CalculusI_Version2.pdf

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Copyright 2014 Gregory Hartman

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PA Note on Using this Text

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This text is Part I of a threetext series on Calculus. The first part covers

material taught in many Calc 1 courses: limits, derivaves, and the basics of

integraon, found in Chapters 1 through 6.1. The second text covers material

oen taught in Calc 2: integraon and its applicaons, along with an introduc-

on to sequences, series and Taylor Polynomials, found in Chapters 5 through

8. The third text covers topics common in Calc 3 or mulvariable calc: para-

metric equaons, polar coordinates, vectorvalued funcons, and funcons of

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Contents

Preface iii

Table of Contents v

1 Limits 1

1.1 An Introducon To Limits . . . . . . . . . . . . . . . . . . . . . 1

1.2 Epsilon-Delta Definion of a Limit . . . . . . . . . . . . . . . . 9

1.3 Finding Limits Analycally . . . . . . . . . . . . . . . . . . . . . 161.4 One Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.5 Connuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.6 Limits involving infinity . . . . . . . . . . . . . . . . . . . . . . 43

2 Derivaves 55

2.1 Instantaneous Rates of Change: The Derivave . . . . . . . . . 55

2.2 Interpretaons of the Derivave . . . . . . . . . . . . . . . . . 69

2.3 Basic Differenaon Rules . . . . . . . . . . . . . . . . . . . . 76

2.4 The Product and Quoent Rules . . . . . . . . . . . . . . . . . 83

2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 93

2.6 Implicit Differenaon . . . . . . . . . . . . . . . . . . . . . . 103

2.7 Derivaves of Inverse Funcons . . . . . . . . . . . . . . . . . 114

3 The Graphical Behavior of Funcons 121

3.1 Extreme Values . . . . . . . . . . . . . . . . . . . . . . . . . . 121

3.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . 129

3.3 Increasing and Decreasing Funcons . . . . . . . . . . . . . . . 134

3.4 Concavity and the Second Derivave . . . . . . . . . . . . . . . 142

3.5 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . 149

4 Applicaons of the Derivave 157

4.1 Newtons Method . . . . . . . . . . . . . . . . . . . . . . . . . 157

4.2 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
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4.3 Opmizaon . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

4.4 Differenals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

5 Integraon 185

5.1 Anderivaves and Indefinite Integraon . . . . . . . . . . . . 185

5.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . 194

5.3 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . 2045.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . 221

5.5 Numerical Integraon . . . . . . . . . . . . . . . . . . . . . . . 233

6 Techniques of Andifferenaon 247

6.1 Substuon . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

A Soluons To Selected Problems A.1

Index A.11
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1: L

Calculusmeans a method of calculaon or reasoning. When one computesthe sales tax on a purchase, one employs a simple calculus. When one finds the

area of a polygonal shape by breaking it up into a set of triangles, one is using

another calculus. Proving a theorem in geometry employs yet another calculus.

Despite the wonderful advances in mathemacs that had taken place into

the first half of the 17th century, mathemacians and sciensts were keenly

aware of what theycould not do. (This is true even today.) In parcular, two

important concepts eluded mastery by the great thinkers of that me: area and

rates of change.

Area seems innocuous enough; areas of circles, rectangles, parallelograms,

etc., arestandard topics of study for students today just as they were then. How-

ever, the areas ofarbitraryshapes could not be computed, even if the boundary

of the shape could be described exactly.

Rates of change were also important. When an object moves at a constantrate of change, then distance = rate me. But what if therate is notconstant

can distance sll be computed? Or, if distance is known, can we discover the

rate of change?

It turns out that these two concepts were related. Two mathemacians, Sir

Isaac Newton and Goried Leibniz, are creditedwith independently formulang

a system of compung that solved the above problems and showed how they

were connected. Their system of reasoning was a calculus. However, as the

power and importance of their discovery took hold, it became known to many

as the calculus. Today, we generally shorten this to discuss calculus.

The foundaon of the calculus is the limit. It is a tool to describe a par-

cular behavior of a funcon. This chapter begins our study of the limit by ap-

proximang its value graphically and numerically. Aer a formal definion of

the limit, properes are established that make finding limits tractable. Once

the limit is understood, then the problems of area and rates of change can be

approached.

1.1 An Introducon To Limits

We begin our study oflimits by considering examples that demonstrate key con-

cepts that will be explained as we progress.


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0.5

1

1.5

0.6

0.8

1

x

y

Figure 1.1: sin(x)/xnearx= 1.

1

1

0.8

0.9

1

x

y

Figure 1.2: sin(x)/xnearx= 0.

Chapter 1 Limits

Consider the funcon y= sinxx

. When xis near the value 1, what value (if

any) isynear?

While our queson is not precisely formed (what constutes near the value

1?), the answer does not seem difficult to find. One might think firstto look at a

graph of this funcon to approximate the appropriateyvalues. Consider Figure

1.1, wherey= sinxx

is graphed. For values ofxnear 1, it seems thatytakes on

values near 0.85. In fact, whenx= 1, theny= sin 1

1 0.84, so it makes sensethat whenxis near 1,ywill be near 0.84.

Consider this again at a different value forx. Whenxis near 0, what value (if

any) isynear? By considering Figure 1.2, one can see that it seems thatytakes

on values near 1. But what happens when x= 0? We have

y sin 00

00

.

The expression 0/0hasno value;it is indeterminate. Such an expression givesno informaon about what is going on with the funcon nearby. We cannot find

out howybehaves nearx= 0 for this funcon simply by leng x= 0.Finding a limitentails understanding how a funcon behaves near a parcu-

lar value ofx. Before connuing, it will be useful to establish some notaon. Let

y= f(x); that is, let ybe a funcon ofxfor some funconf. The expression thelimit ofyas xapproaches 1 describes a number, oen referred to as L, thaty

nears asxnears 1. We write all this as

limx1

y= limx1

f(x) = L.

This is not a complete definion (that will come in the next secon); this is a

pseudo-definion that will allow us to explore the idea of a limit.

Above, wheref(x) = sin(x)/x, we approximated

limx1

sinx

x 0.84 and lim

x0sinx

x 1.

(Weapproximatedthese limits, hence used the

symbol, since we are work-

ing with the pseudo-definion of a limit, not the actual definion.)

Once we have the true definion of a limit, we will find limitsanalycally;

that is, exactly using a variety of mathemacal tools. For now, we will approx-

imatelimits both graphically and numerically. Graphing a funcon can provide

a good approximaon, though oen not very precise. Numerical methods can

provide a more accurate approximaon. We have already approximated limits

graphically, so we now turn our aenon to numerical approximaons.

Consider again limx1sin(x)/x. To approximate this limit numerically, wecan create a table ofxandf(x)values wherexis near 1. This is done in Figure1.3.

Notes:

2


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x sin(x)/x

0.9 0.870363

0.99 0.844471

0.999 0.841772

1 0.841471

1.001 0.841171.01 0.838447

1.1 0.810189

Figure 1.3: Values of sin(x)/xwithxnear1.

x sin(x)/x

-0.1 0.9983341665

-0.01 0.9999833334

-0.001 0.9999998333

0 not defined0.001 0.9999998333

0.01 0.9999833334

0.1 0.9983341665

Figure 1.4: Values of sin(x)/xwithxnear1.

2.5

3

3.5

0.26

0.28

0.3

0.32

0.34

y

Figure 1.5: Graphically approximang a

limit in Example 1.

x x2x6

6x219x+32.9 0.29878

2.99 0.294569

2.999 0.2941633 not defined

3.001 0.294073

3.01 0.293669

3.1 0.289773

Figure 1.6: Numerically approximang a

limit in Example 1.


Noce that forvalues ofxnear1, wehave sin(x)/xnear0.841. Thex= 1rowis in bold to highlightthe fact that when considering limits, we are notconcerned

with the value of the funcon at that parcular xvalue; we are only concerned

with the values of the funcon whenxisnear1.

Now approximate limx0sin(x)/x numerically. We already approximated

the value of this limit as 1 graphically in Figure 1.2. The table in Figure 1.4 showsthe value of sin(x)/xfor values ofxnear 0. Ten places aer the decimal pointare shown to highlight how close to 1 the value of sin (x)/xgets as xtakes onvalues very near 0. We include the x= 0 row in bold again to stress that we arenot concerned with the value of our funcon at x= 0, only on the behavior ofthe funconnear0.

This numerical method gives confidence to say that 1 is a good approxima-

on of limx0sin(x)/x; that is,

limx0

sin(x)/x1.

Later we will be able to prove that the limit is exactly1.

We now consider several examples that allow us explore different aspects

of the limit concept.

Example 1 Approximang the value of a limit

Use graphical and numerical methods to approximate

limx3

x2 x 66x2 19x+3 .

S To graphically approximate the limit, graph

y= (x2 x 6)/(6x2 19x+3)on a small interval that contains 3. To numerically approximate the limit, create

a table of values where thexvalues are near 3. This is done in Figures 1.5 and

1.6, respecvely.The graph shows that when xis near 3, the value ofyis very near 0.3. By

considering values ofxnear 3, we see that y= 0.294 is a beer approximaon.The graph and the table imply that

limx3

x2 x 66x2 19x+3 0.294.

This example may bring up a few quesons about approximang limits (and

the nature of limits themselves).

1. If a graph does not produce as good an approximaon as a table, why

bother with it?

