Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 47
3f(x)
lim [3f(x)]
(d) lim
x→2 g(x) = x→2
lim g(x) [Limit Law 5]
x→2
=
3lim
x→2
f(x)
lim
x→2 g(x) [Limit Law 3]
= 3(4)
−2 = −6
g(x)
(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim , does not exist because the
x→2 h(x)
denominator approaches 0 while the numerator approaches a nonzero number.
g(x) h(x)
lim [g(x) h(x)]
x→2
(f ) lim =
x→2 f(x) lim f(x) [Limit Law 5]
x→2
=
lim g(x) · lim h(x)
x→2 x→2
lim f(x) [Limit Law 4]
x→2
= −2 · 0
4
=0
3. lim
x→−2 (3x4 +2x 2 − x +1)= lim
x→−2 3x4 + lim
x→−2 2x2 − lim x + lim 1 [Limit Laws 1 and 2]
x→−2 x→−2
= 3 lim
x→−2 x4 +2 lim
x→−2 x2 − lim
x→−2 x + lim
x→−2 1 [3]
=3(−2) 4 +2(−2) 2 − (−2) + (1)
[9,8,and7]
=48+8+2+1=59
5. lim (1 + 3√ x )(2− 6x 2 + x 3 ) = lim (1 + 3√ x ) · lim(2 − 6x 2 + x 3 ) [Limit Law 4]
x→8 x→8 x→8
= lim 1 + lim 3√ x · lim 2 − 6lim
x→8 x→8
x→8 x→8 x2 +limx 3 [1,2,and3]
x→8
7. lim
x→1
3
1+3x
=
1+4x 2 +3x 4
=
lim
x→1
= 1+ 3√ 8 · 2
− 6 · 8 2 +8 3 [7, 10, 9]
= (3)(130) = 390
3
1+3x
[6]
1+4x 2 +3x 4 3
[5]
lim (1 + 3x)
x→1
lim (1 +
x→1 4x2 +3x 4 )
lim 1+3limx
3
x→1 x→1
=
lim 1+4lim
[2, 1, and 3]
x→1 x→1 x2 +3limx 4
x→1
3 3 3
1 + 3(1)
4 1
=
= = = 1 [7, 8, and 9]
1 + 4(1) 2 +3(1) 4 8 2 8
9. lim
x→4 − √
16 − x2 = lim
x→4 − (16 − x2 ) [11]
= lim 16 − lim x2 [2]
x→4− x→4 −
= 16 − (4) 2 =0 [7 and 9]