = L

provided that for all numbers ǫ > 0 there is a number δ > 0 such that

|f(x)− L| < ǫ

for all x satisfying 0 < |x− a| < δ.

This definition can be thought of in terms of a factory. The bosssays, “I want f(x) to approximate L to within ±.1. How close to amust x be?”

You respond, “Were O.K. if x is within ±.01 of a.”The boss says,“I changed my mind. Now I want f(x) to approxi-

mate L to within ±.001.”You respond,“We can do it, but now we’ll need x to be within

±.00005 of a.”If you can always satisfy the boss, no matter how many times he

changes his mind, then limx→a f(x) = L. Proving that you can alwaysfind δ usually requires producing a formula for δ as a function of ǫ.

Remark: The steps in doing the typical limit proof are:

(1) Assume that a value of ǫ is given.(2) State an appropriate δ. (It will depend on ǫ.)(3) Prove that the stated value of δ works.

Example 3. Use the definition of limit to prove that

limx→1

x

2x+ 1=

1

3

Solution:

Scratch Work: Let ǫ > 0 be given. We want

∣

∣

x

2x+ 1− 1

3| < ǫ

∣

∣

x− 1

3(2x+ 1)

∣

∣ = |x− 1|∣

∣

1

3(2x+ 1)

∣

∣ < ǫ


The term 1/3(2x+ 1) is the “trash.” To bound it assume that δo = 1so that |x− 1| < δo implies x ∈ [0, 2]. Then 3(2x+ 1) ≥ 3 and

|x− 1|∣

∣

1

3(2x+ 1)

∣

∣ <1

3|x− 1|

This will be < ǫ if |x− 1| < 3ǫ. Thus, we choose δ to be the smaller of1 and 3ǫ.

Our Proof: Let ǫ > 0 be given and let δ = min{1, 3ǫ}. Assume that0 < |x− 1| < δ. Then x ∈ [0, 2].

Hence

∣

∣

x

2x+ 1− 1

3| =

∣

∣

x− 1

3(2x+ 1)

∣

∣

= |x− 1|∣

∣

1

3(2x+ 1)

∣

∣

< |x− 1| 13

(From the scratch work.)

<δ

3≤ 3ǫ

3= ǫ

Hence, 0 < |x − 1| < δ implies |f(x) − 13| < ǫ, proving the limit

statement.

Example 4. Use the definition of limit to prove that

limx→3

√7− 2x = 1

Scratch Work: Let ǫ > 0 be given. We want

|√7− 2x− 1| < ǫ

Rationalizing:

√7− 2x− 1 =

(√7− 2x− 1)(

√7− 2x+ 1)√

7− 2x+ 1

=(7− 2x)− 1√7− 2x+ 1

=−2(x− 3)√7− 2x+ 1

= (x− 3)−2√

7− 2x+ 1


Thus

|x− 3|∣

∣

−2√7− 2x+ 1

∣

∣

= |x− 3| 2√7− 2x+ 1

≤ 2

1|x− 3|

This is < ǫ if |x− 3| < ǫ/2. Hence we set δ = ǫ/2.There is, however, a small problem. The square root is only defined

if7− 2x ≥ 0

7 ≥ 2x

3.5 ≥ x

This is true if x = 3± .5. Hence we also need to choose δ ≤ .5.

Our Proof: Let ǫ > 0 be given and let δ = min{.5, ǫ/2}. Assume that0 < |x− 1| < δ. Then 2.5 < x < 3.5. For such x

|√7− 2x− 1| = |x− 3| 2√

7− 2x+ 1

≤ 2|x− 3| < 2δ ≤ 2ǫ

2= ǫ

Hence, 0 < |x − 3| < δ implies |f(x) − 1| < ǫ, proving the limitstatement.

What’s the point?In the past, students have made comments such as, “I can do this

δ-ǫ stuff, but I really don’t see any value to it. It doesn’t help mecompute the limit–I need to know the limit before I start. In fact,in most of the exercises, I can find the limit by plugging a into theformula.”

The simplest answer is that the δ-ǫ technique is a way of provingthat the limit is what you think it should be. We guess that

limx→1

x

2x+ 1=

1

3

The work in Example 3 proves it.The skeptical student might respond, “I don’t have to guess. As

x → 1, the numerator gets close to 1 and the denominator gets close to3 so the ratio must be getting close to 1/3. Isn’t this proof enough?”

The answer is, “Yes and no.” The term “close” is much too impre-cise to be in a proof. However, here is a valid version of the student’s


proof:

(3)

limx→1

x

2x+ 1=

limx→1 x

limx→1(2x+ 1)

=1

limx→1(2x) + limx→1 1

=1

2(limx→1 x) + 1=

1

2 + 1=

1

3

This argument relies on the Sum, Product, and Quotient Theorems.Once these theorems have been proved, the above argument just asvalid as the δ-ǫ argument from Example 3. However, proving the Sum,Product, and Quotient Theorems requires δ-ǫ arguments. Many, many,other theorems in mathematics are proved using δ − ǫ methods. Thepurpose of using δ-ǫ arguments to prove “obvious” statements such asthose in Examples 3 and 4 above is to get a better understanding of theδ-ǫ technique so as to better understand both what a limit really is andto better understand the proofs of general facts such as the theoremsform this section. In fact, the δ-ǫ idea is essential for proving most ofthe theorems found in any calculus text. It is, in a sense, the cornerstone of calculus.

Theorem 1 (Sum Theorem). Suppose that limx→a f(x) and limx→a g(x)both exist. Then

limx→a

(f(x) + g(x)) = limx→a

f(x) + limx→a

g(x)

Proof Let L = limx→a f(x) and M = limx→a g(x). Let ǫ > 0 be given.There are numbers δ1 and δ2 such that

|f(x)− L| < ǫ/2

whenever 0 < |x− a| < δ1 and

|g(x)−M | < ǫ/2

whenever 0 < |x−a| < δ2. Let δ be smaller than both δ1 and δ2. Then,for 0 < |x− a| < δ,

|f(x) + g(x)− (L+M)| = |(f(x)− L) + (g(x)−M)|< |f(x)− L|+ |g(x)−M | < ǫ

2+

ǫ

2= ǫ

which proves the theorem. �


In preparation for the proof of the product theorem, we prove thefollowing special case:

Lemma 1. Suppose that limx→a f(x) = 0 = limx→a g(x). Thenlimx→a f(x)g(x) = 0.

