Solução_Calculo_Stewart_6e

F.
62 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
√ √
9x6 − x 9x6 − x/x 3
23. lim
= lim
=
x→∞ x 3 +1 x→∞ (x 3 +1)/x 3
=
lim 9 − 1/x
5
x→∞
lim
x→∞
= √ 9 − 0=3
lim
1+ lim
x→∞ (1/x3 ) =
lim (9x6 − x)/x 6
x→∞
lim (1 +
x→∞ 1/x3 )
x→∞ 9 − lim
x→∞ (1/x5 )
1+0
[since x 3 = √ x 6 for x>0]
√
25. lim 9x2 + x − 3x = lim
x→∞
x→∞
= lim
x→∞
= lim
x→∞
√
27. lim x2 + ax − √ x 2 + bx = lim
x→∞
x→∞
√
9x2 + x − 3x √ 9x 2 + x +3x
√
9x2 + x +3x
9x 2 + x − 9x 2
√
9x2 + x +3x = lim
x→∞
x/x
9x2 /x 2 + x/x 2 +3x/x = lim
x→∞
= lim
x→∞
= lim
x→∞
= lim
x→∞
x
√
9x2 + x +3x · 1/x
1/x
1
9+1/x +3
=
√
9x2 + x 2
− (3x)
2
√
9x2 + x +3x
√
x2 + ax − √ x 2 + bx √ x 2 + ax + √ x 2 + bx
√
x2 + ax + √ x 2 + bx
(x 2 + ax) − (x 2 + bx)
√
x2 + ax + √ x 2 + bx = lim
x→∞
a − b
1+a/x +
1+b/x
=
1
√
9+3
= 1
3+3 = 1 6
[(a − b)x]/x
√
x2 + ax + √ x 2 + bx / √ x 2
a − b
√ 1+0+
√ 1+0
= a − b
2
x + x 3 + x 5
29. lim
x→∞ 1 − x 2 + x = lim (x + x 3 + x 5 )/x 4
[divide by the highest power of x in the denominator]
4 x→∞ (1 − x 2 + x 4 )/x 4
= lim
x→∞
1/x 3 +1/x + x
1/x 4 − 1/x 2 +1 = ∞
because (1/x 3 +1/x + x) →∞and (1/x 4 − 1/x 2 +1)→ 1 as x →∞.
31. lim
x→−∞ (x4 + x 5 )= lim
x→−∞ x5 ( 1 +1) [factor out the largest power of x] = −∞ because x x5 →−∞and 1/x +1→ 1
as x →−∞.
Or: lim x 4 + x 5 = lim
x→−∞
x→−∞ x4 (1 + x) =−∞.
33. lim
x→∞
1 − e x
1+2e x = lim
x→∞
(1 − e x )/e x
(1 + 2e x )/e = lim 1/e x − 1
x x→∞ 1/e x +2 = 0 − 1
0+2 = −1 2
35. Since −1 ≤ cos x ≤ 1 and e −2x > 0, wehave−e −2x ≤ e −2x cos x ≤ e −2x .Weknowthat lim
lim
e
−2x
=0,sobytheSqueezeTheorem, lim
x→∞
x→∞ (e−2x cos x) =0.
x→∞ (−e−2x )=0and
37. (a)
x
f(x)
−10,000 −0.4999625
−100,000 −0.4999962
−1,000,000 −0.4999996
From the graph of f(x) = √ x 2 + x +1+x,weestimate
the value of
lim
x→−∞
f(x) to be −0.5.
(b)
Fromthetable,weestimatethelimit
to be −0.5.