Notes:

3


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1 0.5 0.5 1

0.5

1

x

y

Figure 1.7: Graphically approximang a

limit in Example 2.

x f(x)

-0.1 0.9

-0.01 0.99

-0.001 0.9990.001 0.999999

0.01 0.9999

0.1 0.99

Figure 1.8: Numerically approximang a

limit in Example 2.

Chapter 1 Limits

2. How many values ofxin a table are enough? In the previous example,

could we have just usedx= 3.001 and found a fine approximaon?

Graphs are useful since they give a visual understanding concerning the be-

havior of a funcon. Somemes a funcon may act erracally near certain

xvalues which is hard to discern numerically but very plain graphically. Since

graphingulies arevery accessible, it makes sense to make proper useof them.

Since tables and graphs are used only to approximatethe value of a limit,

there is not a firm answer to how many data points are enough. Include

enough so that a trend is clear, and use values (when possible) both less than

and greater than the value in queson. In Example 1, we used both values less

than and greater than 3. Had we used just x = 3.001, we might have beentempted to conclude that the limit had a value of 0.3. While this is not far off,we could do beer. Using values on both sides of 3 helps us idenfy trends.

Example 2 Approximang the value of a limit

Graphically and numerically approximate the limit of f(x) as x approaches 0,where

f(x) = x+1 x0

x2 +1 x> 0

.

S Again we graphf(x) and create a table of its values nearx=0 to approximate the limit. Note that this is a piecewise defined funcon, so it

behaves differently on either side of 0. Figure 1.7 shows a graph off(x), and oneither side of 0 it seems the yvalues approach 1.

The table shown in Figure 1.8 shows values off(x)for values ofxnear 0. Itis clear that as xtakes on values very near 0, f(x)takes on values very near 1.It turns out that if we let x= 0 for either piece off(x), 1 is returned; this issignificant and well return to this idea later.

The graph and table allow us to say that limx0f(x) 1; in fact, we areprobably very sure itequals1.

Idenfying When Limits Do Not Exist

A funcon may not have a limit for all values ofx. That is, we cannot say

limxcf(x) = Lfor some numbers L for all values ofc, for there may not be anumber thatf(x)is approaching. There are three ways in which a limit may failto exist.

1. The funconf(x)may approach different values on either side ofc.

2. The funcon may grow without upper or lower bound asxapproachesc.

Notes:

4


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0.5

1

1.5

2

1

2

3

x

y

Figure 1.9: Observing no limit asx 1 inExample 3.

x f(x)0.9 2.01

0.99 2.0001

0.999 2.000001

1.001 1.0011.01 1.01

1.1 1.1

Figure 1.10: Values off(x)nearx= 1 inExample 3.

0.5

1

1.5

2

50

100

x

y

Figure 1.11: Observing no limit asx 1in Example 4.

x f(x)0.9 100.

0.99 10000.

0.999 1. 1061.001 1. 1061.01 10000.

1.1 100.

Figure 1.12: Values off(x)nearx= 1 inExample 4.


3. The funcon may oscillate asxapproachesc.

Well explore each of these in turn.

Example 3 Different Values Approached From Le and Right

Explore why limx1

f(x)does not exist, where

f(x) =

x2 2x+3 x1

x x> 1 .

S A graph off(x)aroundx= 1 and a table are given Figures1.9 and 1.10, respecvely. It is clear that as xapproaches 1,f(x)does not seemto approach a single number. Instead, it seems as thoughf(x)approaches twodifferent numbers. When considering values ofxless than 1 (approaching 1

from the le), it seems thatf(x)is approaching 2; when considering values ofxgreater than 1 (approaching 1 from the right), it seems that f(x)is approach-ing 1. Recognizing this behavior is important; well study this in greater depth

later. Right now, it suffices to say that the limit does not exist since f(x)is notapproaching one value asxapproaches 1.

Example 4 The Funcon Grows Without Bound

Explore why limx1

1/(x 1)2 does not exist.

S A graph and table off(x) = 1/(x 1)2 are given in Figures1.11 and 1.12, respecvely. Both show that as xapproaches 1,f(x)grows largerand larger.

We can deduce this on our own, without the aid of the graph and table. Ifx

is near 1, then(x 1)2 is very small, and:1

very small number very large number.Sincef(x)is not approaching a single number, we conclude that

limx1

1

(x 1)2

does not exist.

Example 5 The Funcon Oscillates

Explore why limx0sin(1/x)does not exist.

Notes:

5


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2

4

6

10

20

x

f

Figure 1.14: Interpreng a difference

quoent as the slope of a secant line.

Chapter 1 Limits

S Two graphs off(x) = sin(1/x)are given in Figures 1.13. Fig-ure 1.13(a) showsf(x) on the interval [1, 1]; noce howf(x) seems to oscillatenearx = 0. One might think that despite the oscillaon, as xapproaches 0,

f(x)approaches 0. However, Figure 1.13(b) zooms in on sin (1/x), on the inter-val[0.1, 0.1]. Here the oscillaon is even more pronounced. Finally, in thetable in Figure 1.13(c), we see sin(x)/xevaluated for values ofxnear 0. Asx

approaches 0,f(x)does not appear to approach any value.It can be shown that in reality, asxapproaches 0, sin(1/x) takes on all values

between 1 and 1 infinite mes! Because of this oscillaon,limx0sin(1/x)does not exist.

1

0.5

0.5

1

1

0.5

0.5

1

x

y

0.1

5 102

5 102

0.1

1

0.5

0.5

1

x

y

x sin(1/x)0.1 0.544021

0.01 0.5063660.001 0.82688

0.0001 0.3056141. 105 0.03574881.

106

0.349994

1. 107 0.420548(a) (b) (c)

Figure 1.13: Observing thatf(x) = sin(1/x)has no limit asx 0 in Example 5.

Limits of Difference Quoents

We have approximated limits of funcons asxapproached a parcular num-

ber. We will consider another important kind of limit aer explaining a few key

ideas.

Letf(x) represent the posion funcon, in feet, of some parcle that is mov-ing in a straight line, wherexis measured in seconds. Lets say that whenx= 1,the parcle is at posion 10 ., and whenx= 5, the parcle is at 20 . Anotherway of expressing this is to say

f(1) = 10 and f(5) = 20.

Since the parcle traveled 10 feet in 4 seconds, we can say theparcles average

velocitywas 2.5 /s. We write this calculaon using a quoent of differences,

or, adifference quoent:

f(5) f(1)5 1 =

10

4 =2.5/s.

Notes:

6


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(a)

2

4

6

10

20

f

(b)

2

4

6

10

20

f

(c)

2

4

6

10

20

f

Figure 1.15: Secant lines off(x)at x= 1andx= 1+ h, for shrinking values ofh(i.e.,h 0).

h f(1+h)f(1)

h

0.5 9.250.1 8.650.01 8.515

0.01 8.485

0.1 8.35

0.5 7.75

Figure 1.16: The difference quoent eval-uated at values ofhnear 0.


This difference quoent canbe thought of as thefamiliarriseover run used

to compute the slopes of lines. In fact, that is essenally what we are doing:

given two points on the graph off, we are finding the slope of the secant line

through those two points. See Figure 1.14.

Now consider finding the average speed on another me interval. We again

start atx= 1, but consider the posion of the parcle h seconds later. That is,

consider the posions of the parcle when x = 1 and when x = 1+ h. Thedifference quoent is now

f(1+h) f(1)(1+h) 1 =

f(1+h) f(1)h

.

Letf(x) =1.5x2 +11.5x; note thatf(1) = 10 and f(5) = 20, as in ourdiscussion. We can compute this difference quoent for all values ofh (even

negave values!) excepth = 0, for then we get 0/0, the indeterminate formintroduced earlier. For all valuesh = 0, the difference quoent computes theaverage velocity of the parcle over an interval of me of length h starng at

x= 1.For small values ofh, i.e., values ofh close to 0, we get average velocies

over very short me periods and compute secant lines over small intervals. See

Figure 1.15. This leads us to wonder what the limit of the difference quoent is

ashapproaches 0. That is,

limh0

f(1+h) f(1)h

= ?

As we do not yet have a true definion of a limit nor an exact method for

compung it, we sele for approximang the value. While we could graph the

difference quoent (where the x-axis would represent h values and the y-axis

would represent values of the difference quoent) we sele for making a table.

See Figure 1.16. The table gives us reason to assume the value of the limit is

about 8.5.

Proper understanding of limits is key to understanding calculus. With limits,

we can accomplish seemingly impossible mathemacal things, like adding up an

infinite number of numbers (and not get infinity) and finding the slope of a line

between two points, where the two points are actually the same point. These

are not just mathemacal curiosies; they allow us to link posion, velocity and

acceleraon together, connect cross-seconal areas to volume, find the work

done by a variable force, and much more.

In the next secon we give the formal definion of the limit and begin our

study of finding limits analycally. In the following exercises, we connue our

introducon and approximate the value of limits.

Notes:

7


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Exercises 1.1Terms and Concepts

1. In your own words, what does it mean to find the limit of

f(x)asxapproaches 3?

2. An expression of the form 0

0 is called .3. T/F: The limit off(x)asxapproaches 5 is f(5).

4. Describe three situaons where limxc

f(x)does not exist.

5. In your own words, what is a difference quoent?

Problems

In Exercises 6 15, approximate the given limits both numer-

ically and graphically.

6. limx1

x2 +3x 5

7. limx

0x

3

3x

2 +x

5

8. limx0

x+1

x2 +3x

9. limx3

x2 2x 3x2 4x+3

10. l imx1

x2 +8x+7

x2 +6x+5

11. limx2

x2 +7x+10

x2 4x+412. lim

x2f(x), where

f(x) =

x+2 x 23x 5 x> 2 .