Proof Let ǫ > 0 be given. We need to show that there is a δ > 0 suchthat

(4)|f(x)g(x)− 0| = |f(x)g(x)|

= |f(x)| |g(x)| < ǫ

for all 0 < |x− a| < δ.This will be true if both of the following inequalities hold for all

0 < |x− a| < δ:

(5)|f(x)| <

√ǫ

|f(x)| <√ǫ

However, since limx→a f(x) = 0, we know that there is a numberδ1 such that the first inequality in (5) holds for all 0 < |x − a| < δ1.Similarly, there is a number δ2 such that the second inequality in (5)holds for all 0 < |x−a| < δ2. Both inequalities hold for 0 < |x−a| < δwhere δ = min{δ1, δ2}. Hence, inequality (4) holds for all 0 < |x−a| <δ, proving our lemma.

We also note the following simple results which are left as exercises.(Exercises 12,11, and 10 beginning on page 171.)

Lemma 2. If C is any constant, then limx→aC = C.

Lemma 3. If C is any constant, then limx→aCf(x) = C limx→a f(x).

Lemma 4. If f(x) is a function, then limx→a f(x) = L if and onlyif limx→a(f(x)− L) = 0

We are now ready to prove the product theorem.

Theorem 2 (Product Theorem). Suppose that limx→a f(x) andlimx→a g(x) both exist. Then

limx→a

(f(x)g(x)) = (limx→a

f(x))(limx→a

g(x))

Proof Let L = limx→a f(x) and M = limx→a g(x). Our theorem isequivalent with the statement

limx→a

(f(x)g(x)) = LM


However, from Lemma 4,

limx→a

(f(x)− L) = 0

limx→a

(g(x)−M) = 0

Hence, from Lemma 1,

(6) limx→a

(f(x)− L)(g(x)−M) = 0

On the other hand

(f(x)− L)(g(x)−M) = f(x)g(x)−Mf(x)− Lg(x) + LM.

Solving for f(x)g(x), we get

f(x)g(x) = (f(x)− L)(g(x)−M) +Mf(x) + Lg(x)− LM.

Hence, using equation (6), the Sum Theorem, Lemma 2, and Lemma 3,we see

limx→a

(f(x)g(x)) = 0 +M limx→a

f(x) + L limx→a

g(x)− LM

= ML+ LM − LM = LM

proving the Product Theorem. �

Theorem 3 (Inverse). Suppose that limx→a f(x) 6= 0. Then

limx→a

1

f(x)=

1

limx→a f(x)

Case 1: limx→a f(x) = 1.We note that

∣

∣

1

f(x)− 1

∣

∣ =|1− f(x)||f(x)| =

|f(x)− 1||f(x)|

Since limx→a f(x) = 1, for all ǫ > 0, there is a δ > 0 such that

(7) |f(x)− 1| < ǫ

for 0 < |x− a| < δ.Letting ǫ = .5 in (7) we see that there is a δ1 > 0 such that

−.5 < f(x)− 1 < .5

.5 < f(x) < 1.5

for 0 < |x− a| < δ1. Hence, for such x,

|f(x)− 1||f(x)| | < |f(x)− 1|

.5= 2|f(x)− 1|


Now, let ǫ > 0 be given. Replacing ǫ with ǫ/2 in (7) shows that thereis a δ2 > 0 such that

|f(x)− 1| < ǫ/2

for 0 < |x− a| < δ2. Let δ be smaller than both δ1 and δ2. Then

∣

∣

1

f(x)− 1

∣

∣ < 2|f(x)− 1| < 2ǫ

2= ǫ

for 0 < |x− a| < δ. This finishes Case 1.

General Case:Suppose that limx→a f(x) = L 6= 0. Then from Lemma 3

limx→a

f(x)

L= 1

Hence, from Case 1

1 = limx→a

L

f(x)= L lim

x→a

1

f(x)

Hence

limx→a

1

f(x)=

1

L

as desired.

The following follows from the Product Theorem and Theorem 3and is left as an exercise.

Theorem 4 (Quotient Theorem). Suppose that limx→a f(x) andlimx→a g(x) both exist and limx→a g(x) 6= 0. Then

limx→a

f(x)

g(x)=

limx→a f(x)

limx→a g(x)

The following example is similar to many of the exercises in thenotes.

Example 5. Suppose that limx→a f(x) = 3. Prove, using δ and ǫ,that

limx→a

1

f(x)− 5= −1

2


Scratch work: Let ǫ > 0 be given. We want

(8)

∣

∣

1

f(x)− 5+

1

2

∣

∣ < ǫ

|3− f(x)|2|f(x)− 5| <ǫ

|f(x)− 3| 1

2|f(x)− 5| <ǫ

The term on the left is our “gold” since it becomes small as x ap-proaches a. The other term is our “trash” which we will bound. Specif-ically, we reason that for all x sufficiently close to a, f(x) = 3 ± 1.Thus, for such x,

2 < f(x) < 4

−3 < f(x)− 5 < −1

1 < |f(x)− 5| < 3

Hence

(9) |f(x)− 3| 1

2|f(x)− 5| <1

2|f(x)− 3|

This is < ǫ if |f(x) − 3| < 2ǫ, which is true for all x sufficientlyclose to a.

Proof: We know that for all ǫ > 0, there is a δ > 0 such that |f(x)−3| < ǫ for 0 < |x− a| < δ. Specifically, for ǫ = 1, we see that there is aδo such that

(10)

|f(x)− 3| < 1

−1 < f(x)− 3 < 1

−3 < f(x)− 5 < −1

1 < |f(x)− 5| < 3

Now let ǫ > 0 be given and choose δ1 > 0 so that

(11) |f(x)− 3| < 2ǫ

for 0 < |x − a| < δ1. Let δ = min{δ1, δo}. Then both (10) and (11)hold for 0 < |x− a| < δ. Hence, from (8), 0 < |x− a| < δ implies that

∣

∣

1

f(x)− 5+

1

2

∣

∣ < ǫ

proving the limit statement.


