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lim ,
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

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