13. limx3

f(x), where

f(x) =

x2 x+1 x 3

2x+1 x> 3 .

14. limx0f(x), where

f(x) =

cosx x 0

x2 +3x+1 x> 0 .

15. limx/2

f(x), where

f(x) =

sinx x /2cosx x> /2

.

In Exercises 16 24, a funcon f and a value a are

given. Approximate the limit of the difference quoent,

limh0

f(a+h) f(a)h

, usingh= 0.1,0.01.

16. f(x) =

7x+2, a= 3

17. f(x) = 9x+0.06, a= 118. f(x) = x2 +3x 7, a= 1

19. f(x) = 1

x+1, a= 2

20. f(x) = 4x2 +5x 1, a= 321. f(x) = lnx, a= 5

22. f(x) = sinx, a=

23. f(x) = cosx, a=


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1.2 Epsilon-Delta Definion of a Limit


This secon introduces the formal definion of a limit, the epsilondelta, or

, definion.

Before we give the actual definion, lets consider a few informal ways of

describing a limit. Given a funcony= f(x)and anxvalue, call itc, we say thatthe limit of the funconfis a valueL:

1. if ytends toL as xtends toc.

2. if yapproachesL as xapproachesc.

3. ifyisnear Lwheneverxisnear c.

The problem with these definions is that the words tends, approach,

and especially near are not exact. In what way does the variable xtend to or

approachc? How near doxandyhave to be to candL, respecvely?

The definion we describe in this secon comes from formalizing3. A quick

restatement gets us closer to what we want:

3. Ifxis within a certaintolerance levelofc, then the corresponding valuey=f(x)is within a certaintolerance levelofL.

Theacceptednotaonfor thex-tolerance is the lowercase Greek leer delta,

or , and they-tolerance is lowercase epsilon, or . One more rephrasing of3

nearly gets us to the actual definion:

3. Ifxis within units ofc, then the corresponding value ofyis within unitsofL.

Note that this means (let the represent the word implies):

c < x< c + L < y< L + or |x c|< |y L|<

The point is that and , being tolerances, can be any posive (but typically

small) values. Finally, we have theformal definion of thelimit with thenotaon

seen in the previous secon.

Notes:

9


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2

4

6

1

2

} = .5

} = .5

Choose > 0. Then ...

x

y

2

4

6

1

2

} = .5

} = .5

width= 1.75

width= 2.25

... choose smaller

than each of these

x

y

With = 0.5, we pick any < 1.75.

Figure 1.17: Illustrang the process.

Chapter 1 Limits

Definion 1 The Limit of a Funconf

Letfbe a funcon defined on an open interval containing c. The notaon

limxc

f(x) = L,

read as the limit off(x), asxapproachesc, is L, means that given any > 0, there exists > 0 such that whenever|xc| < , we have|f(x) L|< .

(Mathemacians oen enjoy wring ideas without using any words. Here is

the wordless definion of the limit:

limxc

f(x) = L > 0, > 0 s.t.|x c|< |f(x) L|< .)

There is an emphasis here that we may have passed over before. In the

definion, they-tolerance is givenfirstand then the limit will exist ifwe can

find anx-tolerancethat works.

It is me for an example. Note that the explanaon is long, but it will takeyou through all steps necessary to understand the ideas.

Example 6 Evaluang a limit using the definion

Show that limx4

x= 2.

S Before we usethe formal definion, lets trysome numerical

tolerances. What if theytolerance is 0.5, or = 0.5? How close to 4 does xhave to be so that yis within 0.5 units of 2 (or 1.5< y< 2.5)? In this case, wecan just square these values to get 1.52


17/293


What if theytolerance is 0.01, or = 0.01? How close to 4 doesxhave tobe in order for yto be within 0.01 units of 2 (or 1.99 < y< 2.01)? Again, we

just square these values to get 1.992


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Chapter 1 Limits

secon, we will focus on examples where the answer is, frankly, obvious, be-

cause the nonobvious examples are even harder. That is why theorems about

limits areso useful! Aer doing a few more proofs, you will really appreciate

the analycal short cuts found in the next secon.

Example 7 Evaluang a limit using the definion

Show that limx2x2

=4.

S Lets do this example symbolically from the start. Let > 0be given; we want |y 4| < , i.e.,|x2 4| < . How do we find such thatwhen |x 2|< , we are guaranteed that |x2 4|< ?

This is a bit trickier than the previous example, but lets start by nocing that

|x2 4|=|x 2| |x+2|. Consider:

|x2 4|< |x 2| |x+2|< |x 2|< |x+2| . (1.1)

Could we not set =

|x+2|?We are close toan answer, but the catch isthatmustbea constantvalue(so

it cant containx). There is a way to work aroundthis, but wedo have tomake an

assumpon. Remember that is supposed to be a small number, which implies

that will also be a small value. In parcular, we can (probably) assume that

< 1. If this is true, then|x2| < would imply that|x2| < 1, giving1< x< 3.

Now, back to the fracon

|x+2| . If 1< x< 3, then 3 < x+2 < 5. Taking

reciprocals, we have1

5 0 there exists > 0such that whenever |f(x) K| < , we have|x a| < .

2. Which is given first in establishing a limit, the xtolerance

or theytolerance?

3. T/F: must always be posive.

4. T/F:must always be posive.

Problems

Exercises 5 11, prove the given limit using an proof.5. lim

x

5

3 x= 2

6. limx3

x2 3= 6

7. limx4

x2 +x 5= 15

8. limx2

x3 1= 7

9. limx2

5= 5

10. limx0

e2x 1= 0

11. limx0

sinx = 0 (Hint: use the fact that | sinx| |x|, withequality only whenx= 0.)


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Chapter 1 Limits

1.3 Finding Limits Analycally

In Secon 1.1 we explored the concept of the limit without a strict definion,

meaning we could only make approximaons. In the previous secon we gave

the definion of the limit and demonstrated how to use it to verify our approxi-

maons were correct. Thus far, our method of finding a limit is 1) make a really

good approximaon either graphically or numerically, and 2) verify our approx-imaon is correct using a -proof.

Recognizing that proofs are cumbersome, this secon gives a series of

theorems which allow us to find limits much more quickly and intuively.

Suppose that limx2f(x) = 2 and limx2g(x) = 3. What is limx2(f(x) +g(x))? Intuion tells us that the limit should be 5, as we expect limits to behavein a nice way. The following theorem states that already established limits do

behave nicely.

Theorem 1 Basic Limit Properes

Letb, c, L and Kbe real numbers, letn be a posive integer, and let fand g be

funcons with the following limits:

limxc

f(x) = L and limxc

g(x) = K.

The following limits hold.

1. Constants: limxc

b= b

2. Identy limxc

x= c

3. Sums/Differences: limxc

(f(x) g(x)) = L K4. Scalar Mulples: l im

xcb f(x) = bL

5. Products: limxc

f(x) g(x) = LK

6. Quoents: limxcf(x)/g(x) =L/K, (K =0)

7. Powers: limxc

f(x)n =Ln

8. Roots: limxc

n

f(x) = n

L

9. Composions: Adjust our previously given limit situaon to:

limxc

f(x) = L and limxL

g(x) = K.

Then limxc

g(f(x)) = K.

Notes:

16


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24/293

Chapter 1 Limits

Theorem 2 Limits of Polynomial and Raonal Funcons

Letp(x)andq(x)be polynomials andca real number. Then:

1. limxc

p(x) = p(c)

2. limxc

p(x

)q(x) = p

(c

)q(c) , whereq(c)=0.

Example 10 Finding a limit of a raonal funcon

Using Theorem 2, find

limx1

3x2 5x+1x4 x2 +3 .

S Using Theorem 2, we can quickly state that

limx1

3x2 5x+1x4

x2 +3

= 3(1)2 5(1) +1

(

1)4

(

1)2 +3

= 93

=3.

It was likely frustrang in Secon 1.2 to do a lot of work to prove that

limx2

x2 =4

as it seemed fairly obvious. The previous theorems state that many funcons

behave in such an obvious fashion, as demonstrated by the raonal funcon

in Example 10.

Polynomial and raonal funcons are not the only funcons to behave in

such a predictable way. The following theorem gives a list of funcons whose

behavior is parcularly nice in terms of limits. In the next secon, we will give

a formal name to these funcons that behave nicely.

Theorem 3 Special Limits

Letc be a real number in the domain of the given funcon and let n be a posive integer. The

following limits hold:

1. limxc

sinx= sin c

2. limxc

cosx= cos c

3. limxc

tanx= tan c

4. limxc

cscx= csc c

5. limxc

secx= sec c

6. limxc

cotx= cot c

7. limxc

ax =ac (a> 0)

8. limxc

lnx= ln c

9. limxc

n

x= n

c

Notes:

18


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Example 11 Evaluang limits analycally

Evaluate the following limits.

1. limx

cosx

2. limx

3(sec2x tan2x)

3. limx/2

cosxsinx

4. limx1

elnx

5. limx0

sinx

x

S

1. This is a straighorward applicaon of Theorem 3. limx

cosx= cos =

1.2. We can approach this in at least two ways. First, by directly applying The-

orem 3, we have:

limx3

(sec2x tan2x) = sec2 3 tan2 3.

Using the Pythagorean Theorem, this last expression is 1; therefore

limx3

(sec2x tan2x) = 1.

