Solução_Calculo_Stewart_6e

F.
(c) lim
x→0
TX.10 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 49
√
x 1+3x +1 x √ 1+3x +1
x √ 1+3x +1
√ · √ =lim
=lim
1+3x − 1 1+3x +1 x→0 (1 + 3x) − 1 x→0 3x
= 1 3 lim √
1+3x +1
x→0
= 1 3
lim
= 1 3
lim
x→0
(1 + 3x)+lim
x→0
1
x→0
1+3lim
x→0
x +1
= 1 √
1+3· 0+1
3
[Limit Law 3]
[1 and 11]
[1,3,and7]
[7 and 8]
= 1 3 (1 + 1) = 2 3
33. Let f(x) =−x 2 , g(x) =x 2 cos 20πx and h(x) =x 2 .Then
−1 ≤ cos 20πx ≤ 1 ⇒ −x 2 ≤ x 2 cos 20πx ≤ x 2 ⇒ f(x) ≤ g(x) ≤ h(x).
So since lim
x→0
f(x) = lim
x→0
h(x) =0, by the Squeeze Theorem we have
lim g(x) =0.
x→0
35. We have lim
x→4
(4x − 9) = 4(4) − 9=7and lim
x→4
x 2 − 4x +7 =4 2 − 4(4) + 7 = 7. Since4x − 9 ≤ f(x) ≤ x 2 − 4x +7
for x ≥ 0, lim
x→4
f(x) =7by the Squeeze Theorem.
37. −1 ≤ cos(2/x) ≤ 1 ⇒ −x 4 ≤ x 4 cos(2/x) ≤ x 4 .Sincelim
−x
4
=0and lim x 4 =0,wehave
x→0 x→0
lim x 4 cos(2/x) =0by the Squeeze Theorem.
x→0
39. |x − 3| =
41.
x − 3 if x − 3 ≥ 0
−(x − 3) if x − 3 < 0 =
x − 3 if x ≥ 3
3 − x if x