368

Calculus is the branch of mathematics that was developed to analyze and model change such as velocity and acceleration. We can also apply it to study change in the context of slope, area, volume and a wide range of other real-life phenomena. Although mathematical techniques that you have studied previously deal with many of these concepts, the ability to model change was restricted. For example, consider the curve in Figure 11.1. This shows the motion of an object by indicating the distance (y metres) travelled after a certain amount of time (t seconds). Pre-

calculus mathematics will only allow us to compute the average velocity between two different times (Figure 11.2). With calculus specifically, techniques of differential calculus we will be able to find the velocity of the object at a particular instant, known as its instantaneous velocity (Figure 11.3). The starting point for our study of calculus is the idea of a limit.

Introduction

11 Differential Calculus I--: Fundamentals

Assessment statements6.1 Informalideasoflimitsandconvergence.

Limitnotation.Definitionofderivativefromfirstprinciplesf9(x)5lim

h0(f(x1h)2f(x)_____________h ).

Derivativeinterpretedasgradientfunctionandasrateofchange.Tangentsandnormals,andtheirequations.

6.2 Derivativeofxn(n).Thesecondderivative.

6.3 Localmaximumandminimumpoints;testingformaximumorminimum.Pointsofinflexionwithzeroandnon-zerogradients.

Graphicalbehaviouroffunctionsincludingtherelationshipbetweenf,f9andf99.

6.6 Kinematicproblemsinvolvingdisplacements,velocityv,andaccelerationa.

1

2

Dis

tanc

e (m

etre

s)

3

4

5

10 2 3 4 5 6 7Time (seconds)

8 9 10 11 12 t

y

Figure 11.1

369

A limit is one of the ideas that distinguish calculus from algebra, geometry and trigonometry. The notion of a limit is a fundamental concept of calculus. Limits are not new to us. We often use the idea of a limit in many non-mathematical situations. Mathematically speaking, we have encountered limits on at least two occasions previously in this book finding the sum of an infinite geometric series (Section 3.4) and computing the irrational number e (Section 4.3).

Recall from Section 3.4 that we established that if the sequence of partial sums for an infinite series converges to a finite number L we say that the infinite series has a sum of L. Further on in that section, we used limits to

algebraically confirm that the infinite series 2 1 1 1 1 __ 2

1 1 __ 4 1 1 __ 8

1 has a

sum of 4. As part of the algebra for this, we reasoned that as the value of n increases in the positive direction without bound (i.e. n 1`) the

expression ( 1 _ 2 ) n

converges to zero in other words, the limit of ( 1 _ 2 ) n

as

n goes to positive infinity is zero. We express this result more efficiently

using limit notation, as we did in Chapter 3, by writing lim n

(1 _ 2 ) n

5 0.

It is beyond the requirements of this course to establish a precise formal definition of a limit, but a closer look at justifying this limit and a couple of others can lead us to a useful informal definition.

Example 1

Evaluate lim n

(1 _ 2 ) n

by using your GDC to analyze the behaviour of the

function f(x) 5 ( 1 _ 2 ) x

for large positive values of x.

1

2

Dis

tanc

e (m

etre

s)

3

4

5

10 2 3 4

(4, 2)

(10, 4.25)

Average velocity fromt 4 to t 10 seconds

5 6

4.25 210 4

7Time (seconds)

8 9 10 11 12

0.375 m/s2.25 m6 s

t

y

Figure 11.2

1

2

Dis

tanc

e (m

etre

s)

3

4

5

10 2 3 4

(7, 3.875)

Instantaneous velocityat t 7 seconds 0.25 m/s

5 6 7Time (seconds)

8 9 10 11 12 t

y

Figure 11.3

Limits of functions11.1

370

Differential Calculus I--: Fundamentals11

Solution

The GDC screen images show the graph and table of values for y 5 ( 1 __ 2 ) x.

Clearly, the larger the value of x, the closer that y gets to zero. Although there is no value of x that will produce a value of y equal to zero, we can get as close to zero as we wish. For example, if we wish to produce a value of y

within 0.001 of zero, then we could choose x 5 10 and y 5 ( 1 __ 2 ) 10

5 1 ____ 1024

0.000 976 56; and if we want a result within 0.000 0001 of zero, then we

could choose x 5 24 and y 5 ( 1 __ 2 ) 24

5 1 _________ 167 772 16

0.000 000 059 605; and

so on. Therefore, we can conclude that lim n

(1 _ 2 ) n

5 0.