We can also use the Pythagorean Theorem from the start.

limx3

(sec2x tan2x) = limx3

1= 1,

using the Constant limit rule. Either way, we find the limit is 1.

3. Applying Theorem 3 directly gives

lim

x/2

cosxsinx= cos(/2) sin(/2) = 0

1= 0.

4. Again, we can approach this in two ways. First, we can use the exponen-

al/logarithmic identy thatelnx =xand evaluate limx1

elnx = limx1

x= 1.

We can also use the Composion limit rule. Using Theorem 3, we have

limx1

lnx= ln 1= 0. Thus

limx1

elnx = limx0

ex =e0 =1.

Both approaches are valid, giving the same result.

Notes:

19


26/293

Chapter 1 Limits

5. We encountered this limit in Secon 1.1. Applying our theorems, we at-

tempt to find the limit as

limx0

sinx

x sin 0

0 0

0

.

This, of course, violates a condion of Theorem 1, as the limit of the de-

nominator is notallowed to be 0. Therefore, we aresllunable to evaluate

this limit with tools we currently have at hand.

The secon could have been tled Using Known Limits to Find Unknown

Limits. By knowing certain limits of funcons, we can find limits involving sums,

products, powers, etc., of these funcons. We further the development of such

comparave tools with the Squeeze Theorem, a clever and intuive way to find

the value of some limits.

Before stang this theorem formally, suppose we have funcons f,g and h

wheregalways takes on values betweenfandh; that is, for all xin an interval,

f(x)g(x)h(x).Iffand h have the same limit atc, andg is always squeezed between them,

then g must have the same limit as well. That is what the Squeeze Theorem

states.

Theorem 4 Squeeze Theorem

Letf, g and h be funcons on an open interval I containingc such that

for allxinI,

f(x)g(x)h(x).If

limxc

f(x) = L = limxc

h(x),

then

limxc

g(x) = L.

It cantakesomework to figure outappropriate funcons by which to squeeze

the given funcon you are trying to evaluate a limit of. However, that is gener-

ally the only place work is necessary; the theorem makes the evaluang the

limit part very simple.

Notes:

20


27/293


28/293

Chapter 1 Limits

lim0

cos lim0

sin

lim

01

cos0 lim0

sin

1

1 lim0

sin

1

Clearly this means that lim0

sin

=1.

Two notes about the previous example are worth menoning. First, one

might be discouraged by this applicaon, thinking I wouldneverhave come up

with that on my own. This is toohard! Dont be discouraged; withinthis text we

will guide you in your use of the Squeeze Theorem. As one gains mathemacal

maturity, clever proofs like this are easier and easier to create.

Second, this limit tells us more than just that as xapproaches 0, sin(x)/xapproaches 1. Bothxand sinxare approaching 0, but the raoofxand sinx

approaches 1, meaning that they are approaching 0 in essenally the same way.Another way of viewing this is: for small x, the funconsy= xandy= sinxareessenally indisnguishable.

We include this special limit, along with three others, in the following theo-

rem.

Theorem 5 Special Limits

1. limx0

sinx

x =1

2. limx0cosx

1

x =0

3. limx0

(1+x)1x =e

4. limx0ex

1

x =1

A short word on how to interpret the laer three limits. We know that as

xgoes to 0, cosxgoes to 1. So, in the second limit, both the numerator and

denominator are approaching 0. However, since the limit is 0, we can interpret

this as saying that cosxis approaching 1 faster than xis approaching 0.

In the third limit, inside the parentheses we have an expression that is ap-

proaching 1 (though never equaling 1), and we know that 1 raised to any power

is sll 1. At the same me, the power is growing toward infinity. What happens

Notes:

22


29/293

1

2

1

2

3

x

y

Figure 1.20: Graphing fin Example 13 to

understand a limit.


to a number near 1 raised to a very large power? In this parcular case, the

result approaches Eulers number,e, approximately 2.718.In the fourth limit, we see that as x 0, ex approaches 1 just as fast as

x0, resulng in a limit of 1.

Our final theorem for this secon will be movated by the following exam-

ple.

Example 13 Using algebra to evaluate a limit

Evaluate the following limit:

limx1

x2 1x 1 .

S We begin by aempngto apply Theorem 3 andsubstung

1 forxin the quoent. This gives:

limx1

x2 1x 1 =

12 11 1 =

0

0

,

and indeterminate form. We cannot apply the theorem.By graphing the funcon, as in Figure 1.20, we see that the funcon seems

to be linear, implying that the limit should be easy to evaluate. Recognize that

the numerator of our quoent can be factored:

x2 1x 1 =

(x 1)(x+1)x 1 .

The funcon is not defined when x= 1, but for all other x,

x2 1x 1 =

(x 1)(x+1)x 1 =

(x 1)(x+1)x 1 =x+1.

Clearly limx1

x+ 1= 2. Recall that when considering limits, we are not concerned

with the value of the funcon at 1, only the value the funcon approaches as xapproaches 1. Since(x2 1)/(x 1)andx+ 1 are the same at all points except

x= 1, they both approach the same value as xapproaches 1. Therefore we canconclude that

limx1

x2 1x 1 =2.

The key to the above example is that the funconsy= (x2 1)/(x 1)andy= x+ 1 are idencal except atx= 1. Since limits describe a value the funconis approaching, not the value the funcon actually aains, the limits of the two

funcons are always equal.

Notes:

23


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Chapter 1 Limits

Theorem 6 Limits of Funcons Equal At All But One Point

Letg(x) = f(x)for allxin an open interval, except possibly at c, and letlim

xcg(x) = L for some real numberL. Then

limx

cf(x) = L.

The Fundamental Theorem of Algebra tells us that when dealing with a ra-

onal funcon of the form g(x)/f(x)and directly evaluang the limit limxc

g(x)

f(x)returns 0/0, then(x c)is a factor of both g(x)and f(x). One can then usealgebra to factor this term out, cancel, then apply Theorem 6. We demonstrate

this once more.

Example 14 Evaluang a limit using Theorem 6

Evaluatelimx3

x3 2x2 5x+62x3 +3x2 32x+15 .

We begin by applying Theorem 3 and substung 3 forx. This returns the famil-

iar indeterminate form of 0/0. Since thenumerator and denominator areeach

polynomials, we know that(x 3)is factor of each. Using whatever method ismost comfortableto you, factorout (x3) from each (using polynomial division,synthec division, a computer algebra system, etc.). We find that

x3 2x2 5x+62x3 +3x2 32x+15 =

(x 3)(x2 +x 2)(x 3)(2x2 +9x 5) .

We can cancel the (x3) terms as long asx=3. Using Theorem 6 we conclude:

limx3

x3 2x2 5x+62x3 +3x2 32x+15 = limx3

(x 3)(x2 +x 2)(x 3)(2x2 +9x 5)

= limx

3

(x2 +x 2)

(2x2

+9x

5

)=

10

40=

1

4.

S

We end this secon by revising a limit first seen in Secon 1.1, a limit of

a difference quoent. Letf(x) =1.5x2 +11.5x; we approximated the limitlim

h0f(1+h) f(1)

h 8.5.We formally evaluate this limit in the following ex-

ample.

Notes:

24


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Example 15 Evaluang the limit of a difference quoent

Letf(x) =1.5x2 +11.5x; find limh0

f(1+h) f(1)h

.

S Since fis a polynomial, our first aempt should be to em-

ploy Theorem 3 and substute 0 forh. However, we see that this gives us0

0

.

Knowing that we have a raonal funcon hints that some algebra will help. Con-sider the following steps:

limh0

f(1+h) f(1)h

= limh0

1.5(1+h)2 +11.5(1+h) 1.5(1)2 +11.5(1)h

= limh0

1.5(1+2h+h2) +11.5+11.5h 10h

= limh0

1.5h2 +8.5hh

= limh0

(1.5h+8.5) (using Theorem 6, ash =0)=8.5 (using Theorem 3)

This matches our previous approximaon.

This secon contains several valuable tools for evaluang limits. One of the

main results of this secon is Theorem 3; it states that many funcons that we

use regularly behave in a very nice, predictable way. In the next secon we give

a name to this nice behavior; we label such funcons as connuous. Defining

that term will require us to look again at what a limit is and what causes limits

to not exist.

Notes:

25


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1. Explain in your own words, without using formality,why lim

xcb= b.

2. Explain in your own words, without using formality,why lim

xcx= c.

3. What does the text mean when it says that certain func-

ons behavior is nice in terms of limits? What, in par-

cular, is nice?

4. Sketch a graph that visually demonstratesthe Squeeze The-

orem.

5. You are given the following informaon:

(a) limx1

f(x) = 0

(b) limx1

g(x) = 0

(c) limx1

f(x)/g(x) = 2

What can be said about the relave sizes off(x)and g(x)asxapproaches 1?

Problems

Using:

limx9

f(x) = 6 limx6

f(x) =9

limx9

g(x) = 3 limx6

g(x) = 3

evaluate the limits given in Exercises 6 13, where possible.

If it is not possible to know, state so.

6. limx9

(f(x) +g(x))

7. limx9(3f(x)/g(x))

8. limx9

f(x) 2g(x)

g(x)

9. limx6

f(x)

3 g(x)

10. limx9

g(f(x)

)11. lim

x6f(

g(x))

12. limx6

g(f(f(x))

)13. lim

x6f(x)g(x) f2(x) +g2(x)

Using:lim

x1f(x) = 2 lim

x10f(x) = 1

limx1

g(x) = 0 limx10

g(x) =

evaluate the limits given in Exercises 14 17, where possible.

