Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 REVIEW ¤ 217
13. This limit has the form ∞−∞.
x
lim
x→1 + x − 1 − 1
x ln x − x +1
= lim
ln x x→1 + (x − 1) ln x
H
= lim
x→1 + 1/x
1/x 2 +1/x = 1
1+1 = 1 2
H
= lim
x→1 + x · (1/x)+lnx − 1
(x − 1) · (1/x)+lnx = lim
x→1 +
ln x
1 − 1/x +lnx
15. f(0) = 0, f 0 (−2) = f 0 (1) = f 0 (9) = 0, lim
x→∞
f(x) =0, lim
x→6
f(x) =−∞,
f 0 (x) < 0 on (−∞, −2), (1, 6),and(9, ∞), f 0 (x) > 0 on (−2, 1) and (6, 9),
f 00 (x) > 0 on (−∞, 0) and (12, ∞), f 00 (x) < 0 on (0, 6) and (6, 12)
17. f is odd, f 0 (x) < 0 for 0 2,
f 00 (x) > 0 for 0