Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 61
13. lim
x→∞
3x 2 − x +4
2x 2 +5x − 8 = lim
x→∞
(3x 2 − x +4)/x 2
(2x 2 +5x − 8)/x 2 [divide both the numerator and denominator by x 2
(the highest power of x thatappears in the denominator)]
=
lim (3 − 1/x
x→∞ +4/x2 )
lim (2 + 5/x −
x→∞ 8/x2 )
lim 3 − lim (1/x)+ lim
x→∞ x→∞ x→∞ (4/x2 )
=
lim 2 + lim (5/x) − lim
x→∞ x→∞ x→∞ (8/x2 )
=
3 − lim (1/x)+4 lim
x→∞ x→∞ (1/x2 )
2+5 lim(1/x) − 8 lim
x→∞ x→∞ (1/x2 )
= 3 − 0+4(0)
2+5(0)− 8(0)
[Limit Law 5]
[Limit Laws 1 and 2]
[Limit Laws 7 and 3]
[Theorem 5 of Section 2.5]
= 3 2
1
15. lim
x→∞ 2x +3 = lim 1/x
lim (1/x)
lim
x→∞ (2x +3)/x = x→∞
=
lim (2 + 3/x)
1 − x − x 2
17. lim
x→−∞ 2x 2 − 7
x→∞
x→∞ (1/x)
lim 2 + 3 lim (1/x) = 0
x→∞ x→∞
(1 − x − x 2 )/x 2 lim
x→−∞ (1/x2 − 1/x − 1)
= lim
=
x→−∞ (2x 2 − 7)/x 2 lim (2 −
x→−∞ 7/x2 )
lim
x→−∞ (1/x2 ) − lim (1/x) − lim
x→−∞
=
lim 2 − 7 lim
x→−∞ x→−∞ (1/x2 )
x→−∞ 1
= 0 − 0 − 1
2 − 7(0) = −1 2
2 + 3(0) = 0 2 =0
19. Divide both the numerator and denominator by x 3 (the highest power of x that occurs in the denominator).
lim
x→∞
x 3 +5x
2x 3 − x 2 +4 = lim
x→∞
=
x 3 +5x
x 3
2x 3 − x 2 +4
x 3
lim 1+5 lim
x→∞ x→∞
lim 2 − lim
x→∞ x→∞
= lim
x→∞
1
x 2
1
+4 lim
x x→∞
1+ 5 x 2
2 − 1 x + 4 x 3 =
lim
1+ 5
x
2 2 − 1 x + 4
x 3
x→∞
lim
x→∞
= 1 + 5(0)
1 2 − 0+4(0) = 1 2
x 3
21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u 4 .
lim
u→∞
4u 4 +5
(u 2 − 2)(2u 2 − 1) = lim
u→∞
=
lim
u→∞
4u 4 +5
2u 4 − 5u 2 +2 = lim
u→∞
lim
4+ 5
u
2 4 − 5 u + 2 =
2 u 4
u→∞
4u 4 +5
u 4
2u 4 − 5u 2 +2
u 4
= lim
u→∞
lim 4+5 lim
u→∞ u→∞
lim 2 − 5 lim
u→∞ u→∞
1
u 4
1
+2 lim
u2 u→∞
4+ 5 u 4
2 − 5 u 2 + 2 u 4
=
1
u 4
4+5(0)
2 − 5(0) + 2(0) = 4 2 =2