If it is not possible to know, state so.

14. limx1

f(x)g(x)

15. limx10

cos(

g(x))

16. limx1

f(x)g(x)

17. limx1

g(

5f(x))

In Exercises 18 32, evaluate the given limit.

18. limx3

x2 3x+7

19. limx

x 3x 5

720. lim

x/4cosxsinx

21. limx0

lnx

22. limx3

4x38x

23. limx/6

cscx

24. limx0

ln(1+x)

25. limx

x2 +3x+5

5x2 2x 326. lim

x3x+1

1 x

27. limx6

x2 4x 12x2 13x+42

28. limx0

x2 +2x

x2 2x

29. limx2

x2 +6x 16x2 3x+2

30. limx2

x2 10x+16x2 x 2

31. l imx2x2

5x

14

x2 +10x+16

32. l imx1

x2 +9x+8

x2 6x 7Use the Squeeze Theorem in Exercises 33 35, where appro-

priate, to evaluate the given limit.

33. limx0

xsin

1

x

34. limx0

sinxcos

1

x2

35. limx3

f(x), wherex2 f(x) 3xon[0, 3].

Exercises 36 39 challenge your understanding of limits butcan be evaluated using the knowledge gained in this secon.

36. limx0

sin3x

x

37. limx0

sin5x

8x

38. limx0

ln(1+x)

x

39. limx0

sinx

x , wherexis measured in degrees, not radians.


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1.4 One Sided Limits


We introduced the concept of a limit gently, approximang their values graphi-

cally and numerically. Next came the rigorous definion of the limit, along with

an admiedly tedious method for compung them. The previous secon gave

us tools (which we call theorems) that allow us to compute limits with greater

ease. Chief among the results were the facts that polynomials and raonal,trigonometric, exponenal and logarithmic funcons (and their sums, products,

etc.) all behave nicely. In this secon we rigorously define what we mean by

nicely.

In Secon 1.1 we explored the three ways in which limits of funcons failed

to exist:

1. The funcon approached different values from the le and right,

2. The funcon grows without bound, and

3. The funcon oscillates.

In this secon we explore in depth the concepts behind #1 by introducing

the one-sided limit. We begin with formal definions that are very similar to the

definion of the limit given in Secon 1.2, but the notaon is slightly different

and a short phrase has been added to the end.

Definion 2 One Sided Limits

Le-Hand Limit


limxc

f(x) = L,

read as the limit off(x) as xapproaches c from the le is L, or the le-hand limit of f at c is Lmeans that given any > 0, there exists > 0 such that |xc|< implies |f(x)L|< , for allx< c.

Right-Hand Limit


limxc+

f(x) = L,

read as the limit off(x)as xapproachesc from the right is L, or theright-hand limit of f at c is Lmeans that given any > 0, there exists > 0 such that |x c| < implies |f(x) L| < , for all

x> c.

Notes:

27


34/293

1

2

1

2

x

y

Figure 1.21: A graph offin Example 16.

Chapter 1 Limits

Praccally speaking, when evaluang a le-hand limit, we consider only val-

ues ofxto the le ofc, i.e., wherex< c. The admiedly imperfect notaonx c is used to imply that we look at values ofxto the le ofc. The nota-on has nothing to do with posive or negave values of either xorc. A similar

statement holds for evaluang right-hand limits; there we consider only values

ofxto the right ofc, i.e.,x> c. We can use the theorems from previous secons

to help us evaluate these limits; we just restrict our view to one side ofc.We pracce evaluang le and right-hand limits through a series of exam-

ples.

Example 16 Evaluang one sided limits

Letf(x) =

x 0x13 x 1< x< 2 , as shown in Figure 1.21. Find each of the

following:

1. limx1

f(x)

2. limx1+

f(x)

3. limx1f(x)

4. f(1)

5. limx0+

f(x)

6. f(0)

7. limx2f(x)

8. f(2)

S For these problems, the visual aid of the graph is likely more

effecve in evaluang the limits than usingfitself. Therefore we will refer oen

to the graph.

1. Asxgoes to 1from the le, we see thatf(x)is approaching the value of 1.Therefore lim

x1f(x) = 1.

2. Asxgoesto 1from the right, wesee thatf(x) is approachingthe value of 2.Recall that it does not maer that there is an open circle there; we are

evaluang a limit, not the value of the funcon. Therefore limx1+f(x) = 2.

3. Thelimit offas xapproaches 1 does not exist, as discussed in the first

secon. The funcon does not approach one parcular value, but two

different values from the le and the right.

4. Using the definion and by looking at the graph we see thatf(1) = 1.

5. Asxgoes to0 from the right, wesee thatf(x) is also approaching 0. There-fore lim

x0+f(x) = 0. Note we cannot consider a le-hand limit at 0 as fis

not defined for values ofx< 0.

Notes:

28


35/293

1

2

1

2

x

y

Figure 1.22: A graph offfrom Example 17


6. Using the definion and the graph,f(0) = 0.

7. Asxgoes to 2 from the le, we see that f(x)is approaching the value of1. Therefore lim

x2f(x) = 1.

8. The graph and the definion of the funcon show thatf(2) is not defined.

Note howthe le andright-hand limitswere different; this, of course, causes

thelimit to not exist. The following theorem states what is fairly intuive: the

limit exists precisely when the le and right-hand limits are equal.

Theorem 7 Limits and One Sided Limits

Letfbe a funcon defined on an open intervalIcontainingc. Then

limxc

f(x) = L

if, and only if,

limxcf(x) = L and limxc+f(x) = L.

The phrase if, and only if means the two statements are equivalent: they

are either both true or both false. If the limit equalsL, then the le and right

hand limits both equalL. If the limit is not equal to L, then at least one of the

le and right-hand limits is not equal to L(it may not even exist).

One thing to consider in Examples 16 19 is that the value of the funcon

may/may not be equal to the value(s) of its le/right-hand limits, even when

these limits agree.

Example 17 Evaluang limits of a piecewisedefined funcon

Let f(x) = 2 x 0< x< 1(x 2)2 1< x< 2 , as shown in Figure 1.22. Evaluate the

following.

1. limx1

f(x)

2. limx1+

f(x)

3. limx1

f(x)

4. f(1)

5. limx0+

f(x)

6. f(0)

7. limx2

f(x)

8. f(2)

Notes:

29


36/293

1

2

0.5

1

x

y

Figure 1.23: Graphingfin Example 18

1

2

0.5

1

x

y

Figure 1.24: Graphingfin Example 19

Chapter 1 Limits

S Again we will evaluate each using both thedefinionoffand

its graph.

1. Asxapproaches 1 from the le, we see that f(x)approaches 1. Thereforelim

x1f(x) = 1.

2. As xapproaches 1 from the right, we see that again f(x) approaches 1.Therefore limx1+

f(x) = 1.

3. The limit offasxapproaches 1 exists and is 1, asfapproaches 1 from both

the right and le. Therefore limx1

f(x) = 1.

4. f(1)is not defined. Note that 1 is not in the domain offas defined by theproblem, which is indicated on the graph by an open circle when x= 1.

5. Asxgoes to 0 from the right,f(x)approaches 2. So limx0+

f(x) = 2.

6. f(0)is not defined as 0 is not in the domain off.

7. Asxgoes to 2 from the le,f(x)approaches 0. So limx2f(x) = 0.

8. f(2)is not defined as 2 is not in the domain off.


Letf(x) =

(x 1)2 0x2,x=1

1 x= 1 , as shown in Figure 1.23. Evaluate

the following.

1. limx1

f(x)

2. limx

1+

f(x)

3. limx1

f(x)

4. f(1)

S Itis clear by looking atthe graph that both the le and right-

hand limits off, as xapproaches 1, is 0. Thus it is also clear thatthe limit is 0;

i.e., limx1

f(x) = 0. It is also clearly stated thatf(1) = 1.


Letf(x) =

x2 0x12 x 1< x2 , as shown in Figure 1.24. Evaluate the follow-

ing.

Notes:

30


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1. limx1

f(x)

2. limx1+

f(x)

3. limx1

f(x)

4. f(1)

S It is clear from the definion of the funcon and its graph

that all of the following are equal:

limx1

f(x) = limx1+

f(x) = limx1

f(x) = f(1) = 1.

In Examples 16 19 we were asked to find both limx1

f(x)andf(1). Consider

the following table:

limx1

f(x) f(1)

Example 16 does not exist 1

Example 17 1 not defined

Example 18 0 1

Example 19 1 1

Only in Example 19 do both the funcon and the limit exist and agree. This

seems nice; in fact, it seems normal. This is in fact an important situaon

which we explore in the next secon, entled Connuity. In short, a connu-

ous funconis one in which when a funcon approaches a value asxc (i.e.,when lim

xcf(x) = L), it actually aains that value at c. Such funcons behave

nicely as they are very predictable.

Notes:

31


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1. What are the three ways in which a limit may fail to exist?

2. T/F: If limx1

f(x) = 5, then limx1

f(x) = 5

3. T/F: If limx1

f(x) = 5, then limx1+

f(x) = 5

4. T/F: If limx1

f(x) = 5, then limx1

f(x) = 5

Problems

In Exercises 5 12, evaluate each expression using the given

graph off(x).

5.

0.5

1

1.5

2

0.5

1

1.5

2

x

y

(a) l imx1

f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

(e) l imx0

f(x)

(f) l imx0+

f(x)

6.

