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

308 Chapter 4 Applications of Differentiation
ExamplE 5 Find lim
x l 2
sin x
1 2 cos x .
SOLUTION If we blindly attempted to use l’Hospital’s Rule, we would get
lim
x l 2
sin x
1 2 cos x − lim
x l 2
cos x
sin x − 2`
This is wrong! Although the numerator sin x l 0 as x l 2 , notice that the denominator
s1 2 cos xd does not approach 0, so l’Hospital’s Rule can’t be applied here.
The required limit is, in fact, easy to find because the function is continuous at
and the denominator is nonzero there:
lim
x l 2
sin x
1 2 cos x − sin
1 2 cos − 0
1 2 s21d − 0 n
Example 5 shows what can go wrong if you use l’Hospital’s Rule without thinking.
Other limits can be found using l’Hospital’s Rule but are more easily found by other
methods. (See Examples 2.3.3, 2.3.5, and 2.6.3, and the discussion at the beginning of
this section.) So when evaluating any limit, you should consider other meth ods before
using l’Hospital’s Rule.
Indeterminate Products
If lim x l a f sxd − 0 and lim x l a tsxd − ` (or 2`), then it isn’t clear what the value of
lim x l a f f sxd tsxdg, if any, will be. There is a struggle between f and t. If f wins, the
answer will be 0; if t wins, the answer will be ` (or 2`). Or there may be a compromise
where the answer is a finite nonzero number. This kind of limit is called an indeterminate
form of type 0 ? `. We can deal with it by writing the product ft as a quotient:
ft −
f
1yt
or ft − t
1yf
Figure 5 shows the graph of the
function in Example 6. Notice that
the function is undefined at x − 0;
the graph approaches the origin but
never quite reaches it.
y
y=x ln x
This converts the given limit into an indeterminate form of type 0 0 or `y` so that we can
use l’Hospital’s Rule.
ExamplE 6 Evaluate lim x ln x.
x l01 SOLUTION The given limit is indeterminate because, as x l 0 1 , the first factor sxd
approaches 0 while the second factor sln xd approaches 2`. Writing x − 1ys1yxd, we
have 1yx l ` as x l 0 1 , so l’Hospital’s Rule gives
lim x ln x
ln x − lim
x l 0 1 x l 0 1 1yx − lim
x l 0 1
1yx
21yx 2
− lim
x l 0 1 s2xd − 0
n
0
1
FIGURE 5
x
Note In solving Example 6 another possible option would have been to write
x
lim x ln x − lim
x l 01 x l 0 1 1yln x
This gives an indeterminate form of the type 0 0 , but if we apply l’Hospital’s Rule we get
a more complicated expression than the one we started with. In general, when we rewrite
an indeterminate product, we try to choose the option that leads to the simpler limit.
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