Bear H.S. Understanding Calculus

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118

Understanding Calculus

PROBLEMS

20.1 Estimate log2 = fl

using

and

SIO.

20.2 Find

SIO(h

/0)

for fo

tan

20.3 Find

SIO

for fo

.Jf=X2

dx. (Thisis theareaof one-quarter of theunitcircle,so the answer

~.)

20.4 (a) Find an exact answer for fo

,J9 +x

dx, using the substitution fo

,J9 +x

dx =

tan-I

r:;n

fo :J 9 sec!9 d9, and the 1, 3, '" 10 triangleto expresssec9 whentan9 = 3'

(b) Approximate the exactanswerof (a) as a decimal.

20.5 Find

(h = i) and Ss(h =

f6)

for fo

sinx

dx. (Thereis no elementary antiderivative

for sinx

20.6 (a)

Verify

thatthequadratic function p(x) of (20.2)goesthrough

(-h,

Yo),

(0, YI), (h,

Y2).

(b)

Verify

that

f~h

p(x) dx =

~[Yo

+4YI +

Y2].

20.7 ExplainwhySn,foranyn, givestheexactanswerto

f:(Ax

2+Bx+C)

dx, foranyquadratic

function and any interval. Start withn

= 2.

20.8 (i) Checkthat

(withh =

(b;a)

= a, XI =

(a;b)

= b) givesthe exact answerto

dx.

(ii) Showthat Simpson's Rule with n = 2 givesthe exact answerfor

Q(x) dx for any

cubicpolynomial Q(x), and anyinterval[a,b].

Limits at

00;

Sequences

In this chapter we study the limiting behavior of functions f (x) as x

00.

We write

limx~oo

f(x)

= L, or

f(x)

L as x

00,

provided the difference I

.((x)

- L I

becomes arbitrarily small for all sufficiently large x. If the function I

(x)

becomes arbitrarily

large, we write

limx~oo

I(x)

00,

I(x)

as x

00.

The usual rules for limits

apply, so the limit of a sum is the sum of the limits, and so on.

The limit

= 1

x~oo

x + 1

is an obvious example of the kind of behavior we consider. More generally, limits of rational

functions at

are all easy to evaluate simply by dividing both numerator and denominator by

the highest power of

x which occurs in either.

EXAMPLE

21.1

( )

x2-3x

, (b) I' 5x

4+3x

a Im

......

3-2x+l

' Im

......

4+1OOx

In part (a) we divide top and bottom by x

x2 _ 3x I 3

lim = lim

~:;I

1 =

......

oo 4x

2x + 1 x

......

oo 4 -

Since the numerator on the right tends to 0 and the denominator to 4, the limit is 0,

In (b) we divide top and bottom by

+ 3x

5 + 1.. 5

lim = lim

......

oo 7x

lOOx

......

oo 7 +

In (c) we divide top and bottom by

. x

+ 2x , 1+

---

= hm

=00,

......

oo 3x + 5 x

......

oo }, +

x x-

Since the numerator tends to 1 and the denominator to 0, the "limit" is

00.

It is clear from this example that if

P(x)

and

Q(x)

are polynomials, then

~~;~

0 if

Q(x)