0.5

1

1.5

2

0.5

1

1.5

2

x

y

(a) l imx1

f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

(e) l imx2

f(x)

(f) l imx2+

f(x)

7.

0.5

1

1.5

2

0.5

1

1.5

2

x

y

(a) l imx1

f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

(e) l imx2

f(x)

(f) l imx0+

f(x)

8.

0.5

1

1.5

2

0.5

1

1.5

2

x

y

(a) l imx1

f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

9.

0.5

1

1.5

2

0.5

1

1.

5

2

x

y

(a) l imx1

f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

10.

4

3

2

1

1

2

3

4

4

2

2

4

x

y

(a) l imx0

f(x)

(b) l imx0+

f(x)

(c) limx0

f(x)

(d) f(0)


39/293

11.

4

3

2

1

1

2

3

4

4

2

2

4

x

y

(a) limx2

f(x)

(b) limx2+

f(x)

(c) l imx2

f(x)

(d) f(2)

(e) l imx2

f(x)

(f) l imx2+

f(x)

(g) limx2

f(x)

(h) f(2)

12.

4

3

2

1

1

2

3

4

4

2

2

4

x

y

Let 3 a 3 be an integer.

(a) l imxa

f(x)

(b) l imxa+

f(x)

(c) limxa

f(x)

(d) f(a)

In Exercises13 21,evaluate thegiven limits of thepiecewise

defined funconsf.

13. f(x) =

x+1 x 1x2 5 x> 1

(a) l imx1f(x)(b) l im

x1+f(x)

(c) limx1f(x)(d) f(1)

14. f(x) =

2x2 +5x 1 x< 0

sinx x 0(a) l im

x0f(x)

(b) l imx0+

f(x)

(c) limx0

f(x)

(d) f(0)

15. f(x) =

x2 1 x< 1x3 +1 1 x 1x2 +1 x> 1

(a) limx1

f(x)

(b) limx1+

f(x)

(c) l imx1

f(x)

(d) f(1)

(e) l imx1

f(x)

(f) l imx1+

f(x)

(g) limx1

f(x)

(h) f(1)

16. f(x) =

cosx x<

sinx x (a) l im

xf(x)

(b) l imx+

f(x)

(c) limx

f(x)

(d) f()

17. f(x) =

1 cos2x x< a

sin2x x a ,whereais a real number.

(a) l imxa

f(x)

(b) l imx

a+

f(x)

(c) limxa

f(x)

(d) f(a)

18. f(x) =

x+1 x< 11 x= 1

x 1 x> 1(a) l im

x1f(x)

(b) l imx1+

f(x)

(c) limx1

f(x)

(d) f(1)

19. f(x) =

x2 x< 2x+1 x= 2

x2 +2x+4 x> 2(a) l im

x2f(x)

(b) l imx2+

f(x)

(c) limx2

f(x)

(d) f(2)

20. f(x) =

a(x b)2 +c x< b

a(x b) +c x b ,wherea,bandcare real numbers.

(a) l imxb

f(x)

(b) l imxb+

f(x)

(c) limxb

f(x)

(d) f(b)

21. f(x) =

|x|x

x =00 x= 0

(a) l imx0

f(x)

(b) l imx0+f(x)

(c) limx0

f(x)

(d) f(0)

Review

22. Evaluate the limit: limx1

x2 +5x+4

x2 3x 4 .


x2 16x2 4x 32 .


x2 15x+54x2 6x .

25. Evaluate the limit: lim

x2

x2 6x+9x

2

3x .

26. Approximate the limit numerically: limx0.4

x2 4.4x+1.6x2 0.4x .


x2 +5.8x 1.2x2 4.2x+0.8 .

28. Approximate thelimitnumerically: limx0.5

x2 0.5x 0.5x2 +6.5x+3

.


x2 +0.9x 0.1x2 +7.9x 0.8 .


40/293

1

2

3

0.5

1

1.5

x

y

Figure 1.25: A graph offin Example 20.

Chapter 1 Limits

1.5 Connuity

As we have studied limits, we have gained the intuion that limits measure

where a funcon is heading. That is, if limx1

f(x) = 3, then as xis close to 1,

f(x)is close to 3. We have seen, though, that this is not necessarily a good in-dicator of what f(1)actually this. This can be problemac; funcons can tend

to one value but aain another. This secon focuses on funcons thatdo notexhibit such behavior.

Definion 3 Connuous Funcon

Letfbe a funcon defined on an open intervalIcontainingc.

1. fisconnuous atcif limxc

f(x) = f(c).

2. fisconnuous onIiffis connuous atcfor all values ofcinI. Iff

is connuous on(, ), we sayfisconnuous everywhere.

A useful way to establish whether or not a funconfis connuous at cis toverify the following three things:

1. limxc

f(x)exists,

2. f(c)is defined, and

3. limxc

f(x) = f(c).

Example 20 Finding intervals of connuity

Letfbe defined as shown in Figure 1.25. Give the interval(s) on which fis con-

nuous.

S We proceed by examining the three criteria for connuity.

1. The limits limxc

f(x)exists for allcbetween 0 and 3.

2. f(c) is defined for all c between 0 and 3, except for c = 1. We knowimmediately thatfcannot be connuous atx= 1.

3. The limit limxc

f(x) = f(c)for allc between 0 and 3, except, of course, for

c= 1.

We conclude that f is connuous at every point of(0, 3) except at x = 1.Thereforefis connuous on(0, 1)and(1, 3).

Notes:

34


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2

2

2

2

x

y

Figure 1.26: A graph of the step funcon

in Example 21.

1.5 Connuity

Example 21 Finding intervals of connuity

Thefloor funcon,f(x) =x, returns the largest integer smaller than the inputx. (For example,f() = = 3.) The graph offin Figure 1.26 demonstrateswhy this is oen called a step funcon.

Give the intervals on which fis connuous.

S We examine the three criteria for connuity.

1. The limits limxcf(x) do not exist at the jumps from one step to thenext, which occur at all integer values ofc. Therefore the limits exist for

allcexcept whencis an integer.

2. The funcon is defined for all values ofc.

3. The limit limxc

f(x) = f(c) for all values ofc where the limit exist, since each

step consists of just a line.

We conclude thatfis connuous everywhere except at integer values ofc. So

the intervals on whichfis connuous are

. . . ,(2, 1),(1, 0),(0, 1),(1, 2), . . . .

Our definion of connuity on an interval specifies the interval is an open

interval. We can extend the definion of connuity to closed intervals by con-

sidering the appropriate one-sided limits at the endpoints.

Definion 4 Connuity on Closed Intervals

Letfbe defined on the closed interval [a, b]for some real numbersa, b.fisconnuous on[a, b]if:

1. fis connuous on(a, b),

2. limxa+

f(x) = f(a)and

3. limxb

f(x) = f(b).

We can make the appropriate adjustments to talk about connuity on half

open intervals such as[a, b)or(a, b]if necessary.

Notes:

35


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Chapter 1 Limits

Example 22 Determining intervals on which a funcon is connuous

For each of the following funcons, give the domain of the funcon and the

interval(s) on which it is connuous.

1. f(x) = 1/x

2. f(x

) =sinx

3. f(x) =

x

4. f(x) =

1 x2

5. f(x) =|x|

S We examine each in turn.

1. The domain off(x) = 1/xis(, 0) (0, ). As it is a raonal funcon,we apply Theorem 2 to recognize thatfis connuous on all of its domain.

2. The domain off(x) = sinxis all real numbers, or (, ). ApplyingTheorem 3 shows that sinxis connuous everywhere.

3. The domain off(x) =

xis [0, ). Applying Theorem 3 shows thatf(x) =xis connuous on its domain of[0, ).

4. The domain off(x) =

1 x2 is[1, 1]. Applying Theorems 1 and 3shows thatfis connuous on all of its domain,[1, 1].

5. The domain off(x) =|x| is(, ). We can define the absolute valuefuncon as f(x) =

x x< 0x x0 . Each piece of this piece-wise de-

fined funcon is connuous on all of its domain, giving thatfis connuous

on (, 0) and [0, ). As we saw before, we cannot assume this impliesthatfis connuous on(, ); we need to check that lim

x0f(x) = f(0),

asx= 0 is the point whereftransions from one piece of its definionto the other. It is easy to verify that this is indeed true, hence we conclude

thatf(x) =|x| is connuous everywhere.

Connuity is inherently ed to the properes of limits. Because of this, the

properes of limits found in Theorems 1 and 2 apply to connuity as well. Fur-

ther, now knowing the definion of connuity we can reread Theorem 3 as

giving a list of funcons that are connuous on their domains. The following

theorem states how connuous funcons can be combined to form other con-

nuous funcons, followed by a theorem which formally lists funcons that we

know are connuous on their domains.

Notes:

36


43/293


44/293

2

4

1

2

3

x

y

Figure1.27: A graph offin Example 23(a).

Chapter 1 Limits

Example 23 Determining intervals on which a funcon is connuous

State the interval(s) on which each of the following funcons is connuous.

1. f(x) =

x 1+ 5 x2. f(x) = xsinx

3. f(x) = tanx

4. f(x) =

lnx

S We examine each in turn, applying Theorems 8 and 9 as ap-propriate.

1. The squareroot terms are connuous on the intervals [1, ) and (, 5],respecvely. Asfis connuous only where each term is connuous, fis

connuous on[1, 5], the intersecon of these two intervals. A graph offis given in Figure 1.27.

