Solução_Calculo_Stewart_6e

F.
170 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
61. y =(4x +1) cot x ln(4x +1) H
⇒ ln y =cotx ln(4x +1),so lim ln y = lim
= lim
x→0 + x→0 + tan x
x→0 + 4
4x +1
sec 2 x =4
⇒
lim
x→0 +(4x +1)cot x = lim
x→0 + eln y = e 4 .
63. y =(cosx) 1/x2 ⇒ ln y = 1 ln cos x
ln cos x ⇒ lim ln y = lim
x2 x→0 + x→0 + x 2
= H − tan x H − sec 2 x
lim = lim = − 1
x→0 + 2x x→0 + 2 2
⇒
lim
x→0 +(cos x)1/x2 = lim
x→0 + eln y = e −1/2 =1/ √ e
65. From the graph, if x = 500, y ≈ 7.36. The limit has the form 1 ∞ .
Now y = 1+ 2 x
⇒ ln y = x ln 1+ 2
⇒
x x
1
− 2
ln(1 + 2/x) H 1+2/x x
lim ln y = lim
= lim
2
x→∞ x→∞ 1/x x→∞ −1/x 2
=2 lim
x→∞
1
1+2/x =2(1)=2
lim 1+ 2 x
= lim
x→∞ x x→∞ eln y = e 2 [≈ 7.39]
f(x)
67. From the graph, it appears that lim
x→0 g(x) =lim f 0 (x)
x→0 g 0 (x) =0.25.
We calculate lim
x→0
f(x)
g(x) = lim
x→0
e x − 1
x 3 +4x
⇒
H
= lim
x→0
e x
3x 2 +4 = 1 4 .
e x
69. lim
x→∞ x n
71. lim
x→∞
= H e x
lim
x→∞ nx n−1
x
√
x2 +1
H
= lim
x→∞
H = lim
x→∞
e x
n(n − 1)x n−2
1
1
2 (x2 +1) −1/2 (2x) = lim
x→∞
= H ··· H= e x
lim
x→∞ n! = ∞
√
x2 +1
. Repeated applications of l’Hospital’s Rule result in the
x
original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator
by x:
lim
x→∞
x
√
x2 +1 = lim
x→∞
x/x
x2 /x 2 +1/x 2 = lim
x→∞
1
1+1/x
2 = 1 1 =1
73. First we will find lim 1+ r nt,
which is of the form 1
n→∞ n ∞ . y = 1+ r nt
⇒ ln y = nt ln 1+ r
,so
n n
−r/n
2
lim ln y = lim
1+ nt ln r
ln(1 + r/n)
= t lim
n→∞ n→∞ n n→∞ 1/n
H
= t lim
n→∞
(1 + r/n)(−1/n 2 ) = t lim
n→∞
r
1+i/n = tr
⇒
lim y =
n→∞ ert .Thus,asn →∞, A = A 0
1+ r nt
→ A0 e rt .
n