2

(c)

2

1 tan x ’ x cos x dx

2 3

(d)

e 2

(e) 1 (ln x /x) dx

82

4 x · cos x sin x dx

3 3

(f)

5. Calculate the area under the given function and above the x-axis over the

indicated interval.

f (x) = x 2 + x + 6 [2, 5]

(a)

g(x) = sin x cos x [0, π/4]

(b)

2

h(x) = xex [1, 2]

(c)

k(x) = ln x/x [1, e]

(d)

6. Draw a careful sketch of each function on the given interval, indicating

subintervals where area is positive and area is negative.

f (x) = x 3 + 3x [’2, 2]

(a)

g(x) = sin 3x cos 3x [’2π, 2π ]

(b)

h(x) = ln x/x [1/2, e]

(c)

4

m(x) = x 3 ex [’3, 3]

(d)

122 CHAPTER 4 The Integral

7. For each function in Exercise 6, calculate the positive area between the

graph of the given function and the x-axis over the indicated interval.

8. In each part of Exercise 6, calculate the signed area between the graph of

the given function and the x-axis over the indicated interval.

9. Calculate the area between the two given curves over the indicated interval.

f (x) = 2x 2 ’ 4, g(x) = ’3x 2 + 10 ’1 ¤ x ¤ 1

(a)

f (x) = x 2 , g(x) = x 3 0 ¤ x ¤ 1

(b)

f (x) = 2x, g(x) = ’x 2 + 3 ’3 ¤ x ¤ 1

(c)

f (x) = ln x, g(x) = x 1 ¤ x ¤ e

(d)

f (x) = sin x, g(x) = x 0 ¤ x ¤ π/4

(e)

f (x) = ex , g(x) = x 0 ¤ x ¤ 3

(f )

10. Calculate the area enclosed by the two given curves.

f (x) = x, g(x) = x 2

(a)

√

f (x) = x, g(x) = x 2

(b)

f (x) = x 4 , g(x) = 3x 2

(c)

f (x) = x 4 , g(x) = ’2x 2 + 3

(d)

f (x) = x 4 ’ 2, g(x) = ’x 4 + 2

(e)

f (x) = 2x, g(x) = ’x 2 + 3

(f )

CHAPTER 5

Indeterminate

Forms

5.1 l™Hôpital™s Rule

5.1.1 INTRODUCTION

Consider the limit

f (x)

lim (—)

.

x’c g(x)

If limx’c f (x) exists and limx’c g(x) exists and is not zero then the limit

(—) is straightforward to evaluate. However, as we saw in Theorem 2.3, when

limx’c g(x) = 0 then the situation is more complicated (especially when

limx’c f (x) = 0 as well).

For example, if f (x) = sin x and g(x) = x then the limit of the quotient as

x ’ 0 exists and equals 1. However if f (x) = x and g(x) = x 2 then the limit of

the quotient as x ’ 0 does not exist.

In this section we learn a rule for evaluating indeterminate forms of the type (—)

when either limx’c f (x) = limx’c g(x) = 0 or limx’c f (x) = limx’c g(x) = ∞.

Such limits, or “forms,” are considered indeterminate because the limit of the quo-

tient might actually exist and be ¬nite or might not exist; one cannot analyze such

a form by elementary means.

123

Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

124 CHAPTER 5 Indeterminate Forms

5.1.2 L™H”PITAL™S RULE

Theorem 5.1 (l™Hôpital™s Rule)

Let f (x) and g(x) be differentiable functions on (a, c) ∪ (c, b). If

lim f (x) = lim g(x) = 0

x’c x’c

then

f (x) f (x)

= lim

lim ,

x’c g(x) x’c g (x)

provided this last limit exists as a ¬nite or in¬nite limit.

Let us learn how to use this new result.