2. The funcons y= xand y= sinxare each connuous everywhere, hencetheir product is, too.

3. Theorem 9 states thatf(x) = tanxis connuous on its domain. Its do-main includes all real numbers except odd mulples of/2. Thusf(x) =

tanxis connuous on

. . .

3

2 ,

2

,

2,

2

,

2,3

2

, . . . ,

or, equivalently, onD={x R |x=n 2

,n is an odd integer}.

4. The domain ofy=

xis[0, ). The range ofy= lnxis(, ), but ifwe restrict its domain to[1, )its range is[0, ). So restricngy= lnxto the domain of[1, )restricts its output is[0, ), on whichy=xisdefined. Thus the domain off(x) =

lnxis[1, ).

A common way of thinking of a connuous funcon is that its graph can

be sketched without liing your pencil. That is, its graph forms a connuous

curve, without holes, breaks or jumps. While beyond the scope of this text,

this pseudodefinion glosses over some of the finer points of connuity. Very

strange funcons are connuous that one would be hard pressed to actually

sketch by hand.

This intuive noon of connuity does help us understand another impor-

tant concept as follows. Suppose fis defined on [1, 2] and f(1) =10 andf(2) = 5. Iffis connuous on[1, 2](i.e., its graph can be sketched as a conn-uous line from(1, 10)to(2, 5)) then we know intuively that somewhere on[1, 2]fmust be equal to 9, and 8, and 7,6, . . . , 0, 1/2,etc. In short, f

Notes:

38


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1.5 Connuity

takes on allintermediatevalues between 10 and 5. It may take on more val-ues;fmay actually equal 6 at some me, for instance, but we are guaranteed all

values between 10 and 5.While this noon seems intuive, it is not trivial to prove and its importance

is profound. Therefore the concept is stated in the form of a theorem.

Theorem 10 Intermediate Value Theorem

Letfbe a connuous funcon on [a, b]and, without loss of generality,letf(a)< f(b). Then for every valuey, wheref(a)< y< f(b), there is avaluecin[a, b]such thatf(c) = y.

One important applicaon of the Intermediate Value Theorem is root find-

ing. Given a funconf, we are oen interested in finding values ofxwhere

f(x) = 0. These roots may be very difficult to find exactly. Good approximaonscan be found through successive applicaons of this theorem. Suppose through

direct computaon we find thatf(a)< 0 andf(b)> 0, wherea< b. The Inter-mediate Value Theorem states that there is a c in[a, b]such thatf(c) = 0. Thetheorem does not give us any clue as to where that value is in the interval[a, b],

just that it exists.

There is a technique that produces a good approximaon ofc. Letdbe the

midpoint of the interval[a, b]and considerf(d). There are three possibilies:

1. f(d) = 0 we got lucky and stumbled on the actual value. We stop as wefound a root.

2. f(d)< 0 Then we know there is a root offon the interval[d, b] we havehalved the size of our interval, hence are closer to a good approximaon

of the root.

3. f(d)> 0 Thenwe knowthere isa rootoffon the interval [a, d] again,we

have halved the size of our interval, hence are closer to a good approxi-maon of the root.

Successively applying this technique is called the Bisecon Methodof root

finding. We connue unl the interval is sufficiently small. We demonstrate this

in the following example.

Example 24 Using the Bisecon Method

Approximate the root off(x) = xcosx, accurate to three places aer thedecimal.

S Consider the graph off(x) = xcosx, shown in Figure 1.28.It is clear that the graph crosses thex-axis somewhere nearx= 0.8. To start the

Notes:

39


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0.5

1

1

0.5

0.5

x

y

Figure1.28: Graphing a root off(x) = xcosx.

Iteraon # Interval Midpoint Sign

1 [0.7, 0.9] f(0.8) > 02 [0.7, 0.8] f(0.75) > 03 [0.7, 0.75] f(0.725) < 04 [0.725, 0.75] f(0.7375) < 05 [0.7375, 0.75] f(0.7438) > 06 [0.7375, 0.7438] f(0.7407) > 07 [0.7375, 0.7407] f(0.7391) > 08 [0.7375, 0.7391] f(0.7383) < 09 [0.7383, 0.7391] f(0.7387) < 0

10 [0.7387, 0.7391] f(0.7389) < 011 [0.7389, 0.7391] f(0.7390) < 012 [0.7390, 0.7391]

Figure 1.29: Iteraons of the BiseconMethod of Root Finding

Chapter 1 Limits

Bisecon Method, pick an interval that contains 0.8. We choose [0.7, 0.9]. Notethat all we care about are signs off(x), not their actual value, so this is all wedisplay.

Iteraon 1: f(0.7)< 0, f(0.9)> 0, andf(0.8)> 0. So replace 0.9 with 0.8 andrepeat.

Iteraon 2: f(0.7)< 0,f(0.8)> 0, and atthemidpoint, 0.75,we havef(0.75)>0. So replace 0.8 with 0.75 and repeat. Note that we dont need to con-nue to check the endpoints, just the midpoint. Thus we put the rest of

the iteraons in Table 1.29.

Noce that in the 12th iteraon we have the endpoints of the interval each

starng with 0.739. Thus we have narrowed the zero down to an accuracy ofthe first three places aer the decimal. Using a computer, we have

f(0.7390) =0.00014, f(0.7391) = 0.000024.Either endpoint of the interval gives a good approximaon of where fis 0. The

Intermediate Value Theorem states thatthe actualzero is sll within this interval.

While we do not know its exact value, we know it starts with 0.739.This type of exercise is rarely done by hand. Rather, it is simple to program

a computer to run such an algorithm and stop when the endpoints differ by apreset small amount. One of the authors did write such a program and found

the zerooff, accurate to 10 places aer thedecimal, to be 0.7390851332. While

it took a few minutes to write the program, it took less than a thousandth of a

second for the program to run the necessary 35 iteraons. In less than 8 hun-

dredths of a second, the zero was calculated to 100 decimal places (with less

than 200 iteraons).

It is a simplemaer to extendthe Bisecon Methodto solve similar problems

tof(x) = 0. For instance, we can solve f(x) = 1. This may seem obvious, butto many it is not. It actually works very well to define a new funcong where

g(x) = f(x) 1. Then use the Bisecon Method to solve g(x) = 0.Similarly, given two funcons fand g, we can use the Bisecon Method to

solvef(x) = g(x). Once again, create a new funcon h where h(x) = f(x)g(x)and solveh(x) = 0.

In Secon 4.1 another equaon solving method will be introduced, called

Newtons Method. In many cases, Newtons Method is much faster. It relies on

more advanced mathemacs, though, so we will wait before introducing it.

This secon formally defined what it means to be a connuous funcon.

Most funcons that we deal with are connuous, so oen it feels odd to have

to formally define this concept. Regardless, it is important, and forms the basis

of the next chapter.

In the next secon we examine one more aspect of limits: limits that involve

infinity.

Notes:

40


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1. In your own words, describe what it means for a funcon

to be connuous.

2. In your own words, describe what the Intermediate ValueTheorem states.

3. What is a root of a funcon?

4. Given funconsfandgon an intervalI, how can the Bisec-

on Method be used to find a valuecwheref(c) = g(c)?

5. T/F: Iffis defined on an open interval containing c , and

limxc

f(x)exists, thenfis connuous atc.

6. T/F: Iffis connuous atc, then limxc

f(x)exists.

7. T/F: Iffis connuous atc, then limx

c+

f(x) = f(c).

8. T/F: Iffis connuous on[a, b], then limxa

f(x) = f(a).

9. T/F: Iffis connuous on[0, 1)and[1, 2), thenfis connu-ous on[0, 2).

10. T/F: The sum of connuous funcons is also connuous.

Problems

In Exercises11 17, a graph of a funconfis given along with

a valuea. Determine iffis connuous at a; if it is not, state

why it is not.

11. a= 1

0.5

1

1.5

2

0.5

1

1.5

2

x

y

12. a= 1

0.5

1

1.5

2

0.5

1

1.5

2

x

y

13. a= 1

0.5

1

1.5

2

0.5

1

1.5

2

x

y

14. a= 0

0.5

1

1.5

2

0.5

1

1.5

2

x

y

15. a= 1

0.5

1

1.5

2

0.5

1

1.5

2

x

y

16. a= 4

4

3

2

1

1

2

3

4

4

2

2

4

x

y

17. (a) a= 2(b) a= 0

(c) a= 2

4

3

2

1

1

2

3

4

4

2

2

4

x

y


48/293

In Exercises 18 21, determine iffis connuous at the indi-

cated values. If not, explain why.

18. f(x) =

1 x= 0

sinxx

x> 0

(a) x= 0

(b) x=

19. f(x) =

x3 x x< 1x 2 x 1

(a) x= 0

(b) x= 1

20. f(x) =

x2+5x+4

x2+3x+2 x = 1

3 x= 1(a) x= 1(b) x= 10

21. f(x) = x264

x211x+24 x =8

5 x= 8(a) x= 0

(b) x= 8

In Exercises 22 32, give the intervals on which the given

funcon is connuous.

22. f(x) = x2 3x+923. g(x) =

x2 4

24. h(k) =

1 k+ k+125. f(t) =

5t2 30

26. g(t) = 1

1 t227. g(x) =

1

1+x2

28. f(x) = ex

29. g(s) = ln s

30. h(t) = cos t

31. f(k) =

1 ek

32. f(x) = sin(ex +x2)

33. Letfbe connuous on[1, 5]wheref(1) =2 andf(5) =10. Does a value 1 < c < 5 exist such that f(c) =9?Why/why not?

34. Let g be connuous on[3, 7] where g(0) = 0 and g(2) =25. Does a value

3 < c < 7 exist such that g(c) = 15?

