James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

310 Chapter 4 Applications of Differentiation
Indeterminate Powers
Several indeterminate forms arise from the limit
tsxd
lim − f f sxdg
x la
Although forms of the type 0 0 , `0,
and 1` are indeterminate, the form
0` is not indeterminate. (See
Exercise 86.)
1. lim f sxd − 0
x l a
and lim tsxd − 0
x l a
type 0 0
2. lim f sxd − `
x l a
and lim tsxd − 0
x l a
type ` 0
3. lim f sxd − 1
x l a
and lim tsxd − 6` type 1`
x l a
Each of these three cases can be treated either by taking the natural logarithm:
let y − f f sxdg tsxd , then ln y − tsxd ln f sxd
or by writing the function as an exponential:
f f sxdg tsxd tsxd ln f sxd
− e
(Recall that both of these methods were used in differentiating such functions.) In either
method we are led to the indeterminate product tsxd ln f sxd, which is of type 0 ? `.
ExamplE 9 Calculate lim
x l 0 1 s1 1 sin 4xd cot x .
SOLUTION First notice that as x l 0 1 , we have 1 1 sin 4x l 1 and cot x l `, so the
given limit is indeterminate (type 1`). Let
Then
y − s1 1 sin 4xd cot x
ln y − lnfs1 1 sin 4xd cot x g − cot x lns1 1 sin 4xd −
so l’Hospital’s Rule gives
lns1 1 sin 4xd
lim ln y − lim
− lim
x l 0 1 xl 0 1
tan x
x l 0 1
4 cos 4x
1 1 sin 4x
sec 2 x
lns1 1 sin 4xd
tan x
− 4
The graph of the function y − x x ,
x . 0, is shown in Figure 6. Notice
that although 0 0 is not defined, the
values of the function approach 1 as
x l 0 1 . This confirms the result of
Example 10.
2
So far we have computed the limit of ln y, but what we want is the limit of y. To find
this we use the fact that y − e ln y :
ExamplE 10 Find lim
x l 0 1 x x .
lim s1 1 sin
x l 0 1 4xdcot x − lim y − lim e ln y − e 4
x l 0 1 x l 0 1
SOLUTION Notice that this limit is indeterminate since 0 x − 0 for any x . 0 but
x 0 − 1 for any x ± 0. (Recall that 0 0 is undefined.) We could proceed as in Example 9
or by writing the function as an exponential:
n
x x − se ln x d x − e x ln x
_1
Figure 6
0
2
In Example 6 we used l’Hospital’s Rule to show that
lim x ln x − 0
x l 0 1
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