Solução_Calculo_Stewart_6e

F.
168 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
13. This limit has the form 0 . lim tan px H p sec 2 px
=lim
0 x→0 tan qx x→0 q sec 2 qx = p(1)2
q(1) = p 2 q
15. This limit has the form ∞ . lim ln x
√
∞
= H lim
x→∞ x
x→∞
1
2
1/x
= lim
x−1/2 x→∞
2
√
x
=0
17. lim
x→0 + [(ln x)/x] =−∞ since ln x →−∞as x → 0+ anddividingbysmallvaluesofx just increases the magnitude of the
quotient (ln x)/x. L’Hospital’s Rule does not apply.
19. This limit has the form ∞ . lim e x
∞ x→∞ x 3
21. This limit has the form 0 . lim e x − 1 − x
0 x→0 x 2
= H e x
lim
x→∞ 3x 2
= H e x
lim
x→∞ 6x
H e x
= lim
x→∞ 6 = ∞
=lim
H e x − 1 H e x
=lim
x→0 2x x→0 2 = 1 2
23. This limit has the form 0 . lim tanh x H sech 2 x
=lim
0 x→0 tan x x→0 sec 2 x = sech2 0
sec 2 0 = 1 1 =1
25. This limit has the form 0 . lim 5 t − 3 t
0 t→0 t
H 5 t ln 5 − 3 t ln 3
=lim
=ln5− ln 3 = ln 5
t→0 3
1
27. This limit has the form 0 . lim sin −1 x H 1/ √ 1 − x
= lim
2
1
=lim√ 0 x→0 x x→0 1
= 1 x→0 1 − x
2 1 =1
29. This limit has the form 0 . lim 1 − cos x
0
x→0 x 2
=lim
H sin x H cos x
= lim = 1
x→0 2x x→0 2 2
x +sinx
31. lim
x→0 x +cosx = 0+0
0+1 = 0 =0. L’Hospital’s Rule does not apply.
1
33. This limit has the form 0 . lim 1 − x +lnx
0 x→1 1+cosπx
H
=lim
x→1
−1+1/x
−π sin πx
H −1/x 2
=lim
x→1 −π 2 cos πx =
−1
−π 2 (−1) = − 1 π 2
35. This limit has the form 0 . lim x a − ax + a − 1 H ax a−1 − a H a(a − 1)x a−2 a(a − 1)
=lim
= lim
=
0 x→1 (x − 1) 2 x→1 2(x − 1) x→1 2
2
37. This limit has the form 0 . lim cos x − 1+ 1 2 x2
0 x→0 x 4
39. This limit has the form ∞ · 0.
H =lim
x→0
− sin x + x
4x 3
H = lim
x→0
− cos x +1
12x 2
sin(π/x) H cos(π/x)(−π/x 2 )
lim x sin(π/x) = lim = lim
= π lim cos(π/x) =π(1) = π
x→∞ x→∞ 1/x x→∞ −1/x 2
x→∞
41. This limit has the form ∞ · 0. We’ll change it to the form 0 0 .
sin 6x H 6cos6x
lim cot 2x sin 6x =lim =lim
x→0 x→0 tan 2x x→0 2sec 2 2x = 6(1)
2(1) =3 2
=lim
H sin x H cos x
= lim
x→0 24x x→0 24 = 1
24
x 3
43. This limit has the form ∞ · 0. lim
x→∞ x3 e −x2 = lim
x→∞ e x2
= H 3x 2
lim
x→∞ 2xe x2
= lim
x→∞
3x
2e x2
= H 3
lim =0
x→∞ x2
4xe