Solução_Calculo_Stewart_6e

F.
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 167
(b) f(0) = 0 and since sin 1 x and hence x4 sin 1 is both positive and negative inifinitely often on both sides of 0,and
x
arbitrarily close to 0, f has neither a local maximum nor a local minimum at 0.
Since 2+sin 1
x ≥ 1, g(x) =x4 2+sin 1
> 0 for x 6= 0,sog(0) = 0 is a local minimum.
x
Since −2+sin 1
x ≤−1, h(x) =x4 −2+sin 1
< 0 for x 6= 0,soh(0) = 0 is a local maximum.
x
4.4 Indeterminate Forms and L'Hospital's Rule
Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign:
1. (a) lim
x→a
f(x)
g(x) is an indeterminate form of type 0 0 .
f(x)
(b) lim =0because the numerator approaches 0 while the denominator becomes large.
x→a p(x)
h(x)
(c) lim =0because the numerator approaches a finite number while the denominator becomes large.
x→a p(x)
H
=
p(x)
(d) If lim p(x) =∞ and f(x) → 0 through positive values, then lim
x→a x→a f(x) = ∞. [Forexample,takea =0, p(x) =1/x2 ,
and f(x) =x 2 p(x)
.] If f(x) → 0 through negative values, then lim
x→a f(x) = −∞. [Forexample,takea =0, p(x) =1/x2 ,
and f(x) =−x 2 .] If f(x) → 0 through both positive and negative values, then the limit might not exist. [For example,
take a =0, p(x) =1/x 2 ,andf(x) =x.]
p(x)
(e) lim
x→a q(x) is an indeterminate form of type ∞ ∞ .
3. (a) When x is near a, f(x) is near 0 and p(x) is large, so f(x) − p(x) is large negative. Thus, lim
x→a
[f(x) − p(x)] = −∞.
(b) lim
x→a
[ p(x) − q(x)] is an indeterminate form of type ∞−∞.
(c) When x is near a, p(x) and q(x) are both large, so p(x)+q(x) is large. Thus, lim
x→a
[ p(x)+q(x)] = ∞.
5. This limit has the form 0 0 .Wecansimplyfactorandsimplifytoevaluatethelimit.
x 2 − 1
lim
x→1 x 2 − x =lim (x +1)(x − 1) x +1
= lim = 1+1 =2
x→1 x(x − 1) x→1 x 1
7. This limit has the form 0 . lim x 9 − 1 H 9x 8
0
=lim
x→1 x 5 − 1 x→1 5x = 9 4 5 lim
x→1 x4 = 9 5 (1) = 9 5
9. This limit has the form 0 0 . lim
x→(π/2) +
11. This limit has the form 0 . lim e t − 1
0 t→0 t 3
cos x
1 − sin x
H
=lim
t→0
H − sin x
= lim
x→(π/2) + − cos x =
lim
x→(π/2)
+
tan x = −∞.
e t
3t 2 = ∞ since et → 1 and 3t 2 → 0 + as t → 0.