5128_Ch08_434-471

Section 8.2 L’Hôpital’s Rule 447
[/4, 3/4] by [–2, 4]
Figure 8.8 The graph of
y sec x1 tan x. (Example 4)
SOLUTION
The numerator and denominator are discontinuous at x p/2, so we investigate the
one-sided limits there. To apply l’Hôpital’s Rule we can choose I to be any open interval
containing x p/2.
lim sec x
from the left
x→p2 1 tan x
Next differentiate the numerator and denominator.
sec x
x n x
lim
x→p2 lim
1 tan x x→p2 sec
sec tax
2 lim sin x 1
x→p2
The right-hand limit is 1 also, with ()() as the indeterminate form. Therefore,
the two-sided limit is equal to 1. The graph of (sec x)(1 tan x) in Figure 8.8 appears
to pass right through the point (p2, 1) and supports the work above.
Now try Exercise 13.
EXAMPLE 5 Working with Indeterminate Form
Identify the indeterminate form and evaluate the limit using l’Hôpital’s Rule. Support
your answer graphically.
lim ln x
x→ 2x
SOLUTION
lim ln x 1x
lim lim 0
x→ 2x
x→ 1x x→
1x
[0, 1000] by [–2, 2]
Figure 8.9 A graph of
y (ln x)(2x). (Example 5)
The graph in Figure 8.9 supports the result. Now try Exercise 15.
We can sometimes handle the indeterminate forms • 0 and by using algebra to
get 00 or instead. Here again we do not mean to suggest that there is a number • 0
or any more than we mean to suggest that there is a number 00 or . These
forms are not numbers but descriptions of function behavior.
EXAMPLE 6 Working With Indeterminate Form • 0
Find (a) lim
x→ ( x sin 1 x ) (b) lim
x→ ( x sin 1 x ) .
SOLUTION
Figure 8.10 suggests that the limits exist.
[–5, 5] by [–1, 2]
Figure 8.10 The graph of
y x sin 1x. (Example 6)
(a)
(b) Similarly,
lim
x→ (
lim
h→0 (
1
x sin 1 x ) • 0
1 h )
sin h Let h 1x.
lim
x→ (
x sin 1 x ) 1. Now try Exercise 17.