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sin(3x + 1) dx = ’ cos(3x + 1) + C.
3
We invite the reader to verify that the derivative of the answer on the right-hand
side gives sin(3x + 1).
EXAMPLE 4.2
Calculate
x
dx.
x2 + 3

SOLUTION
We notice that the numerator of the fraction is nearly the derivative of the
denominator. Put in other words, if we were asked to integrate
2x
x2 + 3
102 CHAPTER 4 The Integral

then we would see that we are integrating an expression of the form
• (x)
•(x)
(which we in fact encountered among our differentiation rules in Section 2.5).
As we know, expressions like this arise from differentiating log •(x).
Returning to the original problem, we pose our initial guess as log[x 2 + 3].
Differentiation of this expression gives the answer 2x/[x 2 + 3]. This is close
to what we want, but we must adjust by a factor of 1/2. We write our ¬nal
answer as
1
x
dx = log[x 2 + 3] + C.
x2 + 3 2
You Try It: Calculate the inde¬nite integral
2 +5
xe3x dx.

EXAMPLE 4.3
Calculate the inde¬nite integral

(x 3 + x 2 + 1)50 · (6x 2 + 4x) dx.

SOLUTION
We observe that the expression 6x 2 + 4x is nearly the derivative of x 3 +
x 2 + 1. In fact if we set •(x) = x 3 + x 2 + 1 then the integrand (the quantity
that we are asked to integrate) is
[•(x)]50 · 2• (x).
It is natural to guess as our antiderivative [•(x)]51 . Checking our work, we
¬nd that
([•(x)]51 ) = 51[•(x)]50 · • (x).
We see that the answer obtained is quite close to the answer we seek; it is off by
a numerical factor of 2/51. With this knowledge, we write our ¬nal answer as
2
(x 3 + x 2 + 1)50 · (6x 2 + 4x) dx = · [x 3 + x 2 + 1]51 + C.
51
You Try It: Calculate the inde¬nite integral
x2
dx.
x3 + 5
103
CHAPTER 4 The Integral

4.2 Area
Consider the curve shown in Fig. 4.1. The curve is the graph of y = f (x). We set
for ourselves the task of calculating the area A that is (i) under the curve, (ii) above
the x-axis, and (iii) between x = a and x = b. Refer to Fig. 4.2 to see the geometric
region we are considering.

y = f (x)




Fig. 4.1




y = f (x)




a b




Fig. 4.2

We take it for granted that the area of a rectangle of length and width w is
— w. Now our strategy is to divide the base interval [a, b] into equal subintervals.
Fix an integer k > 0. We designate the points
P = {x0 , x1 , x2 , . . . , xk },
with x0 = a and xk = b. We require that |xj ’ xj ’1 | = |b ’ a|/k ≡ x for
j = 1, . . . , k. In other words, the points x0 , x1 , . . . , xk are equally spaced. We call
104 CHAPTER 4 The Integral

the set P a partition. Sometimes, to be more speci¬c, we call it a uniform partition
(to indicate that all the subintervals have the same length). Refer to Fig. 4.3.

b_a
k


x0 = a xj xk = b

Fig. 4.3

The idea is to build an approximation to the area A by erecting rectangles over
the segments determined by the partition. The ¬rst rectangle R1 will have as base
the interval [x0 , x1 ] and height chosen so that the rectangle touches the curve at its
upper right hand corner; this means that the height of the rectangle is f (x1 ). The
second rectangle R2 has as base the interval [x1 , x2 ] and height f (x2 ). Refer to
Fig. 4.4.



y = f (x)




x0 = a x1 x2 xk = b


Fig. 4.4

Continuing in this manner, we construct precisely k rectangles, R1 , R2 , . . . , Rk ,
as shown in Fig. 4.5. Now the sum of the areas of these rectangles is not exactly
equal to the area A that we seek. But it is close. The error is the sum of the little
semi-triangular pieces that are shaded in Fig. 4.6. We can make that error as small
as we please by making the partition ¬ner. Figure 4.7 illustrates this idea.
Let us denote by R(f, P) the sum of the areas of the rectangles that we created
from the partition P. This is called a Riemann sum. Thus
k
R(f, P) = f (xj ) · x ≡ f (x1 ) · x + f (x2 ) · x + · · · + f (xk ) · x.
j =1

