Calculus 2nd Edition Rogawski

384 CHAPTER 7 EXPONENTIAL FUNCTIONS
−2 cos x sin x
Note that the quotient is still indeterminate at x = π/2. We removed this
− cos x
indeterminacy by cancelling the factor − cos x.
EXAMPLE 4 The Form 0 · ∞
Evaluate lim x ln x.
x→0+
Solution This limit is one-sided because f (x) = x ln x is not defined for x ≤ 0. Furthermore,
as x → 0+,
• x approaches 0
• ln x approaches −∞
So f (x) presents an indeterminate form of type 0 · ∞. To apply L’Hôpital’s Rule we
rewrite our function as f (x) = (ln x)/x −1 so that f (x) presents an indeterminate form
of type −∞/∞. Then L’Hôpital’s Rule applies:
ln x (ln x) ′
lim x ln x = lim = lim
x→0+ x→0+ x−1 x→0+
(x −1 ) ′ = lim
x→0+
( x
−1
−x −2 )
} {{ }
L’Hôpital’s Rule
= lim (−x) = 0
x→0+
} {{ }
Simplify
EXAMPLE 5 Using L’Hôpital’s Rule Twice Evaluate lim
x→0
e x − x − 1
cos x − 1 .
Solution For x = 0, we have
e x − x − 1 = e 0 − 0 − 1 = 0, cos x − 1 = cos 0 − 1 = 0
A first application of L’Hôpital’s Rule gives
e x − x − 1
lim
x→0 cos x − 1 = lim (e x − x − 1) ′ ( e x )
− 1 1 − e x
x→0 (cos x − 1) ′ = lim
= lim
x→0 − sin x x→0 sin x
This limit is again indeterminate of type 0/0, so we apply L’Hôpital’s Rule again:
1 − e x
lim
x→0 sin x
= lim −e x
x→0 cos x = −e0
cos 0 = −1
EXAMPLE 6 Assumptions Matter Can L’Hôpital’s Rule be applied to lim
x→1
x 2 + 1
2x + 1 ?
Solution The answer is no. The function does not have an indeterminate form because
x 2 + 1
2x + 1∣ = 12 + 1
x=1 2 · 1 + 1 = 2 3
x 2 + 1
However, the limit can be evaluated directly by substitution: lim
x→1 2x + 1 = 2 . An incorrect
application of L’Hôpital’s Rule gives the wrong
3
answer:
(x 2 + 1) ′ 2x
lim
x→1 (2x + 1) ′ = lim
x→1 2
= 1 (not equal to original limit)