IMPORTANT!: Unlike the limit problems done earlier, we neverstated an explicit value for δ in Example 5. All we said was choose δso that (10) and (11) both hold for 0 < |x − a| < δ. The value 2ǫ in(11) is not δ. We would need to solve (11) for x to find δ.

|x−a| Knob

δ

|f(x)−L| meter

2ε 1

0

0

Figure 3. The “ delta machine”

This can be explained in terms of a machine. The output of themachine is values of |f(x)−L| which are measured using the “|f(x)−L| meter.” (See Figure 3). The input of the machine is values of|x − a| which we adjust by turning the “|x − a| Knob.” The solutionto Example 5 says that we need to turn the |x − a| knob untile thevalue of |f(x)− L| is below both 2ǫ and 1. The value the |x− a| knobpoints to is δ. We know that there is such a δ since the definition oflimx→a f(x) = L tells us, in effect, that we can make the |f(x)−L|meterpoint as close to 0 as we wish by turning the |x − a| knob sufficientlyfar.

Exercises

(1) We wish to produce a cube with volume 27± .01. Find a valueof δ such any cube with side length 3± δ will have volume inthe stated interval.

(2) A farmer bought exactly 10 feet of fencing wire to fence in a2× 6 plot of land beside his barn. (Figure 3). How accuratelymust he cut the short side of his fence if he wants the areato be 12± .1 square feet. (Why he wants this is beyond me!)Solve both graphically and analytically. Hint: Call the actuallength of the short side x.


Barn

2 2

6

Figure 3(3) For each of the following functions:

(a) Find a value of δ for which f(x) approximates f(a) towithin ±ǫ for all x within ±δ of a where a and δ are asstated.

(b) Give a general formula for δ in terms of ǫ.

(i) f(x) = 5x+ 3 a = 2 ǫ = .003

(ii) f(x) = 3x− 1 a = −1 ǫ = 10−2

(iii) f(x) = x2 − 4x+ 5 a = 3 ǫ = .02

(iv) f(x) = x3 − 7x2 + 3 a = −1 ǫ = .01

(v) f(x) =1

2x+ 3a = 3 ǫ = .025

(vi) f(x) =1

1− 2xa = 0 ǫ = .04

(vii) f(x) =√3 + 4x a = 1 ǫ = .01

(viii) f(x) =1√

5 + 2xa = 2 ǫ = .05

(4) Find a value of δ such that sinxx

approximates 1 to within ±.001for all x within ±δ of 0, x 6= 0. Repeat for ǫ = .0005. Solvethis problem graphically–not analytically.

(5) Obtain evidence for the following limit statement by finding anappropriate δ for (i) ǫ = .5, (ii) ǫ = .1, (iii) ǫ = .01. ReasonGraphically.

limx→0

sin x

x= 1


(6) Find the following limits and use the definition of limit toprove your answer.

(a) limx→2

(5x+ 3)

(b) limx→3

7

(c) limx→1

1

x2

(d) limx→.1

1

x2

(e) limx→−.05

1

x3

(f) limx→−1

(3x− 1)

(g) limx→3

(x2 − 4x+ 5)

(h) limx→−1

(x3 − 7x2 + 3)

(i) limx→3

1

2x+ 3

(j) limx→0

1

1− 2x

(k) limx→1

√3 + 4x

(l) limx→2

1√5 + 2x

(7) Evaluate each of the limits (a)-(d) using the limit theoremsfrom the notes, together with the following results:(a) For n ∈ N, limx→a x

n = an. (Power Theorem)

(b) If limx→a f(x) > 0, then limx→a

√

f(x) =√

limx→a f(x).(Square Root Theorem)Do your proof in a step-by-step manner, stating the main

theorems being used. We have done (a) as an example.


Solution of (a):

limx→2

3x2 −√x+ 2

x2 + 1

=limx→2(3x

2 + (−1)√x+ 2)

limx→2(x2 + 1)Quotient Theorem

=limx→2(3x

2) + limx→2(−1)(√x+ 2)

limx→2(x2) + limx→2 1Sum Theorem

=3 limx→2(x

2) + (−1)√

limx→2(x+ 2))

4 + 1Constant, Power, and Root Theorems

=3 · 4 + (−1)

√2 + 2

5= 2 Power and Sum Theorems

(a) limx→2

3x2 −√x+ 2

x2 + 1

(b) limx→1

x2 + 2x+ 1

x3 + 1

(c) limx→3

(

x2 +x

x2 + 1

)

(

x3 − 7x+ 2)

(d) limx→−1

√

(x2 + 5)(x2 + 2)

(8) Prove the following:(a) limx→a x = a. (Use a δ-ǫ argument.)(b) limx→a x

2 = a2. (Use the Product Theorem and x2 = x·x.(c) limx→a x

3 = a3. (Use the Product Theorem, part (b), andx3 = x · (x2).)

(d) limx→a x4 = a4. (Use the Product Theorem, part (c), and

x4 = x · (x3).)(e) Suppose that limx→a x

n = an has been proven for some n.Use this and the Product Theorem to show that limx→a x

n+1 =an+1. It follows that limx→a x

n = an for all n.


(9) Suppose that limx→a f(x) = 3. Use a δ-ǫ argument to prove:

(a) limx→a

3

f(x) + 1=

3

4

(b) limx→a

√

f(x) + 1 = 2 Hint: Rationalize

(c) limx→a

3√

f(x) + 1=

3

2Hint: Rationalize

(d) limx→a

f(x)2 = 9

(e) limx→a

f(x)3 = 27

(f) limx→a

f(x)

f(x) + 1=

3

4

(10) Use a δ-ǫ argument to prove Lemma 2.(11) Use a δ-ǫ argument to prove Lemma 3.(12) Use a δ-ǫ argument to prove Lemma 4.(13) Suppose that limx→a f(x) and limx→a g(x) both exist. Use a

δ-ǫ argument to prove that

limx→a

(f(x)− g(x)) = limx→a

f(x)− limx→a

g(x)

Your proof will be very similar to that of the Sum Theorem.(14) Suppose that limx→a f(x) and limx→a g(x) both exist. Use a

the Sum Theorem and the Constant Theorem to prove that

limx→a

(f(x)− g(x)) = limx→a

f(x)− limx→a

g(x)

(15) Suppose that limx→a f(x), limx→a g(x), and limx→a h(x) all ex-ist. Use a δ-ǫ argument to prove that

limx→a

(f(x) + g(x) + h(x)) = limx→a

f(x) + limx→a

g(x) + limx→a

h(x)

Your proof will be very similar to that of the Sum Theorem.(16) Suppose that for all x, f(x) < g(x) < h(x) and limx→a f(x) =

L = limx→a h(x). Use a δ-ǫ argument to prove that limx→a g(x) =L. Hint: Write |g(x)− L| < ǫ in the equivalent form L− ǫ <g(x) < L+ ǫ. Similarly for f(x) and h(x).

