<< . .

. 14
( : 41)



. . >>

1 (ln x/x) + x sin x dx
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 )

<< . .

. 14
( : 41)



. . >>