EXAMPLE 5.1

Evaluate

ln x

lim .

x2 + x ’ 2

x’1

SOLUTION

We ¬rst notice that both the numerator and denominator have limit zero

as x tends to 1. Thus the quotient is indeterminate at 1 and of the form 0/0.

l™Hôpital™s Rule therefore applies and the limit equals

(d/dx)(ln x)

lim ,

x’1 (d/dx)(x 2 + x ’ 2)

provided this last limit exists. The last limit is

1/x 1

= lim

lim .

x’1 2x + 1 x’1 2x 2 + x

Therefore we see that

ln x 1

=.

lim

x’1 x 2 + x ’ 2 3

You Try It: Apply l™Hôpital™s Rule to the limit limx’2 sin(π x)/(x 2 ’ 4).

You Try It: Use l™Hôpital™s Rule to evaluate limh’0 (sin h/ h) and limh’0 (cos h’

1/ h). These limits are important in the theory of calculus.

EXAMPLE 5.2

Evaluate the limit

x3

lim .

x ’ sin x

x’0

125

CHAPTER 5 Indeterminate Forms

SOLUTION

As x ’ 0 both numerator and denominator tend to zero, so the quotient is

indeterminate at 0 of the form 0/0. Thus l™Hôpital™s Rule applies. Our limit

equals

(d/dx)x 3

lim ,

x’0 (d/dx)(x ’ sin x)

provided that this last limit exists. It equals

3x 2

lim .

x’0 1 ’ cos x

This is another indeterminate form. So we must again apply l™Hôpital™s Rule.

The result is

6x

lim .

x’0 sin x

This is again indeterminate; another application of l™Hôpital™s Rule gives us

¬nally

6

= 6.

lim

x’0 cos x

We conclude that the original limit equals 6.

You Try It: Apply l™Hôpital™s Rule to the limit limx’0 x/[1/ ln |x|].

Indeterminate Forms Involving ∞ We handle indeterminate forms involv-

ing in¬nity as follows: Let f (x) and g(x) be differentiable functions on (a, c) ∪

(c, b). If

lim f (x) and lim g(x)

x’c x’c

both exist and equal +∞ or ’∞ (they may have the same sign or different signs)

then

f (x) f (x)

= lim

lim ,

x’c g(x) x’c g (x)

provided this last limit exists either as a ¬nite or in¬nite limit.

Let us look at some examples.

EXAMPLE 5.3

Evaluate the limit

lim x 3 · ln |x|.

x’0

126 CHAPTER 5 Indeterminate Forms

SOLUTION

This may be rewritten as

ln |x|

lim .

x’0 1/x 3

Notice that the numerator tends to ’∞ and the denominator tends to ±∞ as

x ’ 0. Thus the quotient is indeterminate at 0 of the form ’∞/ + ∞. So we

may apply l™Hôpital™s Rule for in¬nite limits to see that the limit equals

1/x

= lim ’x 3 /3 = 0.

lim ’4

x’0 ’3x x’0

Yet another version of l™Hôpital™s Rule, this time for unbounded intervals, is

this: Let f and g be differentiable functions on an interval of the form [A, +∞).

If limx’+∞ f (x) = limx’+∞ g(x) = 0 or if limx’+∞ f (x) = ±∞ and

limx’+∞ g(x) = ±∞, then

f (x) f (x)

= lim

lim

x’+∞ g(x) x’+∞ g (x)

provided that this last limit exists either as a ¬nite or in¬nite limit. The same result

holds for f and g de¬ned on an interval of the form (’∞, B] and for the limit as

x ’ ’∞.

EXAMPLE 5.4

Evaluate

x4

lim .

ex

x’+∞

SOLUTION

We ¬rst notice that both the numerator and the denominator tend to +∞ as

x ’ +∞. Thus the quotient is indeterminate at +∞ of the form +∞/ + ∞.