(17) Suppose that limx→a f(x) = L where L > 0. Prove that thereis a δ > 0 such that f(x) > 0 for all 0 < |x− a| < δ.

(18) Suppose that limx→a f(x) = L where L < 0. Prove that thereis a δ > 0 such that f(x) < 0 for all 0 < |x− a| < δ.

(19) Suppose that xn is a sequence such that limn→∞ xn = a andxn 6= a for all n. Suppose also that limx→a f(x) = L. Provethat limn→∞ f(xn) = L.


(20) Let f(x) = sin(1/x).(a) Find a sequence xn, limn→∞ xn = 0, such that f(xn) = 1

for all n.(b) Find a sequence xn, limn→∞ xn = 0, such that f(xn) = −1

for all n.(c) Explain how it now follows from the result of Exercise 19

that limx→0 f(x) does not exist.(21) Repeat Exercise 20 for (a) f(x) = cos(1/x), (b) f(x) = sin(1/x2)

(c) f(x) = cos(1/x2).(22) Use a δ-ǫ argument to prove the following formulas. Note: If

|x− a| < 1, then |x| = |(x− a) + a| < |x− a|+ |a| < 1 + |a|.

(a) limx→a

x2 = a2

(b) limx→a

x3 = a3

(c) limx→a

x4 = a4

(d) limx→a

xn = an n ∈ N

(e) limx→.1

1

x= 10

(f) limx→a

1

x=

1

aa 6= 0 Hint: Assume δo = a± |a|/2

(g) limx→.01

1

x2= 10000

(f) limx→a

1

x2=

1

a2a 6= 0

(h) limx→a

1

x3=

1

a3a 6= 0

(i) limx→a

1

xn=

1

ana 6= 0

(j) limx→a

Cx+D = Ca+D C,D ∈ R

(k) limx→a

Cx2 +D = Ca2 +D C,D ∈ R

(23) Suppose that limx→a f(x) = L where L > 0. Use a δ-ǫ argu-

ment to prove that limx→a

√

f(x) =√L. Hint: Rationalize.

(24) Suppose that limx→a f(x) = L. Use a δ-ǫ argument to provethat limx→a(f(x))

2 = L2.(25) Suppose that limx→a f(x) = L. Use a δ-ǫ argument to prove

that limx→a(f(x))3 = L3.

(26) (a) Find an example of functions f(x) and g(x) such thatlimx→0 f(x) = 0 but limx→0 f(x)g(x) = 1.


(b) Suppose that there is a number M such that |g(x)| < Mfor all x. Use a δ-ǫ argument to prove that if limx→a f(x) =0 then limx→a f(x)g(x) = 0.

(27) Suppose that limx→a g(x) exists. Show that there is a δ > 0and a constant M such that |g(x)| < M for 0 < |x − a| < δ.Hint: |g(x)| = |(g(x)− L) + L|.

CHAPTER 11

Continuity

Consider the problem of measuring the side length of a square andthen using the measured date to compute the area of the square. Forexample, if one side is measured to be 2.74 inches, then the area wouldbe computed as (2.74)2 = 7.5076 square inches. We know, of course,that no measurement is precise. There will undoubtedly be small errorsin the measurement of the side length. This will result in errors inthe computed area. We also know from experience that the errorsin the area computation will be small as long as the errors in the sidemeasurement are also small; values of s close to 2.74 will produce valuesof A close to (2.74)2. In mathematical terms, this amounts to sayingthat

lims→2.74

s2 = (2.74)2.

This is a statement of the fact that the area function A(s) = s2 iscontinuous at the point s = 2.74. In general, we define continuity asfollows:

Definition 1. A function f is continuous at a provided that

limx→a

f(x) = f(a).

Implicit in the above definition is the requirement that a belongto the domain of f . It is also implicit that f(x) be defined for all xsufficiently close to a since otherwise, the limit would not exist. (Wemust be able to get |f(x)−f(a)| < ǫ for all x between a− δ and a+ δ.)Thus, the above definition requires that there is a δ > 0 such that theinterval (a− δ, a+ δ) belongs to the domain of f .

This, however, presents us with a difficulty. According to this defi-nition, the function y =

√x is not continuous at x = 0:

limx→0

√x

does not exist since√x is not defined for x < 0. The best we can say

is thatlimx→0+

√x = 0.

175

176 11. CONTINUITY

Because of such examples, we are forced to amend our definition inthe case that the domain of f(x) is a closed (or half-closed) interval.Before doing so, however, we first give the “official” definition of leftand right hand limits:

Definition 2. Let f(x) be a function. We say that

limx→a+

f(x) = L

provided that for all ǫ > 0 there is a δ > 0 such that

|f(x)− L| < ǫ

for all x satisfying0 < |x− a| < δ, x > a.

The definition of limx→a− f(x) = L is identical, except that “x > a”in the last inequality above is replaced by “x < a.”

Now our amended definition of continuity states:

Definition 3. Suppose that the domain of f(x) is the interval[a, b]. We say that f(x) is continuous at a if

limx→a+

f(x) = f(a)

We say that f(x) is continuous at b if

limx→b−

f(x) = f(b).

.a

.x

f(a)f(x)

x near a

f(x) near f(a)

Figure 1

Intuitively, continuity may be described in the same terms as we didfor the square function: values of x near a produce values of f(x) nearf(a). (See Figure 1). It is very fortunate that most of the functionswhich arise in the real world are continuous. Otherwise, we would neverbe able to calculate anything!

11. CONTINUITY 177

You use continuity almost every time you evaluate a limit. Forexample, if you were to compute limx→2 x

2, you would probably just‘plug 2 in,’ finding the answer 22 = 4. What you are saying is thatfor the function f(x) = x2, the limit as x approaches 2 of f(x) is f(2).Thus, another way of describing what continuity at a means is to saythat the limit as x approaches a may be computed by ‘plugging a intof ’.

Figure 2 below show a few common types of discontinuities.

.a

.x

f(x)

x near a

o

.f(a)

f(x) not near f(a)

.

o

a

f(a)

f(x)

x

x near a

f(x) not near f(a)

Figure 2

The first graph is an example of what is called a removable dis-continuity. From the graph, L = limx→a f(x) exists, but just does nothappen to equal f(a). If we were to redefine f(a) by setting f(a) = L,then we would produce a new function which is continuous.

