x’0 1/ ln |x|

so that both the numerator and denominator tend to 0. So l™Hôpital™s Rule

applies and we have

2x/(1 + x 2 ) 2x 2 ln2 (|x|)

lim ln f (x) = lim = lim ’ .

(1 + x 2 )

x’0 ’1/[x ln 2 (|x|)]

x’0 x’0

The numerator tends to 0 (see Example 5.3) and the denominator tends to 1.

Thus

lim ln f (x) = 0.

x’0

130 CHAPTER 5 Indeterminate Forms

But the only way that ln f (x) can tend to zero is if f (x) tends to 1. We conclude

that

lim (1 + x 2 )ln |x| = 1.

x’0

You Try It: Evaluate the limit limx’0+ (1/x)x .

You Try It: Evaluate the limit limx’0+ (1 + x)1/x . In fact this limit gives an

important way to de¬ne Euler™s constant e (see Sections 1.9 and 6.2.3).

5.2.4 PUTTING TERMS OVER A COMMON

DENOMINATOR

Many times a simple algebraic manipulation involving fractions will put a limit

into a form which can be studied using l™Hôpital™s Rule.

EXAMPLE 5.10

Evaluate the limit

1 1

’

lim .

sin 3x 3x

x’0

SOLUTION

We put the fractions over a common denominator to rewrite our limit as

3x ’ sin 3x

lim .

3x · sin 3x

x’0

Both numerator and denominator vanish as x ’ 0. Thus the quotient has

indeterminate form 0/0. By l™Hôpital™s Rule, the limit is therefore equal to

3 ’ 3 cos 3x

lim .

x’0 3 sin 3x + 9x cos 3x

This quotient is still indeterminate; we apply l™Hôpital™s Rule again to obtain

9 sin 3x

= 0.

lim

x’0 18 cos 3x ’ 27x sin 3x

EXAMPLE 5.11

Evaluate the limit

1 1

’

lim .

e4x ’ 1

4x

x’0

131

CHAPTER 5 Indeterminate Forms

SOLUTION

The expression is indeterminate of the form ∞’∞. We put the two fractions

over a common denominator to obtain

e4x ’ 1 ’ 4x

lim .

x’0 4x(e4x ’ 1)

Notice that the numerator and denominator both tend to zero as x ’ 0, so this

is indeterminate of the form 0/0. Therefore l™Hôpital™s Rule applies and our

limit equals

4e4x ’ 4

lim .

x’0 4e4x (1 + 4x) ’ 4

Again the numerator and denominator tend to zero and we apply l™Hôpital™s

Rule; the limit equals

16e4x 1

=.

lim

x’0 16e4x (2 + 4x) 2

1 2

+

You Try It: Evaluate the limit limx’0 .

cos x ’ 1 x2

5.2.5 OTHER ALGEBRAIC MANIPULATIONS

Sometimes a factorization helps to clarify a subtle limit:

EXAMPLE 5.12

Evaluate the limit

x 2 ’ (x 4 + 4x 2 + 5)1/2 .

lim

x’+∞

SOLUTION

The limit as written is of the form ∞ ’ ∞. We rewrite it as

1 ’ (1 + 4x ’2 + 5x ’4 )1/2

’2 ’4 1/2

lim x 1 ’ (1 + 4x + 5x ) = lim

2

.

x ’2

x’+∞ x’+∞

Notice that both the numerator and denominator tend to zero, so it is now

indeterminate of the form 0/0. We may thus apply l™Hôpital™s Rule. The result

is that the limit equals

(’1/2)(1 + 4x ’2 + 5x ’4 )’1/2 · (’8x ’3 ’ 20x ’5 )

lim

’2x ’3

x’+∞

= lim ’(1 + 4x ’2 + 5x ’4 )’1/2 · (2 + 5x ’2 ).

x’+∞

132 CHAPTER 5 Indeterminate Forms

Since this last limit is ’2, we conclude that

x 2 ’ (x 4 + 4x 2 + 5)1/2 = ’2.

lim

x’+∞

EXAMPLE 5.13

Evaluate

e’x ’ (e’3x ’ x 4 )1/3 .

lim

x’’∞

SOLUTION

First rewrite the limit as

1 ’ (1 ’ x 4 e3x )1/3

’x

1 ’ (1 ’ x e ) = lim

4 3x 1/3

lim e .

ex

x’’∞ x’’∞

Notice that both the numerator and denominator tend to zero (here we use the

result analogous to Example 5.7 that x 4 e3x ’ 0). So our new expression is

indeterminate of the form 0/0. l™Hôpital™s Rule applies and our limit equals

’(1/3)(1 ’ x 4 e3x )’2/3 · (’4x 3 · e3x ’ x 4 · 3e3x )

lim

ex

x’’∞

= lim (1/3)(1 ’ x 4 e3x )’2/3 (4x 3 · e2x + 3x 4 · e2x ).

x’’∞

Just as in Example 5.7, x 4 · e3x x 3 · e2x , and x 4 · e2x all tend to zero. We conclude

that our limit equals 0.

√ √

You Try It: Evaluate limx’+∞ [ x + 1 ’ x].

5.3 Improper Integrals: A First Look

5.3.1 INTRODUCTION

The theory of the integral that we learned earlier enables us to integrate a continuous

function f (x) on a closed, bounded interval [a, b]. See Fig. 5.1. However, it is

frequently convenient to be able to integrate an unbounded function, or a function

de¬ned on an unbounded interval. In this section and the next we learn to do so, and

we see some applications of this new technique. The basic idea is that the integral

of an unbounded function is the limit of integrals of bounded functions; likewise,

the integral of a function on an unbounded interval is the limit of the integral on

bounded intervals.