Here the symbol k =1 denotes the sum of the expression to its right for each of
j
the instances j = 1 to j = k.
105
CHAPTER 4 The Integral


y = f (x)




x0 = a x1 x2 xk = b


Fig. 4.5



y = f (x)




x0 = a x1 x2 xk = b


Fig. 4.6



y = f (x)




a b

Fig. 4.7

The reasoning just presented suggests that the true area A is given by
lim R(f, P).
k’∞
We call this limit the integral of f from x = a to x = b and we write it as
b
f (x) dx.
a
106 CHAPTER 4 The Integral

Thus we have learned that
b
area A = f (x) dx.
a
It is well to take a moment and comment on the integral notation. First, the
integral sign



is an elongated “S,” coming from “summation.” The dx is an historical artifact,
coming partly from traditional methods of developing the integral, and partly from
a need to know explicitly what the variable is. The numbers a and b are called the
limits of integration”the number a is the lower limit and b is the upper limit. The
function f is called the integrand.
Before we can present a detailed example, we need to record some important
information about sums:
N
I. We need to calculate the sum S = 1 + 2 + · · · + N = j =1 j . To achieve this
goal, we write
S =1+2 + · · · + (N ’ 1) + N
S = N + (N ’ 1) + · · · + 2 +1
Adding each column, we obtain
2S = (N + 1) + (N + 1) + · · · + (N + 1) + (N + 1) .
N times

Thus
2S = N · (N + 1)
or
N · (N + 1)
S= .
2
This is a famous formula that was discovered by Carl Friedrich Gauss (1777“1855)
when he was a child.
n
II. The sum S = 12 + 22 + · · · + N 2 = 2 is given by
j =1 j

2N 3 + 3N 2 + N
S= .
6
We shall not provide the details of the proof of this formula, but refer the interested
reader to [SCH2].
107
CHAPTER 4 The Integral

For our ¬rst example, we calculate the area under a parabola.
EXAMPLE 4.4
Calculate the area under the curve y = x 2 , above the x-axis, and between
x = 0 and x = 2.
SOLUTION
Refer to Fig. 4.8 as we reason along. Let f (x) = x 2 .




Fig. 4.8

Consider the partition P of the interval [0, 2] consisting of k + 1 points
x0 , x1 , . . . , xk . The corresponding Riemann sum is
k
R(f, P) = f (xj ) · x.
j =1

Of course
2’0 2
x= =
k k
and
2
xj = j · .
k
In addition,
2
4j 2
2
f (xj ) = j · = 2.
k k
108 CHAPTER 4 The Integral

As a result, the Riemann sum for the partition P is
k
4j 2 2
R(f, P) = ·
k2 k
j =1
k k
8j 2 8
= =3 j 2.
k3 k
j =1 j =1

Now formula II above enables us to calculate the last sum explicitly. The result
is that
8 2k 3 + 3k 2 + k
R(f, P) = 3 ·
6
k
84 4
= + + 2.
3 k 3k
In sum,


Y
2 84 4 8
x 2 dx = lim R(f, P) = lim ++ 2=.
FL
3 k 3k 3
k’∞ k’∞
0
We conclude that the desired area is 8/3.
AM

You Try It: Use the method presented in the last example to calculate the area
under the graph of y = 2x and above the x-axis, between x = 1 and x = 2. You
should obtain the answer 3, which of course can also be determined by elementary
TE



considerations”without taking limits.

The most important idea in all of calculus is that it is possible to calculate an
integral without calculating Riemann sums and passing to the limit. This is the
Fundamental Theorem of Calculus, due to Leibniz and Newton. We now state the
theorem, illustrate it with examples, and then brie¬‚y discuss why it is true.
Theorem 4.1 (Fundamental Theorem of Calculus)
Let f be a continuous function on the interval [a, b]. If F is any antiderivative of
f then
b
f (x) dx = F (b) ’ F (a).
a

EXAMPLE 4.5
Calculate
2
x 2 dx.
0
109
CHAPTER 4 The Integral

SOLUTION
We use the Fundamental Theorem. In this example, f (x) = x 2 . We need to
¬nd an antiderivative F . From our experience in Section 4.1, we can determine
that F (x) = x 3 /3 will do. Then, by the Fundamental Theorem of Calculus,
2 23 03 8
x dx = F (2) ’ F (0) = ’ =.
2
3 3 3
0

Notice that this is the same answer that we obtained using Riemann sums in
Example 4.4.