In the second graph, the continuity is not removable since limx→a f(x)does not exist: we get different answers for the limit, depending uponwhether we approach a from the left or from the right.

Example 1. Let f(x) be

f(x) =

√x− 2

x− 4.

Then f is not continuous at x = 4 because x = 4 does not belong tothe domain of f . Show that x = 4 is a removable discontinuity. Howshould f(4) be defined so as to make f continuous?

Solution: To be continuous at x = 4 we require

(1) f(4) = limx→4

f(x) = limx→4

√x− 2

x− 4.

If we attempt to evaluate this limit by “plugging” x = 4 into thefraction, we get the indeterminant form 0/0, which does not help.

178 11. CONTINUITY

There are several correct ways to evaluate this limit. Our favorite,perhaps, is to rationalize:

√x− 2

x− 4=

(√x− 2)(

√x+ 2)

(x− 4)(√x+ 2)

=x− 4

(x− 4)(√x+ 2)

=1√x+ 2

.

The limit is 1/4. Hence, we should define f(4) = 1/4.

The next example exhibits a much more ‘serious’ discontinuity.

Example 2. Let f(x) be

f(x) = sin1

xx 6= 0

= 0 x = 0.

Show that f is discontinuous at x = 0.

Solution: The function f(x) is zero whenever 1/x = kπ which is equiv-alent with x = π

k. Hence, f has an infinite number of zeros between π

and 0. Between these zeros, f oscillates between 1 and −1. Thus, thegraph of f looks something like that shown in Figure 3 below:

.

Figure 3

As x → 0, f(x) does not approach any single value, showing thatf has a non-removable discontinuity at x = 0.

Proving that a given function is continuous at a given value is oftenquite easy.

11. CONTINUITY 179

Example 3. Prove that the function f(x) = x2 is continuous at alla ∈ R.

Solution: Let a be some real number. Then, from the product theoremfor limits,

limx→a

f(x) = limx→a

x2 = (limx→a

x)(limx→a

x) = aa = a2 = f(a).

This simple problem illustrates the following theorem which is adirect consequence of the Product Theorem for limits of functions.

Theorem 1 (Product). Suppose that f(x) and g(x) are both con-tinuous at a. Then h(x) = f(x)g(x) is continuous at a.

It follows from the product theorem that the function f(x) = x3

is continuous for every a since x3 = x(x2). The following propositionfollows by similar reasoning:

Proposition 1. For n ∈ N, function f(x) = xn is continuous atevery a ∈ R.

Once we know the continuity of such functions, it is easy to provethe continuity of many other functions as well.

Example 4. Prove that the function f below is continuous at everya in its domain:

f(x) =x2 + 1

x2 − 3x+ 2.

Solution: Let us first note that the denominator of f factors as (x−1)(x− 2). Hence the domain of f is all real x, x 6= 1 and x 6= 2. Let abe an element of the domain of f . Since a is not equal to either 1 or 2,we see that as x approaches a, the denominator in f will not approach0. This allows us to apply the quotient rule for limits:

limx→a

x2 + 1

x2 − 3x+ 2=

limx→a x2 + limx→a 1

limx→a x2 − 3 limx→a x+ 2 limx→a

=a2 + 1

a2 − 3a+ 2.

Since the final answer is what would have been obtained by plugginga into the formula for f , the continuity is proved.

The above example illustrates the following theorem which is adirect consequence of the quotient theorem for limits of functions.

180 11. CONTINUITY

Theorem 2 (Quotient). Suppose that f(x) and g(x) are both con-tinuous at a and that g(a) 6= 0. Then h(x) = f(x)/g(x) is continuousat a.

In a calculus class, one might compute a limit such as

limn→∞

√

n

2n+ 1

as follows:Let xn = n

2n+1. Since

limn→∞

xn =1

2we see that

limn→∞

√

n

2n+ 1= lim

n→∞

√xn

= limx→1/2

√x =

√

1

2.

This method is based on the continuity of y =√x at x = 1/2.

Specifically, it uses the following theorem:

Theorem 3 (Sequence). Let f(x) be continuous at a and let xn bea sequence such that limn→∞ xn = a. Then

limn→∞

f(xn) = f(a).

Proof Let ǫ > 0 be given. Since limx→a f(x) = f(a), there is a δ > 0such that

(2) |f(x)− f(a)| < ǫ.

for |x− a| < δ, x 6= a. This inequality holds even if x = a since in thiscase the left hand quantity is zero.

But, since limn→∞ xn = a, there is an N such that

|xn − a| < δ

for all n > N . Replacing x with xn in (2) shows that

|f(xn)− f(a)| < ǫ

for n > N , which proves our theorem.

Continuity is important for solving equations.

Example 5. Show that the following equation has a solution x ∈[0, 1].

2x3 + x2 − 1 = 0

11. CONTINUITY 181

Solution: We compute that f(0) = −1 and f(1) = 2 which suggeststhat f has a zero somewhere in the interval [0, 1]. As a check we graphf over [0, 1]. The graph certainly seems to confirm the existence of azero.

x0 10.90.80.70.60.50.40.30.20.10

0

2

1.5

1

0.5

0

-0.5

-1

Figure 4

We stress, however, that the graph only seems to cross the axis.Indeed, on many graphing calculators, if you trace the graph, you willnot find a value of x for which y is exactly zero. This is becausethe calculator only plots a finite number of points. That the graphactually does cross the x-axis is a consequence of Proposition 2 below.The last sentence in the statement of this proposition says that thereis a smallest a such that f(a) = 0. Theorem 4 below, which is animmediate consequence of Proposition 2, is one of the fundamentalresults in analysis.

Proposition 2. Let f be continuous at every x in a closed interval[b, c]. Suppose that f(b) < 0 and f(c) > 0. Then there is an a ∈ [b, c]such that f(a) = 0. We may choose a so that f(x) < 0 for all x < a,x ∈ [b, c].

Proof Let

S = {x ∈ [b, c] | f(x) ≥ 0}and let a = inf S. (See Figure 5 below.) We will show that f(a) = 0.This will finish our proof since if x ∈ [b, c] satisfies x < a, then x /∈ S,showing that f(x) < 0.

From Exercise 9 on page 94 in Chapter 6, there is a sequence xn ∈ Swith limn→∞ xn = a. From Theorem 3

f(a) = limn→∞

f(xn).

