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

Section 4.4 Indeterminate Forms and l’Hospital’s Rule 307
The graph of the function of Example
3 is shown in Figure 3. We have discussed
previously the slow growth of
logarithms, so it isn’t surprising that
this ratio approaches 0 as x l `. See
also Exercise 74.
0
2
_1
FIGURE 3
y= ln x
œ„x
10,000
ln x
ExamplE 3 Calculate lim .
x l ` sx
SOLUtion Since ln x l ` and sx l ` as x l `, l’Hospital’s Rule applies:
ln x
lim − lim
x l ` sx x l `
1
2
1yx
− lim
x21y2 xl`
1yx
1y(2sx )
Notice that the limit on the right side is now indeterminate of type 0 0 . But instead of
applying l’Hospital’s Rule a second time as we did in Example 2, we simplify the
expression and see that a second application is unnecessary:
ln x
lim − lim
x l ` sx x l `
1yx
1y(2sx ) − lim
x l `
2
sx − 0
In both Examples 2 and 3 we evaluated limits of type `y`, but we got two different
results. In Example 2, the infinite limit tells us that the numerator e x increases significantly
faster than the denominator x 2 , resulting in larger and larger ratios. In fact, y − e x
grows more quickly than all the power functions y − x n (see Exercise 73). In Example
3 we have the opposite situation; the limit of 0 means that the denominator outpaces the
numerator, and the ratio eventually approaches 0.
n
ExamplE 4 Find lim
x l 0
tan x 2 x
x 3 . (See Exercise 2.2.50.)
SOLUtion Noting that both tan x 2 x l 0 and x 3 l 0 as x l 0, we use l’Hospital’s
Rule:
tan x 2 x sec 2 x 2 1
lim − lim
x l 0 x 3 x l 0 3x 2
Since the limit on the right side is still indeterminate of type 0 0 , we apply l’Hospital’s
Rule again:
The graph in Figure 4 gives visual
confirmation of the result of Example 4.
If we were to zoom in too far, however,
we would get an inaccurate graph
because tan x is close to x when x is
small. See Exercise 2.2.50(d).
1
sec 2 x 2 1 2 sec 2 x tan x
lim − lim
x l 0 3x 2 x l 0 6x
Because lim x l 0 sec 2 x − 1, we simplify the calculation by writing
2 sec 2 x tan x
lim
− 1
x l 0 6x 3 lim
tan x
x l 0 sec2 x ? lim − 1
x l 0 x 3 lim tan x
xl 0 x
We can evaluate this last limit either by using l’Hospital’s Rule a third time or by
writing tan x as ssin xdyscos xd and making use of our knowledge of trigonometric
limits. Putting together all the steps, we get
tan x-x
y=
˛
_1 1
0
FIGURE 4
lim
x l 0
tan x 2 x
x 3
− lim
x l 0
− 1 3 lim
x l 0
sec 2 x 2 1
3x 2
tan x
x
− lim
x l 0
− 1 3 lim
x l 0
2 sec 2 x tan x
6x
sec 2 x
1
− 1 3
n
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