Solução_Calculo_Stewart_6e

F.
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 169
45. This limit has the form 0 · (−∞).
lim ln x tan(πx/2) = lim
x→1 + x→1 +
47. This limit has the form ∞−∞.
x
lim
x→1 x − 1 − 1
ln x
ln x
cot(πx/2)
=lim
x→1
x ln x − (x − 1)
(x − 1) ln x
H
= lim
x→1 + 1/x
(−π/2) csc 2 (πx/2) = 1
(−π/2)(1) 2 = − 2 π
H
= lim
x→1
x(1/x)+lnx − 1
(x − 1)(1/x)+lnx = lim
x→1
H 1/x
=lim
x→1 1/x 2 +1/x · x2
x = lim x
2 x→1 1+x = 1
1+1 = 1 2
ln x
1 − (1/x)+lnx
49. We will multiply and divide by the conjugate of the expression to change the form of the expression.
√
lim x2 + x − x √ √
x2 + x − x x2 + x + x
x 2 + x − x 2
= lim
· √ = lim √
x→∞
x→∞ 1 x2 + x + x x→∞ x2 + x + x
x
= lim √
x→∞ x2 + x + x = lim 1
1
= √ = 1
x→∞ 1+1/x +1 1+1 2
As an alternate solution, write √ x 2 + x − x as √ x 2 + x − √ x 2 ,factorout √ x 2 ,rewriteas( 1+1/x − 1)/(1/x),and
apply l’Hospital’s Rule.
51. The limit has the form ∞−∞and we will change the form to a product by factoring out x.
lim (x − ln x) = lim x 1 − ln x
ln x H 1/x
= ∞ since lim = lim
x→∞ x→∞ x
x→∞ x x→∞ 1 =0.
53. y = x x2 ⇒ ln y = x 2 ln x
ln x,so lim ln y = lim x2 ln x = lim
H 1/x
= lim
x→0 + x→0 + x→0 + 1/x 2 x→0 + −2/x = lim − 1 3 x→0 + 2 x2 =0 ⇒
lim
x→0 + xx2 = lim eln y = e 0 =1.
x→0 +
55. y =(1− 2x) 1/x ⇒ ln y = 1 x
lim
x→0 (1 − 2x)1/x =lim
x→0
e ln y = e −2 .
57. y =
1+ 3 x + 5 x
⇒ ln y = x ln
1+ 3 x 2 x + 5
x 2
ln
1+ 3 x + 5
x
lim ln y = lim
2
x→∞ x→∞ 1/x
ln(1 − 2x) H −2/(1 − 2x)
ln(1 − 2x), solim ln y =lim
= lim
= −2 ⇒
x→0 x→0 x
x→0 1
H
= lim
x→∞
so lim 1+ 3
x→∞ x + 5 x
= lim
x 2 x→∞ eln y = e 3 .
⇒
− 3 x 2 − 10
x 3
1+ 3 x + 5 x 2
−1/x 2
59. y = x 1/x ln x
⇒ ln y =(1/x) lnx ⇒ lim ln y = lim
x→∞ x→∞ x
lim
x→∞ x1/x = lim
x→∞ eln y = e 0 =1
= lim
x→∞
H 1/x
= lim
x→∞ 1 =0 ⇒
3+ 10 x
1+ 3 x + 5 x 2 =3,