182 11. CONTINUITY

S

a

Figure 5

Since xn ∈ S, f(xn) ≥ 0. Hence f(a) ≥ 0. (Exercise 29 on page 72 inChapter 4.)

Suppose that f(a) 6= 0. Then f(a) > 0. We claim that it followsthat there is a δ > 0 such that f(x) > 0 for all x ∈ [b, c] satisfying|x− a| < δ. If our claim is true, then we have reached a contradiction,since it follows that f(x) is positive on the interval a− δ < x < a+ δ,which denies our observation that f(x) < 0 for x < a.

To prove the claim, let ǫ > 0 be chosen so that

0 < ǫ < f(a).

Since

limx→a

f(x) = f(a)

there is a δ > 0 such that for 0 < |x− a| < δ,

|f(x)− f(a)| < ǫ

−ǫ < f(x)− f(a) < ǫ

f(a)− ǫ < f(x) < f(a) + ǫ

It is clear that the above inequalities are also valid for x = a. Thisproves or claim since f(a)− ǫ > 0. Hence, our proposition follows. �

The following result follows by applying Proposition 2 to the func-tion f(x) = g(x)−D. The details are left to the reader. (Exercise 11)As before, the last sentence in the statement of this theorem says thatthere is a smallest a such that g(a) = D.

Theorem 4 (Intermediate Value (IVT)). Let g be continuous atevery x in a closed interval [b, c]. Suppose that g(b) < D and g(c) > D.Then there is an a in the interval [b, c] such that g(a) = D. This a maybe chosen so that g(x) < D for x ∈ [b, c], x < a.

11. CONTINUITY 183

Remark: The conclusion of the Theorem 4 holds if we assume insteadthat g(b) > D and g(c) < D. This result follows by applying Proposi-tion 2 to the function f(x) = D − g(x). Again, we leave the details tothe reader. (Exercise 12)

Remark: By definition,√2 is that positive number a such that

a2 = 2.

How do we know that such a number exists? None of the axiomsfrom Chapters 1 and 2 state that such a number exists. In fact, since

√2

is irrational, the axioms from Chapters 1 and 2 cannot, by themselves,be used to prove the existence of

√2: if we could prove its existence

using these axioms then our proof would prove the existence of√2 in

the rational numbers since these axioms all hold for both the real andthe rational number systems. Thus, in the context of these notes, wecannot prove the existence of

√2 without using either the Least Upper

Bound Axiom or one of its consequences, such as the IVT.

. .1 2

1

3

4

0

d=2

f(2)

f(0)

y=x 2

The square root of 2

Figure 6

In fact, the existence of√2 is a simple consequence of the IVT.

Consider the function f(x) = x2 on the interval [0, 2] shown in Figure 6.Proposition 1 Shows that f(x) is continuous for all x. Also, f(0) = 0and f(2) = 4. Since 0 < 2 < 4, it follows from the IVT that thereis a number a ∈ [0, 2] such that 2 = f(a) = a2, proving the existenceof

√2. In fact, in precisely the same manner we can prove that every

positive number d has a positive square root.There is a deep and important difference between the way contin-

uous functions behave on closed intervals and other types of intervals.Consider for example the function f(x) = x2 on the interval (0, 2). The

184 11. CONTINUITY

maximum value of f(x) appears to 4; except 4 is not a value of f(x) atall since 2 /∈ (0, 2). Rather 4 is a sup. This function has no maximumover (0, 2). Similarly f has only an inf over (0, 2) since 0 /∈ (0, 2). Evenworse, consider the function f(x) = 1/x on the same interval. Thisfunction isn’t even bounded on this interval.

On the other hand, our intuition tells us that this kind of “misbe-havior” cannot happen for a continuous function over a closed interval.Such a function should have both a maximum and a minimum.

Theorem 5. Let f be continuous at every x in a closed interval[a, b]. Then there is a value c ∈ [a, b] such that f(c) ≥ f(x) for allx ∈ [a, b].

Proof The proof breaks down into two steps:

(1) Prove that there is a number M such that f(x) ≤ M for allx ∈ [a, b]. (We say that f(x) is bounded from above.)

(2) Prove the existence of c.

1

2

3

4

5

6

7

8

x x x x x3 4 5 6 7

[ ]a b

Figure 7

To prove (1), assume that it is false. Then for each M ∈ R, thereis an x ∈ [a, b] such that f(x) > M . In particular, for each n ∈ N,n > f(a), there is an x ∈ [a, b] such that

f(x) > n > f(a).

It then follows from the MVT that there is a smallest value xn ∈ [a, b]such that

f(xn) = n.

(See Figure 7.)Figure 7 suggests that the xn are increasing. This is indeed true:

Since f(xn+1) = n + 1 > n > f(a), there is a value of x between a

11. CONTINUITY 185

and xn+1 such that f(x) = n. Since xn is the first such x, we see thatxn ≤ x ≤ xn+1, as claimed.

From the Bounded Increasing Theorem, x = limn→∞ xn exists. Butthen

f(x) = limn→∞

f(xn) = limn→∞

n = ∞which is nonsense, proving that f is bounded.

Now letymax = sup{f(x) | x ∈ [a, b]}.

This exists since, as we just showed, f is bounded from above. Wewant to prove that there is a c ∈ [a, b] such that

f(c) = ymax.

Suppose that this is false. Then the function

g(x) = ymax − f(x)

is positive on [a, b]. Hence, from Theorem 2

h(x) =1

ymax − f(x)

is continuous on [a, b]. Thus, from the argument done to prove part(1), h(x) is bounded from above–i.e. there is a number M ′ such that

(3) h(x) ≤ M ′

for all x ∈ [a, b].On the other hand, since ymax is the sup of the y-values of f(x)

over [a, b], there is a sequence xn ∈ [a, b] such that

ymax = limn→∞

f(xn).

(Exercise 9 on page 94 in Chapter 6.)There is then an N such that for all n ≥ N ,

|ymax − f(xn)| <1

M ′ .

This implies that h(xn) > M ′, contradicting inequality (3). This fin-ishes the proof of our theorem. �

Remark: There is of course a Min Theorem. Being lazy, and notliking to type, we shall leave both the statement and the proof to thereader.

Exercises:

(1) Let f(x) = (sin x)/x of x 6= 0 and let f(0) = 1. Why is fcontinuous at x = 0?

186 11. CONTINUITY

(2) Let g be the function defined by

g(x) =x100 − 2100

x− 2x 6= 2.