EXAMPLE 4.6
Calculate
π
sin x dx.
0

SOLUTION
In this example, f (x) = sin x. An antiderivative for f is F (x) = ’ cos x.
Then
π
sin x dx = F (π) ’ F (0) = (’ cos π ) ’ (’ cos 0) = 1 + 1 = 2.
0

EXAMPLE 4.7
Calculate
2
ex ’ cos 2x + x 3 ’ 4x dx.
1

SOLUTION
In this example, f (x) = ex ’ cos 2x + x 3 ’ 4x. An antiderivative for f is
F (x) = ex ’ (1/2) sin 2x + x 4 /4 ’ 2x 2 . Therefore
2
ex ’ cos 2x + x 3 ’ 4x dx = F (2) ’ F (1)
1
24
1
= e ’ sin(2 · 2) + ’ 2 · 22
2
2 4
14
1
’ e ’ sin(2 · 1) + ’ 2 · 12
1
2 4
1 9
= (e2 ’ e) ’ [sin 4 ’ sin 2] ’ .
2 4
110 CHAPTER 4 The Integral

You Try It: Calculate the integral
’1
x 3 ’ cos x + x dx.
’3

Math Note: Observe in this last example, in fact in all of our examples, you can
use any antiderivative of the integrand when you apply the Fundamental Theorem
of Calculus. In the last example, we could have taken F (x) = ex ’ (1/2) sin 2x +
x 4 /4 ’ 2x 2 + 5 and the same answer would have resulted. We invite you to provide
the details of this assertion.
Justi¬cation for the Fundamental Theorem Let f be a continuous function on
the interval [a, b]. De¬ne the area function F by
F (x) = area under f , above the x-axis, and between 0 and x.




Fig. 4.9

Let us use a pictorial method to calculate the derivative of F . Refer to Fig. 4.9
as you read on. Now
F (x + h) ’ F (x) [area between x and x + h, below f ]
=
h h
f (x) · h

h
= f (x).
As h ’ 0, the approximation in the last display becomes nearer and nearer to
equality. So we ¬nd that
F (x + h) ’ F (x)
= f (x).
lim
h
h’0

But this just says that F (x) = f (x).
What is the practical signi¬cance of this calculation? Suppose that we wish to
calculate the area under the curve f , above the x-axis, and between x = a and
x = b. Obviously this area is F (b) ’ F (a). See Fig. 4.10. But we also know that
111
CHAPTER 4 The Integral

b
that area is We conclude therefore that
a f (x) dx.
b
f (x) dx = F (b) ’ F (a).
a




y = f (x)




a b




F (b) _ F (a)



Fig. 4.10

Finally, if G is any other antiderivative for f then G(x) = F (x) + C. Hence
b
G(b) ’ G(a) = [F (b) + C] ’ [F (a) + C] = F (b) ’ F (a) = f (x) dx.
a
That is the content of the Fundamental Theorem of Calculus.

You Try It: Calculate the area below the curve y = ’x 2 + 2x + 4 and above the
x-axis.



4.3 Signed Area
Without saying so explicitly, we have implicitly assumed in our discussion of area in
the last section that our function f is positive, that is its graph lies about the x-axis.
But of course many functions do not share that property. We nevertheless would
like to be able to calculate areas determined by such functions, and to calculate the
corresponding integrals.
This turns out to be simple to do. Consider the function y = f (x) shown in
Fig. 4.11. It is negative on the interval [a, b] and positive on the interval [b, c].
112 CHAPTER 4 The Integral

Suppose that we wish to calculate the shaded area as in Fig. 4.12. We can do so by
breaking the problem into pieces.