has larger degree than

P(x),

and

~~~~

P(x)

has larger degree than

Q(x),

and

119

120

Understanding

Calculus

the lead coefficients of

P(x)

and Q(x) havethe same sign. If

P(x)

and

Q(x)

have the same

degree,then

~~:~

where

and b; are the lead coefficients of

P(x)

and Q(x).

If a limit isn't obvious, then it is probably one of the following indeterminate forms:

§,

·0,

, or

00°.

For example,

. logx

(i) hm

has the form

X--1>oo

(ii)

(iii)

(iv)

(v)

1 0

lim + has the form

X--1>oo

- 0

logx

lim x log

+~)

has the form

X--1>oo

lim

~)X

has the form 1

;

X--1>oo

lim x

has the form

00°.

X--1>oo

Notice that (i) and (ii) are exactlythe same limit, and this illustrateshow §forms can

be rewritten as

and vice

versa.

Our standard weapon against the indeterminate forms is

l'Hospital's Rule. Since I'Hospital's Rule appliesonly to the forms

§and

the othersmust

firstbe put in one of these

forms.

Forexample,the

. 0 form of (iii) can be put in the §form

follows:

(

1) log (1

+ 1)

lim x log 1 + - = lim 1 x

X--1>oo

The forms 1

and

00°

are treated by considering the limit of the logarithm.

Taking

the

logarithmchanges both 1

and

00°

to the form

. 0 (or 0 .

00),

which can then be treated

like (iii)

above.

I'Hospital's Rule for §or

as x

----+

00:

If both

!(x)

and

g(x)

approach zero

Of b h f . h 0

fi·

. h 1·

f(x)

f'(x)

as x

----+

00,

or I sot unctions ave an In mte nmt, t en Im

--1>OO

g(x)

= Im

--1>OO

g'(x)

providedthat

f:~~

approaches a

finite

limit or tends to

infinity,

that is, that lim

--1>

f:g~

= L

f'(x)

or hm

--1>

g'(x)

000

EXAMPLE 21.2

0 logx

Im--.

x~oo

This has the form

so l'Hospital's Rule will apply if

has a limit. Here

lim

!,(x)

= 1=

x~oo

g'(x)

logx

lim

= lim - =

x~oo

x~oc

It is not really necessary to check first that

limx~oo

makes sense, for that becomes clear in the

computation; just write

lim

f<x) = lim

f'(x)

x~oo

g(x)

x~oo

g'(x)

Chapter 21 • Limits at 00; Sequences

EXAMPLE

21.3

lim x log (1 +

~).

X-H'C)

First put this in the form

§:

(

1) log (1

lim x log 1 + - = lim 1 x

x.-+oo X x.-+oo -

d-r

(-~)

=lim

x.-+oo

-;!

= lim

_1_

= 1.

x.-+oo 1+ x

121

A sequence is a function that is defined only for positive integers, and we will be

interested principally in limits at

sequences. We use the notation {x

}

to denote the

sequence whose values are

Xl,

X2, X3,

••••

For sequences, x; is used in place

the usual

functional notation

x(n).

A sequential limit will be indicated by lim

= L, or x;

asn

00.

f(x)

L as X

00,

and we let X

f(n),

then clearly X

L as n

00.

For example, from Example 21.1, we have

lim =0,

n-.oo

2n + 1

and Example 21.2 shows that

logn

lim

=0.

n-.oo

From Example 21.3 we get

lim

n log

= lim log

!)n

n-.oo

and since the logarithm

(1 +

~)n

approaches 1, we have the important limit

( 1 +

=e.

Now consider the following four sequences, which we will show represent different

orders

growth as n

00:

{tlog n)"}, {n

}, {an}, tnt}; (21.1)

here k > 0, P > 0, and a > 1 are fixed positive numbers. The notation nt, read n-factorial,

denotes the product

the first n positive integers:

n! = 1

·2

3·

....

Each

the sequences (21.1) approaches

as n

00,

but they grow at very different rates,

so that for any

k > 0, P > 0, and a > 1,

(log

n)k n

--~o·

-~o·

-~o.

(21.2)

'an

'n!

For example,

(log

n)5

.;n

l000n

122

UnderstandingCalculus

We use l'Hospital's Rule to show that !¥

----+

0 if p > 0, and this of course shows

that

----+

logx

= li

xII

= lim - - =

x~oo

pxr:'

x~oo

P x

We again use I'Hospital's Rule to show that

----+

0 for p =

1,2,3,

...

and hence

for any

p >