How should g(2) be defined so as to make g continuous for allreal numbers x.

(3) Let f(x) be differentiable at x = a. How should the followingfunction be defined at x = a to make it continuous.

g(x) =f(x)− f(a)

x− a

(4) Let f(x) = x sin(1/x). How should f(0) be defined so as tomake f continuous.

(5) Let f be the function defined by

f(x) = x x < 1

f(x) = a− x x ≥ 1

where a is some real number. How should a be chosen so asto make f continuous for all real x?

(6) Find an example (reader’s choice) of a function which in notcontinuous at

1,1

2,

1

3

1

4, . . . ,

1

n, . . .

but is continuous for all other values of x, including x = 0.Hint: Let f(x) = 0 if x 6= 1/n. How should f(1/n) be defined?

(7) Suppose that we define a function f by saying that f(x) = 1if x is rational and f(x) = −1 if x is irrational. Thus, f(π) =−1 and f(2/3) = 1. Graph f . For which values of x is fcontinuous? Explain.

(8) Define a function f as follows. Suppose first that x is rational,x = p/q where p and q are integers with q > 0 and p and qhave no common factors. In this case, we define f(x) = 1/q.If x is irrational, we define f(x) = 0. Thus

f(3

4) = f(

1

4) =

1

4

f(4

18) = f(

2

9) = fn19f(

32

17) =

1

17

f(√2) = f(π) = 0.

(a) Compute f(3), f(3.1), f(3.14) and f(3.141). Do youthink f is continuous at x = π? Explain.

11. CONTINUITY 187

(b) Compute f(3.1), f(3.01), f(3.001) and f(3.0001). Do youthink f is continuous at x = 3? Explain.

(c) For which values of x do you think f(x) continuous? Ex-plain.

(9) Consider the function f(x) = 1/x. Then f(1) = 1 > 0 andf(−1) = −1 < 0. Theorem 1 would seem to say that thereis an a ∈ [−1, 1] such that f(a) = 0. This, of course, is false.Why is this not a contradiction to the IVT?

(10) Prove that there is a value of x such that x3 − x = 10. Findthe value of x to within ±.005. Prove your answer.

(11) Write a careful proof of the IVT (Theorem 4) using Proposi-tion 2.

(12) Write a complete statement of the theorem implied by theremark immediately following the statement of the IVT (The-orem 4). Then use Proposition 2 to prove this theorem.

(13) Prove that there is an x ∈ [0, 1] such that cos x = x and findx to within ±.001. Hint: Let f(x) = cos x− x. Consider f(0)and f(1).

(14) Suppose that f is continuous at every x in [0, 1] and that forall x in this interval, 0 ≤ f(x) ≤ 1. Prove that there isan x ∈ [0, 1] such that f(x) = x. Hint: This is similar toExercise 13.

(15) Suppose that f is continuous at every x in [0, 1] and that forall x in this interval, 0 ≤ f(x) ≤ 1. Prove that there is anx ∈ [0, 1] such that f(x) = 1 − x. Hint: This is similar toExercise 13.

(16) Find both points of intersection of the curves curves y = ex andy = 3x + 1. Give an answer accurate to within ±.01. (Note:If you were asked to find the area between these curves, youwould need to find these points before integration. There isno algebraic way to solve for these points.)

(17) Prove than any cubic polynomial f(x) = ax3+bx2+cx+d hasat least one real zero. For this you should consider limx→∞ f(x)and limx→−∞ f(x).

(18) Draw a graph which represents a one-to-one function f(x)which is defined for all real numbers x and which is increas-ing for some values of x and decreasing for other values of x.What ‘bad’ property does your graph exhibit. Prove that anysuch example must necessarily have this same ‘bad’ property.

(Recall that one-to-one means that for each y-value thereis at most one x such that f(x) = y.)

188 11. CONTINUITY

(19) Let f be a continuous at every x in a closed interval [a, b].Prove that the range of f is also a closed interval. Hint: Provethat the range is [c, d] where c = min f(x) and d = max f(x)

(20) Let f be a one-to-one function and let g be the inverse function.(Hence, g(f(x)) = x for all x in the domain of f .) Prove thatf(g(y)) = y for all y in the range of f . Hint: Since y is in therange of f , y = f(x) for some x.

(21) Let f be a one-to-one function which is increasing. Prove thatf−1 is also increasing. Hint Suppose that there are numbersa < b such that g(a) ≥ g(b). What do you know about theeffect of applying f to inequalities?

(22) Let f be a continuous, increasing function defined for all realnumbers and let g(x) = f−1(x). Below is a rather poorlywritten proof of the continuity of g(x). Rewrite this proof ina more acceptable form. Specifically(a) You will need to begin with a statement defining ǫ fol-

lowed by a statement defining δ.(b) You will need to prove that the value of δ defined in (5)

is positive. Hint: Apply f(x) to the inequality g(a)+ ǫ >g(a) > g(a)− ǫ.

(c) The given proof is a “backwards” proof. You will need toreverse it.

(d) You will need to include a statement between (4) and (5)defining x such as “Let 0 < |x− a| < δ.”

(e) You will need to explain how (4) follows from your defi-nition of δ.

(f) You will need to explain how (2) follows from (3).(g) You will need to explain how (1) follows from (2). Hint:

See Exercise 22 above.(h) You will need to put in a “bottom line” statement indi-

cating that you have done what was necessary.Proof

|g(x)− g(a)| < ǫ

g(a)− ǫ < g(x) < g(a) + ǫ (1)

f(g(a)− ǫ) < f(g(x)) < f(g(a) + ǫ) (2)

f(g(a)− ǫ) < x < f(g(a) + ǫ) (3)

f(g(a)− ǫ)− a < x− a < f(g(a) + ǫ)− a (4)

Let δ be the smaller of

a− f(g(a)− ǫ) and f(g(a) + ǫ)− a. (5)

11. CONTINUITY 189

(23) State carefully a theorem relating to minimums which is analo-gous to Theorem 5. Give a careful proof of your theorem usinga similar line of reasoning as was used in proving Theorem 5.

(24) Letf(x) =2 x > 3

f(x) =1 x ≤ 3

Graph f . What is the largest x such that f(x) ≤ 1? What isthe smallest x such that f(x) ≥ 2?