Fig. 4.11




Fig. 4.12

Of course, because f ≥ 0, the area between x = b and x = c is given by the
c
integral b f (x) dx, just as we have discussed in the last section. But our discussions
do not apply directly to the area between x = a and x = b. What we can do is
instead consider the function g = ’f . Its graph is shown in Fig. 4.13. Of course
g is a positive function on [a, b], except at the endpoints a and b; and the area
under g”between x = a and x = b”is just the same as the shaded area between
113
CHAPTER 4 The Integral




Fig. 4.13


x = a and x = b in Fig. 4.14. That area is
b b
g(x) dx = ’ f (x) dx.
a a




Fig. 4.14

In total, the aggregate shaded area exhibited in Fig. 4.15, over the entire interval
[a, c], is
b c
’ f (x) dx + f (x) dx.
a b
114 CHAPTER 4 The Integral




Fig. 4.15

What we have learned is this: If f (x) < 0 on the interval under discussion, then
the integral of f will be a negative number. If we want to calculate positive area
then we must interject a minus sign.
Let us nail down our understanding of these ideas by considering some examples.
EXAMPLE 4.8
Calculate the (positive) area, between the graph of f (x) = x 3 ’ 2x 2 ’
11x + 12 and the x-axis, between x = ’3 and x = 4.
SOLUTION
Consider Fig. 4.16. It was drawn using the technique of Section 3.1, and it
plainly shows that f is positive on [’3, 1] and negative on [1, 4]. From the




Fig. 4.16
115
CHAPTER 4 The Integral

discussion preceding this example, we know then that
1 4
Area = f (x) dx ’ f (x) dx
’3 1
1
= x 3 ’ 2x 2 ’ 11x + 12 dx
’3
4
’ x 3 ’ 2x 2 ’ 11x + 12 dx
1
1
4 2x 3 11x 2
x
= ’ ’ + 12x
4 3 2 ’3
4
x 4 2x 3 11x 2
’ ’ ’ + 12x (—)
.
4 3 2 1
Here we are using the standard shorthand
F (x)|b
a

to stand for
F (b) ’ F (a).
Thus we have
160 297
(—) = + .
3 12
Notice that, by design, each component of the area has made a positive
contribution to the ¬nal answer. The total area is then
937
Area = .
12
EXAMPLE 4.9
Calculate the (positive) area between f (x) = sin x and the x-axis for
’2π ¤ x ¤ 2π.
SOLUTION
We observe that f (x) = sin x ≥ 0 for ’2π ¤ x ¤ ’π and 0 ¤ x ¤ π.
Likewise, f (x) = sin x ¤ 0 for ’π ¤ x ¤ 0 and π ¤ x ¤ 2π. As a result
’π 0
Area = sin x dx ’ sin x dx
’2π ’π

π
+ sin x dx ’ sin x dx.
0 π
116 CHAPTER 4 The Integral

This is easily calculated to equal
2 + 2 + 2 + 2 = 8.

You Try It: Calculate the (positive) area between y = x 3 ’ 6x 2 + 11x ’ 6 and
the x-axis.
EXAMPLE 4.10
Calculate the signed area between the graph of y = cos x + 1/2 and the
x-axis, ’π/2 ¤ x ¤ π .

SOLUTION
This is easy, because the solution we seek is just the value of the integral:
π 1
Area = cos x + dx
2
’π/2
π
x
= sin x +
2 ’π/2
’π
π
= 0+ ’ ’1 +
2 4

= + 1.
4
Math Note: In the last example, we have counted positive area as positive and
negative area as negative. Our calculation shows that the aggregate area is positive.
We encourage the reader to draw a graph to make this result plausible.

You Try It: Calculate the actual positive area between the graph of y = x 2 ’ 4,
’5 ¤ x ¤ 5 and the x-axis.

You Try It: Calculate the signed area between the graph of y = x 2 ’ 4 and the
x-axis, ’4 ¤ x ¤ 5.