If a > 1, then

· x 1

hm - = lim = 0; (21.3)

x~oo

log a

· x

2x 2 x

hm - = hm =

lim -

=0'

(21.4)

x~oo

aXlog a log a

x~oo

aX '

· x

3 x

hm - = lim =

lim -

=0.

(21.5)

x~oo

aXlog a log a

x~oo

In equations (21.3)-(21.5) we used the result (21.3) in the last step of (21.4), and the result of

(21.4) in the last step of (21.5). The process obviously can be continued to show that f;

----+

for all positive integers

n, and hence for arbitrarily large numbers p, and hence for any positive

number p.

EXAMPLE 21.4

(logn)3

Show

---

0 for all p >

Solution

We write n'' = n! .n! .n! . Since f > 0,

(I:r

---+

Therefore,

(lO:p

n)3

c:~n

y---+

The argument of Example21.4 will workfor any powerof logn, so for any k and any p > 0,

(log

n)k

--~o.

EXAMPLE 21.5

· d I' (log

n)2

In im

---.

n-HX)

Solution

Since 3

dominates n" and n" dominates (logn)2 for any p, it follows that 3

dominates (log

nr'.

can makean explicit comparisonas follows:

(logn)2

(IOgn)2

---+

0.0=0.

n 3"

EXAMPLE 21.6

, . n

410gn

Find lim

--,

n~oo

Solution

Since logn < n for all large n,

410gn

lim

< lim - =

,,~oo

en -

n~oo

The eyeball rule for fractional sequences like those in Examples 21.5 and 21.6 is to

search out the dominant term - 3

in Example 21.5 and en in Example 21.6. If the dominant

term is in the denominator, the fraction tends to zero; if the dominant term is in the numerator,

the fraction tends to

00.

Chapter 21 • Limits at 00; Sequences

123

EXAMPLE

21.7

. . n

+,Jnlogn

FInd hm .

n~oo

n! +n

2log

Solution

The dominanttermis n !,so thefraction tendsto zero. Wecan makeanexplicitestimate by using,Jn < n

and logn < n to increase the numerator, and dropping the term n

logn to decrease the denominator.

Thus,

nlO+,Jnlogn

lO+n

9.n

-----

< = 2 . - ----+

n!+n

2logn

n! n!

EXAMPLE 21.8

lim n

n~oo

This has the indeterminate form

00°.

For any such indeterminate exponential expression, we first take

the logarithm

lim

logn~

= lim

logn =

n~oo

ll~OO

Since the logarithm tends to zero, the function tends to eO = 1;that is,

lim nn = 1.

n~oo

To see why

0, no matter how large a is, let N be some number larger than 2a,

so N<

Then for n = N + k,

.a.

a·

.....

( a ) ( a )

)

n! =

1·2·3·····

N . N + 1 N + 2

...

N + k

[~]

(~y.

Since N is fixed,

is a fixed number, and

(~)k

°as n = N +k

00.

In our later works we will be concerned with functions of the form a.x", For these

functions, the critical question will be for what values of

x is lim

)o anx" = 0, and for what

values of

x is lim

)

anx" =

00.

EXAMPLE 21.9

Find the values of x such that

2:211XIII

----+ 0, and the values of x for which the limit is

00.

ogn

Solution

The dominant terms are the exponentials Ix

III

and 2

so we write

21xl"

_ n

logn - logn

If I

I < 1, the terms approach zero since the sequence is the same as

(Io~l~)all

with a = I

I > 1. If

I > 1, the terms tend to

00,

since

~:~/:

----+

if a

The same sort of argument shows that

(logn) Ix

-;;2"2

----+ 0

if]r]

::; 2,

(logn) Ix

III

-;;2

----+

[x]

> 2.

124

Understanding

Cal

cuI

finish

witha

computation

that

shows

howl'Hospital'sRulefor §asx

follows

fromI'Hospital'sRulefor §as x

0+:

lim

I(x)

= lim I

(~)

x-+oo

g(x)

x-+O+ g

(~)

= lim

(~)

(-~)

x-+O+ g'

(~)

(-*)

= lim I'

(~)

x-+O+ g'

(~)

= lim

f'(x).

x-+oo

g'(x)

PROBLEMS

21.1 lim

--3

x-+oo 1 - x

. 2x + 1

21.2 lim

x-+oo x +

21.3 lim 2-

x-+oo

21.4 lim sinx

x-+oo 5x + 1

21.5 lim -

x-+oo

21.6 lim (logx )2

x-+oo X

21.7 lim

-02

n-+oo

(1.

21.8 lim

n-+oo

21.9 lim 10"

n-+oo

. n logn

21.10 lim

-2-

n-+oo

21.11 lim n sin

n-+oo

21.12 lim

n-+oo

21.13 lim n

2e-

n-+oo

21.14 lim 2

•

n-+oo