(25) Suppose that f is continuous on the closed interval [a, b] anddifferentiable on the open interval (a, b). Suppose also thatf(a) = f(b). Prove that there is a point xo , a < xo < b, suchthat f ′(xo) = 0. For your proof you may assume the theoremthat states that f(x) has either a max or a min at xo ∈ (a, b),then f ′(xo) = 0.

f(a)=f(b)

x xo oa b

y=f(x)

Rolle’s Theorem

Figure 8. Exercise 25

(26) Rolle’s theorem is important for one, and only one, reason: Itis used in proving the Mean Value Theorem. The Mean ValueTheorem is pictured below. Pictorially, it says that given asecant line for some differentiable curve, there is a point atwhich the slope of the tangent line is equal to that of thesecant line.

In writing, the MVT says

Theorem 6 (Mean Value Theorem MVT). Let f be con-tinuous on the closed interval [a, b] and differentiable on theopen interval (a, b). Then there is a value c, a < c < b, suchthat

f(b)− f(a)

b− a= f ′(c).

190 11. CONTINUITY

a b

(b,f(b))

c c

(a,f(a))

The Mean Value Theorem

Figure 9. Exercise 26

In this problem, we request that you answer the questionsbelow and, hence, prove the MVT.(a) Let l be the line which passes through the points (a, f(a))

and (b, f(b)) in the figure above. This is the secant line.Compute a formula for l. Express your formula in theform y = mx+ B.

(b) Let h(x) = f(x) − (mx + B) where mx + B is from (a)above. Indicate on a graph similar to the one above whatquantity h(x) measures.

(c) Let h be as in (b) above. Show that h(a) = h(b) andh′(x) = f ′(x)−m. What, explicitly, does Rolle’s Theoremtell you about h? The MVT should drop out!

(27) If f is a continuous function defined over a closed interval [a, b],we define the ‘average value’ of f to be

A =1

b− a

∫ b

a

f(x)d x.

The reason that this is thought of as an average is that theintegral is thought of as summing the values of f(x) for x ∈[a, b] and (b− a), in some sense, represents the number of x in[a, b].

Now, suppose that f is increasing (and continuous) over[a, b]. We expect that f(a) is ‘below average’ and f(b) is ‘aboveaverage’. There should, then, exist some value c between aand b where f(c) is exactly average. This is the content of the‘Mean Value Theorem for Integrals’.

Theorem 7 (Integral Mean Value). Let f be a continuousfunction over the interval [a, b]. Then there is a c between a

11. CONTINUITY 191

and b such that

f(c) =1

b− a

∫ b

a

f(x)d x.

In this exercise, you are asked to prove this important the-orem in the case where f(a) > 0 and f is increasing over theinterval [a, b]. Four your proof, you should use geometric rea-soning involving area to prove that

f(a) ≤ 1

b− a

∫ b

a

f(x)d x

(f(a) is ‘below average’) and

f(b) ≥ 1

b− a

∫ b

a

f(x)d x.

(f(b) is above average.) How does the Integral Mean ValueTheorem follow? How have you used the continuity of f?Hint: Put a rectangle of height f(a) under the curve and arectangle of height f(b) over the curve.

(28) In the above exercise, you needed to assume that f was in-creasing. You can avoid this if you deal with the minimumand maximum values of f instead of f(a) and f(b). Explicitly,use geometric reasoning to prove that the minimum value of fis ‘below average’ and the maximum is ‘above average’. Howdoes the theorem follow?

(29) One of the most important uses of the Integral Mean ValueTheorem is to prove the Fundamental Theorem of Calculus.Let f be a continuous function defined for all real numbers.Let

F (x) =

∫ x

0

f(t)d t.

The Fundamental Theorem says that for all a,

F ′(a) = f(a).

In this exercise, we request that you use the Mean ValueTheorem for Integrals to prove the Fundamental Theorem.The proof is based upon

F ′(a) = limx→a

F (x)− F (a)

x− a.

The most important step is to prove that

F (x)− F (a)

x− a=

1

x− a

∫ x

a

f(t)d t.

178 11. CONTINUITY

Once you have done this, you apply the Mean Value Theoremfor Integrals and let x → a. Write out in detail how this allworks.

CHAPTER 12

Taylor (Maclaurin) Approximations

Our goal in this section is to describe one way that calculators(and computers) can use to compute complicated functions such assin, cosine and exponentials. Before doing so, however, we need todiscuss integration of inequalities. Suppose that f(t) and g(t) are twointegrable functions defined over an interval [a, b] such that

f(t) ≤ g(t)

Then their graphs might look something like those shown in Figure 1below. Since the integral represents area under the curve, we expectthat

∫ b

a

f(t) dt ≤∫ b

a

g(t) dt

g(x)

f(x)

a b

Figure 1

This is indeed true. We say that integration preserves inequalities.

Theorem 1. Suppose that f an g are integrable functions over [a, b]and,

f(t) ≤ g(t)

179

180 12. TAYLOR (MACLAURIN) APPROXIMATIONS

for all t ∈ [a, b]. Then∫ b

a

f(t) dt ≤∫ b

a

g(t) dt

Proof The above inequality is equivalent with

0 ≤∫ b

a

g(t) dt−∫ b

a

f(t) dt

=

∫ b

a

(g(t)− f(t)) dt

This is positive because 0 < g(t)− f(t).

An important consequence of Theorem 1 is the observation that:integration over short intervals improves the accuracy of certain ap-proximations. To be precise, suppose that we are approximating afunction f by the function fo and that

−ǫ < f(x)− fo(x) < ǫ

for all x in, say, [0, .5]. Then∫ .5

0

−ǫ dx <

∫ .5

0

(f(x)− fo(x)) dx <

∫ .5

0

ǫ

−.5ǫ <

∫ .5

0

f(x) dx−∫ .5

0

fo(x) dx < .5ǫ

Thus, in this case, the integral of fo approximates the integral of fwith half the maximum error in the approximation to fo to f .

Now, let us return to calculators. Calculators are essentially arith-metic machines: although they might have a few values of a givenfunction stored in their memory, most values must be computed usingonly addition, subtraction, multiplication and division. We will beginby describing how to compute e.4321 using a calculator that can onlydo arithmetic. We assume that our calculator displays 6 digits so thatwe will want an answer accurate to within, say, ±10−7.

Let f(x) be a function which is continuous at x = 0. Then, for xnear 0,

f(x) ≈ f(0)

This is refered to as the horizontal line approximation since y = f(0)describes a horizontal line. For y = ex, the horizontal line approxima-tion is

(1) ex ≈ e0 = 1

(See Figure 2.)

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