4.4 The Area Between Two Curves
Frequently it is useful to ¬nd the area between two curves. See Fig. 4.17. Following
the model that we have set up earlier, we ¬rst note that the intersected region has
left endpoint at x = a and right endpoint at x = b. We partition the interval [a, b]
as shown in Fig. 4.18. Call the partition
P = {x0 , x1 , . . . , xk }.
117
CHAPTER 4 The Integral




Fig. 4.17




Fig. 4.18

Then, as usual, we erect rectangles over the intervals determined by the partition
(Fig. 4.19).
Notice that the upper curve, over the interval [a, b], is y = f (x) and the lower
curve is y = g(x) (Fig. 4.17). The sum of the areas of the rectangles is therefore
k
[f (xj ) ’ g(x)] · x.
j =1

But of course this is a Riemann sum for the integral
b
[f (x) ’ g(x)] dx.
a
118 CHAPTER 4 The Integral




Fig. 4.19




Y
We declare this integral to be the area determined by the two curves.
FL
EXAMPLE 4.11
Find the area between the curves y = x 2 ’ 2 and y = ’(x ’ 1)2 + 3.
AM

SOLUTION
We set the two equations equal and solve to ¬nd that the curves intersect
at x = ’1 and x = 2. The situation is shown in Fig. 4.20. Notice that y =
TE



’(x ’ 1)2 + 3 is the upper curve and y = x 2 ’ 2 is the lower curve. Thus the
desired area is
2
Area = [’(x ’ 1)2 + 3] ’ [x 2 ’ 2] dx
’1
2
= ’2x 2 + 2x + 4 dx
’1
2
’2x 3
= + x + 4x
2
3 ’1
’16 2
= +4+8 ’ +1’4
3 3
= 9.
The area of the region determined by the two parabolas is 9.
EXAMPLE 4.12
Find the area between y = sin x and y = cos x for π/4 ¤ x ¤ 5π/4.
119
CHAPTER 4 The Integral

_2
y = x2




_ (x _ 1)2 + 3
y=


Fig. 4.20

SOLUTION
On the given interval, sin x ≥ cos x. See Fig. 4.21. Thus the area we wish to
compute is
5π/4
Area = [sin x ’ cos x] dx
π/4
x=5π/4
= [’ cos x ’ sin x]x=π/4
√ √ √ √
2 2 2 2
=’’ ’’ ’’ ’
2 2 2 2

= 2 2.

y




y = sin x

x
y = cos x




Fig. 4.21


You Try It: Calculate the area between y = sin x and y = cos x, ’π ¤ x ¤ 2π.

You Try It: Calculate the area between y = x 2 and y = 3x + 4.
120 CHAPTER 4 The Integral

4.5 Rules of Integration
We have been using various rules of integration without enunciating them explicitly.
It is well to have them recorded for future reference.


4.5.1 LINEAR PROPERTIES
I. If f, g are continuous functions on [a, b] then
b b b
f (x) + g(x) dx = f (x) dx + g(x) dx.
a a a
II. If f is a continuous function on [a, b] and c is a constant then
b b
cf (x) dx = c f (x) dx.
a a



4.5.2 ADDITIVITY
III. If f is a continuous on [a, c] and a < b < c then
b c c
f (x) dx + f (x) dx = f (x) dx.
a b a

You Try It: Calculate
3
4x 3 ’ 3x 2 + 7x ’ 12 cos x dx.
1




Exercises
1. Calculate each of the following antiderivatives:
Antiderivative of x 2 ’ sin x
(a)
Antiderivative of e3x + x 4 ’ 2
(b)
Antiderivative of 3t 2 + (ln t/t)
(c)
Antiderivative of tan x ’ cos x + sin 3x
(d)
Antiderivative of cos 3x + sin 4x + 1
(e)
Antiderivative of (cos x) · esin x
(f)
121
CHAPTER 4 The Integral

2. Calculate each of the following inde¬nite integrals:
x sin x 2 dx
(a)
(3/x) ln x 2 dx
(b)
sin x · cos x dx
(c)
tan x · ln cos x dx
(d)
sec2 x · etan x dx
(e)
(2x + 1) · (x 2 + x + 7)43 dx
(f)
3. Use Riemann sums to calculate each of the following integrals:
22
1 x + x dx
(a)
1 2
(b) ’1 (’x /3) dx
4. Use the Fundamental Theorem of Calculus to evaluate each of the following
integrals:
32
1 x ’ 4x + 7 dx
3
(a)
2
6
xex ’ sin x cos x dx
(b) 2
4
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 )
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

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