Why/why not?

35. Let fbe connuous on [1, 1] where f(1) =10 andf(1) = 10. Does a value1 < c < 1 exist such thatf(c) = 11? Why/why not?

36. Let h be a funcon on [1, 1] where h(1) =10 andh(1) = 10. Does a value1 < c < 1 exist such thath(c) = 0? Why/why not?

In Exercises 37 40, use the Bisecon Method to approxi-

mate, accurate to two decimal places, the value of the root

of the given funcon in the given interval.

37. f(x) = x2 +2x 4 on [1, 1.5].38. f(x) = sinx 1/2 on [0.5, 0.55]39. f(x) = ex 2 on [0.65, 0.7].40. f(x) = cosx sinxon[0.7, 0.8].

Review

41. Letf(x) =

x2 5 x< 5

5x x 5 .

(a) l imx5

f(x)

(b) l imx5+

f(x)

(c) limx5

f(x)

(d) f(5)

42. Numerically approximate the following limits:

(a) limx4/5+

x2 8.2x 7.2x2 +5.8x+4

(b) limx4/5

x2 8.2x 7.2x2 +5.8x+4

43. Give an example of funconf(x) for which limx0

f(x) does not

exist.


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1

0.5

0.5

1

50

100

x

y

Figure 1.30: Graphingf(x) =1/x2 for val-ues ofxnear 0.

0.5

1

1.5

2

50

100

x

y

Figure 1.31: Observing infinite limit as

x 1 in Example 25.

1.6 Limits involving infinity


In Definion 1 we stated that in the equaon limxc

f(x) = L, both c and L were

numbers. In this secon we relax that definion a bit by considering situaons

when it makes sense to let cand/orLbe infinity.

As a movang example, consider f(x) = 1/x2, as shown in Figure 1.30.

Note how, asxapproaches 0,f(x)grows very, very large. It seems appropriate,and descripve, to state that

limx0

1

x2 =.

Also note that asxgets very large,f(x) gets very, very small. We could representthis concept with notaon such as

limx

1

x2 =0.

We explore both types of use of in turn.

Definion 5 Limit of Infinity, We say lim

xcf(x) = if for everyM > 0 there exists > 0 such that if

0


50/293

1

0.5

0.5

1

50

50

x

y

Figure 1.32: Evaluang limx0

1

x.

5

5

10

10

x

y

Figure 1.33: Graphingf(x) = 3x

x2 4 .

Chapter 1 Limits

number. But the limit could be infinite. And in fact, we see that the funcon

does appear to be growing larger and larger, as f(.99) = 104, f(.999) = 106,f(.9999) = 108. A similar thing happens on the other side of 1. In general,let a large value M be given. Let = 1/

M. Ifx is within of 1, i.e., if

|x 1|< 1/M, then:

|x 1|< 1

M(x 1)2 < 1

M1

(x 1)2 >M,

which is what we wanted to show. So we may say limx1

1/(x 1)2 =.

Example 26 Evaluang limits involving infinity

Find limx0

1

x, as shown in Figure 1.32.

S It is easy to see that the funcon grows without bound near

0, but it does so in different ways on different sides of 0. Since its behavior is notconsistent, we cannot say that lim

x01

x =. However, we can make a statement

about onesided limits. We can state that limx0+

1

x = and lim

x01

x =.

Vercal asymptotes

If the limit off(x)as xapproachesc from either the le or right (or both) is or , we say the funcon has a vercal asymptoteatc.

Example 27 Finding vercal asymptotes

Find the vercal asymptotes off(x) = 3x

x2

4.

S Vercal asymptotes occur where the funcon grows without

bound; this occurs at values ofcwhere the denominator is 0. The denominator

is small near x = c, which in turn can make the funcon overall take on largevalues. In the case of thegiven funcon, thedenominatoris 0 atx=2. Subs-tung in values ofxclose to 2 and 2 seems to indicate that the funcon tendstoward or at those points. We can graphically confirm this by looking atFigure 1.33.

Notes:

44


51/293

1 1 2

1

2

3

x

y

Figure 1.34: Graphically showing that

f(x) = x2 1

x 1 does not have an asymptote atx= 1.


When a funcon has a vercal asymptote, we can conclude that the de-

nominator is 0 for some part of that funcon. However, just because the de-

nominator is 0 at a certain point does not mean there is a vercal asymptote

there. For instance,f(x) = (x2 1)/(x 1)does not have a vercal asymptoteatx= 1, as shown in Figure 1.34. While the denominator does get small near

x= 1, the numerator gets small too, matching the denominator step for step.

In fact, factoring the numerator, we get

f(x) =(x 1)(x+1)

x 1 .

Canceling the common term, we get that f(x) = x+ 1 for x= 1. So there isclearly no asymptote, rather a hole exists in the graph at x= 1.

The above example may seem a lile contrived. Another example demon-

strang this important concept is f(x) = (sinx)/x. We have considered this

funcon several mes in the previous secons. We found that limx0

sinx

x = 1;

i.e., there is no vercal asymptote. No simple algebraic cancellaon makes this

fact obvious; we used the Squeeze Theorem in Secon 1.3 to prove this.

If the denominator is 0 at a certain point but the numerator is not, then

there will usually be a vercal asymptote at that point. On the other hand, if the

numerator and denominator are both zero at that point, then there may or may

not be a vercal asymptote at that point. This case where the numerator and

denominator are both zero returns us to an important topic.

Indeterminate Forms

We have seen how the limits

limx0

sinx

x and lim

x1x2 1x

1

each return the indeterminate form0

0

when we blindly plug in x = 0 and

x= 1, respecvely. However, 0/0 is not a valid arithmecal expression. It givesno indicaon that the respecve limits are 1 and 2.

With a lile cleverness, one can come up 0/0 expressions which have a limitof, 0, or any other real number. That is why this expression is called indeter-minate.

A key concept to understand is that such limits do not really return 0 /0.Rather, keep in mind that we are taking limits. What is really happening is that

the numerator is shrinking to 0 while the denominator is also shrinking to 0.

Notes:

45


52/293

Chapter 1 Limits

The respecve rates at which they do this are very important and determine the

actual value of the limit.

An indeterminate form indicates that one needs to do more work in order to

compute the limit. That work may be algebraic (such as factoring and canceling)

or it may require a tool such as the Squeeze Theorem. In a later secon we will

learn a technique called lHospitals Rule that provides another way to handle

indeterminate forms.Some other common indeterminate forms are, 0, /, 00, 0

and 1. Again, keep in mind that these are the blind results of evaluang alimit, and each, in and of itself, has no meaning. The expression doesnot really mean subtract infinity from infinity. Rather, it means One quanty

is subtracted from the other, but both are growing without bound. What is the

result? It is possible to get every value between and Note that 1/0 and/0 are not indeterminate forms, though they are not

exactly valid mathemacal expressions, either. Rather they indicate that the

limit will be , , or not exist.

Limits at Infinity and Horizontal Asymptotes

Atthe beginningof this seconwe brieflyconsidered what happens tof(x) =1/x2 asxgrew very large. Graphically, it concerns the behavior of the funcon tothe far right of the graph. We make this noon more explicit in the following

definion.

Definion 6 Limits at Infinity and Horizontal Asymptote

1. We say limx

f(x) = Lif for every > 0 there exists M > 0 such

that ifxM, then |f(x) L|< .

2. We say limx

f(x) = L if for every > 0 there existsM < 0 such

that ifxM, then |f(x) L|< .

3. If limx

f(x) = L or limx

f(x) = L, we say that y= L is a horizontal

asymptoteoff.

We can also define limits such as limx

f(x) =by combining this definionwith Definion 5.

Notes:

46


53/293

20

10

10

20

0.5

1

x

y

(a)

x f(x)

10 0.9615

100 0.9996

10000 0.999996

10 0.9615100 0.9996

10000 0.999996(b)

Figure 1.36: Using a graph and a table

to approximate a horizontal asymptote in

Example 28.


Example 28 Approximang horizontal asymptotes

Approximate the horizontal asymptote(s) off(x) = x2

x2 +4.

S We will approximate the horizontal asymptotes by approxi-

mang the limits

limx

x2

x2 +4 and lim

xx2

x2 +4.

Figure 1.36(a) shows a sketch off, and part(b) gives valuesoff(x) for large mag-nitude values ofx. It seems reasonable to conclude from both of these sources

thatfhas a horizontal asymptote aty= 1.

Later, we will show how to determine this analycally.

Horizontal asymptotes can take on a variety of forms. Figure 1.35(a) shows

that f(x) = x/(x2 + 1) has a horizontal asymptote ofy = 0, where 0 is ap-proached from both above and below.

Figure 1.35(b) shows thatf(x) = x/

x2 +1 has two horizontal asymptotes;

one aty= 1 and the other at y=1.Figure 1.35(c) shows that f(x) = (sinx)/xhas even more interesng behav-

ior than at justx= 0; asxapproaches ,f(x)approaches 0, but oscillates asit does this.

20

10

10

20

1

0.5

0.5

1

x

y

20

10

10

20

1

0.5

0.5

1

x

y

20

10

10

20

0.5

1

x

y

(a) (b) (c)

Figure 1.35: Considering different types of horizontal asym

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