133

CHAPTER 5 Indeterminate Forms

Fig. 5.1

5.3.2 INTEGRALS WITH INFINITE INTEGRANDS

Let f be a continuous function on the interval [a, b) which is unbounded as

x ’ b’ (see Fig. 5.2). The integral

b

f (x) dx

a

Fig. 5.2

is then called an improper integral with in¬nite integrand at b. We often just say

“improper integral” because the source of the improperness will usually be clear

from context. The next de¬nition tells us how such an integral is evaluated.

If

b

f (x) dx

a

is an improper integral with in¬nite integrand at b then the value of the integral is

de¬ned to be

b’

lim f (x) dx,

’0+ a

provided that this limit exists. See Fig. 5.3.

134 CHAPTER 5 Indeterminate Forms

Fig. 5.3

EXAMPLE 5.14

Evaluate the integral

8

4(8 ’ x)’1/3 dx.

2

SOLUTION

The integral

8

4(8 ’ x)’1/3 dx

2

is an improper integral with in¬nite integrand at 8. According to the de¬nition,

the value of this integral is

8’

4(8 ’ x)’1/3 dx,

lim

’0+ 2

provided the limit exists. Since the integrand is continuous on the interval

[2, 8 ’ ], we may calculate this last integral directly. We have

8’

’ 6(8 ’ x)2/3 = lim ’6 ’ 62/3 .

2/3

lim 2

’0+ ’0+

This limit is easy to evaluate: it equals 65/3 . We conclude that the integral is

convergent and

8

4(8 ’ x)’1/3 dx = 65/3 .

2

135

CHAPTER 5 Indeterminate Forms

EXAMPLE 5.15

Analyze the integral

3

(x ’ 3)’2 dx.

2

SOLUTION

This is an improper integral with in¬nite integrand at 3. We evaluate this

integral by considering

3’

3’

’2 ’1

(x ’ 3) dx = lim ’(x ’ 3)

lim

’0+ 2 ’0+ 2

’1

’ 1’1 .

= lim

’0+

This last limit is +∞. We therefore conclude that the improper integral diverges.

’1

+ 1)4/5 ) dx.

You Try It: Evaluate the improper integral ’2 (dx/(x

Improper integrals with integrand which is in¬nite at the left endpoint of

integration are handled in a manner similar to the right endpoint case:

EXAMPLE 5.16

Evaluate the integral

1/2

1

dx.

x · ln2 x

0

SOLUTION

This integral is improper with in¬nite integrand at 0. The value of the integral

is de¬ned to be

1/2 1

lim dx,

x · ln x 2

’0+

provided that this limit exists.

Since 1/(x ln2 x) is continuous on the interval [ , 1/2] for > 0, this last

integral can be evaluated directly and will have a ¬nite real value. For clarity,

write •(x) = ln x, • (x) = 1/x. Then the (inde¬nite) integral becomes

• (x)

dx.

• 2 (x)

Clearly the antiderivative is ’1/•(x). Thus we see that

1/2

1/2 1 1 1 1

dx = lim ’ = lim ’ ’’

lim .

x ·ln2 x

’0+ ’0+ ln x ’0+ ln(1/2) ln

136 CHAPTER 5 Indeterminate Forms

Now as ’ 0+ we have ln ’ ’∞ hence 1/ ln ’ 0. We conclude that

the improper integral converges to 1/ ln 2.

+ 2)’1/2 dx.

0

You Try It: Evaluate the improper integral ’2 1/(x

Many times the integrand has a singularity in the middle of the interval of inte-

gration. In these circumstances we divide the integral into two pieces for each of

which the integrand is in¬nite at one endpoint, and evaluate each piece separately.

EXAMPLE 5.17

Evaluate the improper integral

4

4(x + 1)’1/5 dx.

’4

SOLUTION

The integrand is unbounded as x tends to ’1. Therefore we evaluate

separately the two improper integrals

’1 4

’1/5

4(x + 1)’1/5 dx.

4(x + 1) dx and

’4 ’1

The ¬rst of these has the value

’1’

’1’

’1/5

4(x + 1) dx = lim 5(x + 1) 4/5

lim

’0+ ’4 ’0+ ’4

= lim 5 (’ )4/5 ’ (’3)4/5

’0+

= ’5 · 34/5

The second integral has the value

4

4

’1/5

4(x + 1) dx = lim 5(x + 1) 4/5

lim

’0+ ’1+ ’0+ ’1+

= lim 5 54/5 ’ 4/5

’0+

=5 9/5

.

We conclude that the original integral converges and

’1

4 4

’1/5 ’1/5

4(x + 1)’1/5 dx

4(x + 1) dx = 4(x + 1) dx +

’4 ’4 ’1

= ’5 · 3 +5

4/5 9/5

.

’1 dx.

3

You Try It: Evaluate the improper integral ’4 x

137

CHAPTER 5 Indeterminate Forms

It is dangerous to try to save work by not dividing the integral at the singularity.

The next example illustrates what can go wrong.

EXAMPLE 5.18

Evaluate the improper integral

2

x ’4 dx.

’2

SOLUTION

What we should do is divide this problem into the two integrals

0 2

’4

x ’4 dx.

dx and (—)

x

’2 0

Suppose that instead we try to save work and just antidifferentiate:

2

2 1 1

’4

dx = ’ x ’3 =’

x .

3 12

’2 ’2

A glance at Fig. 5.4 shows that something is wrong. The function x ’4 is

positive hence its integral should be positive too. However, since we used

an incorrect method, we got a negative answer.

Fig. 5.4

In fact each of the integrals in line (—) diverges, so by de¬nition the improper

integral

2

x ’4 dx

’2

diverges.

138 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.19

Analyze the integral