Therefore the new version of l™Hôpital™s Rule applies and our limit equals

4x 3

lim .

x’+∞ ex

Again the numerator and denominator tend to +∞ as x ’ +∞, so we once

more apply l™Hôpital™s Rule. The limit equals

12x 2

= 0.

lim

x’+∞ ex

We must apply l™Hôpital™s Rule two more times. We ¬rst obtain

24x

lim

x’+∞ ex

127

CHAPTER 5 Indeterminate Forms

and then

24

lim .

x’+∞ ex

We conclude that

x4

= 0.

lim

x’+∞ ex

ex

You Try It: Evaluate the limit limx’+∞ .

x ln x

You Try It: Evaluate the limit limx’’∞ x 4 · ex .

EXAMPLE 5.5

Evaluate the limit

sin(2/x)

lim .

sin(5/x)

x’’∞

SOLUTION

We note that both numerator and denominator tend to 0, so the quotient is

indeterminate at ’∞ of the form 0/0. We may therefore apply l™Hôpital™s Rule.

Our limit equals

(’2/x 2 ) cos(2/x)

lim .

x’’∞ (’5/x 2 ) cos(5/x)

This in turn simpli¬es to

2 cos(2/x) 2

=.

lim

x’’∞ 5 cos(5/x) 5

l™Hôpital™s Rule also applies to one-sided limits. Here is an example.

EXAMPLE 5.6

Evaluate the limit

√

sin

x

√.

lim

x’0+ x

SOLUTION

Both numerator and denominator tend to zero so the quotient is indeterminate

at 0 of the form 0/0. We may apply l™Hôpital™s Rule; differentiating numerator

and denominator, we ¬nd that the limit equals

√

[cos x] · (1/2)x ’1/2 √

= lim cos x

lim

(1/2)x ’1/2

x’0+ x’0+

= 1.

You Try It: How can we apply l™Hôpital™s Rule to evaluate limx’0+ x · ln x?

128 CHAPTER 5 Indeterminate Forms

5.2 Other Indeterminate Forms

5.2.1 INTRODUCTION

By using some algebraic manipulations, we can reduce a variety of indeterminate

limits to expressions which can be treated by l™Hôpital™s Rule. We explore some of

these techniques in this section.

5.2.2 WRITING A PRODUCT AS A QUOTIENT

The technique of the ¬rst example is a simple one, but it is used frequently.

EXAMPLE 5.7

Evaluate the limit

lim x 2 · e3x .

x’’∞

Y

SOLUTION

Notice that x 2 ’ +∞ while e3x ’ 0. So the limit is indeterminate of the

FL

form 0 · ∞. We rewrite the limit as

x2

AM

lim .

x’’∞ e’3x

Now both numerator and denominator tend to in¬nity and we may apply

l™Hôpital™s Rule. The result is that the limit equals

TE

2x

lim .

x’’∞ ’3e’3x

Again the numerator and denominator both tend to in¬nity so we apply

l™Hôpital™s Rule to obtain:

2

lim .

9e’3x

x’’∞

It is clear that the limit of this last expression is zero. We conclude that

lim x · e3x = 0.

x’’∞

√

You Try It: Evaluate the limit limx’+∞ e’ · x.

x

5.2.3 THE USE OF THE LOGARITHM

The natural logarithm can be used to reduce an expression involving exponentials

to one involving a product or a quotient.

129

CHAPTER 5 Indeterminate Forms

EXAMPLE 5.8

Evaluate the limit

lim x x .

x’0+

SOLUTION

We study the limit of f (x) = x x by considering ln f (x) = x · ln x. We

rewrite this as

ln x

lim ln f (x) = lim .

x’0+ x’0+ 1/x

Both numerator and denominator tend to ±∞, so the quotient is indeterminate

of the form ’∞/∞. Thus l™Hôpital™s Rule applies. The limit equals

1/x

= lim ’x = 0.

lim

’1/x 2 x’0+

x’0+

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

conclude that

lim x x = 1.

x’0+

EXAMPLE 5.9

Evaluate the limit

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

x’0

SOLUTION

Let f (x) = (1 + x 2 )ln |x| and consider ln f (x) = ln |x| · ln(1 + x 2 ). This

expression is indeterminate of the form ’∞ · 0.

We rewrite it as

ln(1 + x 2 )