In calculus we are interested in limits of functions of real numbers. Although many of the limits of functions that we will encounter can only be approached and not actually reached (as in Example 1), this is not always the case. For example, if asked to evaluate the limit of the function

f (x) 5 x__ 2

2 1 as x approaches 6, we simply need to evaluate the function

for x 5 6. Since f (6) 5 2, then lim n 6

( x__ 2 2 1 ) 5 2. However, it is more common that we are unable to evaluate the limit of f (x) as x approaches some number c because f (c) does not exist.

Example 2

Evaluate lim x 0

sin x____ x .

Solution

We are not able to evaluate this limit by direct substitution because when

x5 0, sin x____ x 5 0 __ 0

and is therefore undefined. Lets use our GDC again to

analyze the behaviour of the function f (x) 5 sin x____ x as x approaches zero

from the right side and the left side.

Hint: xmustbeinradiansbecauseincalculusweareinterestedinfunctionsofrealnumbers.

Plot1

Y1=(1/2)XPlot2 Plot3

Y2=Y3=Y4=Y5=Y6=Y7=

WINDOWXmin=-1Xmax=8Xscl=1Ymin=-. 1Ymax=1 Yscl=1Xres=1

Y1

Y1=1.22070313E-4

78910111213

X.00781.00391.001959.8E-44.9E-42.4E-41.2E-4

Y1

X=0

X1.5.25.125.0625.03125.01563

TABLE SETUPTblStart=0

Indpnt: Auto AskDepend: Auto Ask

Tbl=10123456

Theliney5cisahorizontal asymptoteofthegraphofafunctiony5f(x)ifeitherlimx

f(x)5cor limx2`

f(x)5c.Forexample,theliney50(x-axis)isahorizontalasymptoteofthegraphof

y5(1__2)xbecauselim

n(1__2)

n50.

371

Although there is no point on the graph of y 5 sin x____ x corresponding to

x 5 0, it is clear from the graph that as x approaches zero (from either

direction) the value of sin x____ x converges to one. We can get the value of sin x____ x

arbitrarily close to 1 depending on our choice of x. If we want sin x____ x to be

within 0.001 of 1, we choose x 5 0.05 giving sin 0.05 _______ 0.05

0.999 583 and

1 2 0.999 583 5 0.000 417 , 0.001; and if we want sin x____ x to be within

0.000 001 of 1, then we choose x 5 0.002 giving sin 0.02 _______ 0.02

0.999 999 3333

and 1 2 0.999 999 3333 5 0.000 000 6667 , 0.000 001; and so on.

Therefore, lim x 0

sin x____ x 5 1.

Functions do not necessarily converge to a finite value at every point its possible for a limit not to exist.

Example 3

Find lim x 0

1 __ x2

, if it exists.

SolutionAs x approaches zero, the value of 1 ___

x2 becomes increasingly large in the

positive direction. The graph of the function (left) seems to indicate that

we can make the values of y 5 1 __ x2

arbitrarily large by choosing x close

enough to zero. Therefore, the values of y 5 1 __ x2

do not approach a finite

number, so lim x 0

1 __ x2

does not exist.

Although we can describe the behaviour of the function y 5 1 __ x2

by writing

lim x 0

1 __ x2

5 `, this does not mean that we consider ` to represent a number

it does not. This notation is simply a convenient way to indicate in what manner the limit does not exist.

Limit of a functionIff(x)becomesarbitrarilyclosetoauniquefinitenumberLasxapproachescfromeitherside,thenthelimitoff(x)asxapproachescisL.Thenotationforindicatingthisislim

xcf(x)5L.

Whenafunctionf(x)becomesarbitrarilyclosetoafinitenumberL,wesaythatf(x)convergestoL.

For our purposes in this course, it is also important to be able to apply some basic algebraic manipulation in order to evaluate the limits of some functions algebraically, rather than by conjecturing from a graph or table.

x

y

y

0.5

sin xx

1

0 2 44 2

x

y

y 1x2

0

Thelinex5cisavertical asymptoteofthegraphofafunctiony5f(x)ifeitherlimxc

f(x)5`orlimxc

f(x)52`.Forexample,thelinex50(y-axis)isaverticalasymptoteofthegraphofy51___

x2because

limx0

1___x2

5`.

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Differential Calculus I--: Fundamentals11

Example 4

Evaluate each limit algebraically.

a) lim x

5x 2 3 ______ x

b) lim p 0

(3x2 2 4px 1 p2)

c) lim h 0

[(x 1 h)2 2 6] 2 (x2 2 6)

______________________ h

Solution

a) lim x

5x 2 3 ______ x 5 lim x ( 5x__ x 2 3 __ x) Split the fraction into two terms and 5 lim

x 5 2 lim

x 3 _ x evaluate the limit of each term separately.

5 5 2 0 5 5 Therefore, lim x

5x 2 3 ______ x 5 5.

b) lim p 0

(3x2 2 4px 1 p2) 5 lim p 0