21.15 lim -4

n-+oo

21.16 lim (I +

~)n

n-+oo

Hint: Consider sinx as x

---+

(logn)lOn100

loon

21.17 lim

------

n-+oo

Chapter 21 • Limits at

00;

Sequences

. n!

21.18 hm -

n-HXJ

125

•

21.19 lim

---

n~oo

21.20 lim log n +n

n~oo

(~)n

+ Jil

Find the values of x for which the following sequences approach zero, and the values of x for

which they approach

00.

21.21 -Ixl

21.22 n

31x

21.23 (log

Ix I"

lx l

21.24

21.25 (n

+ 1)lxl

= (n

+ 1) I

+ n 1+

f;;

21.26

n!lxl

21.27 n!Ix

(n + 2)! (n + 2)(n + 1)

21.28 (n + 2)!lxl

(22.1)

(22.2)

Improper

Integrals

So far all our integrals

f (x) dx involve bounded functions f (x) on bounded intervals [a, b].

Now we extend the definition to include some unbounded functions and some unbounded

intervals. Such integrals are called

improper

integrals.

The integral

4 1

-dx

is an example

an improper integral with an unbounded integrand, and

-dx

1 x

is an example

an improper integral over an unbounded interval.

For integrals over unbounded intervals (i.e., intervals

the form [a,

(0)

(-00,

b]),

we make the definition:

j(x)

dx = lim r

j(x)

dx,

(22.3)

b-e-co

j(x)dx

aJlr:!ool

j(x)dx.

(22.4)

Here we assume that f

(x)

is integrable over every bounded subinterval

the interval

integration, so all the integrals on the right in (22.3) and (22.4) make sense. If the limit on the

right in (22.3) or (22.4) exists, we say the improper integral on the left converges.

Consider, for example, the improper integral (22.2); by definition,

roo

= lim t' -;dx

b~OOJI

= lim

_~]b

b-s

oo X 1

= lim

[-~

= 1.

b-rsx; b 1

127

128 Understanding Calculus

Since the limit exists, the improper integral fl

converges, and its value is 1.

If the integrand

I (x) is unbounded at one end of an interval [a, b], we again use a

limit to define

I(x)

dx.

I(x)

is unbounded at b, but

I(x)

exists for every c with

a < c < b, then we define

f(x)

dx = lim l

f(x)

dx.

c~b-

The notation c

b-: means that c approaches b from the left side. If

I(x)

is unbounded at

a but integrable on [c, b] for every c between a and b, then we define

f(x)

= lim 1

f(x)

dx,

c~a+

where the limit is taken as c approaches a from the right side

a. If the limit that defines the

integral exists, we say the integral

converges; otherwise, the integral diverges, and we attach

no meaning to it.

As an example of this second type of improper integral, consider the integral

~dx,

that is improper since

as x

0+. From the definition we have

4 1 1

-dx

= lim

-dx

c~O+

= lim

2.JX]4

c~O+

= lim

(2J4

2~)

=4.

c~O+

The most significant improper integrals are those with positive integrands, and we will

stick to that case. If

f (x)

0, then

f (x )

or L

f (x)

represents the area

the

region under the curve

Y = f

(x),

and convergence of the integral means that this area is finite

even though the region is unbounded. It is clear from this area interpretation that the following

comparison theorem holds: If 0

f(x)

kg(x)

for

a.and

g(x)

converges (so

that

kg(x)

converges), then

f(x)

converges. In other words, if there is finite

area under

y =

kg(x),

then there is finite area under the lower curve y =

f(x).

A similar

statement,

course, holds for improper integrals like

2~x

where the integrand is

unbounded at one end of the interval.

We will pay particular attention to integrals

the form

I(x)

dx,

with

I(x)

2: 0,

since these integrals figure importantly in the study

infinite series.

EXAMPLE

22.1

1 1

(a)

dx; (b) 2 dx.

I x

Using the definition, we calculate as follows:

dx = lim

b-+oo I

= lim

2~Jb

b-s

co I

= lim (2v'b - 2) =

00.

b-+oo

The limit does not exist, and the integral diverges.