Solução_Calculo_Stewart_6e

F.
80 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
1. (a) (i) lim f(x) =3 (ii) lim f(x) =0
x→2 + x→−3 +
(iii)
lim f(x) does not exist since the left and right limits are not equal. (The left limit is −2.)
x→−3
(iv) lim
x→4
f(x) =2
(v) lim f(x) =∞ (vi) lim f(x) =−∞
x→0 −
(vii) lim f(x) =4 (viii) lim f(x) =−1
x→∞ x→−∞
(b) The equations of the horizontal asymptotes are y = −1 and y =4.
(c) The equations of the vertical asymptotes are x =0and x =2.
(d) f is discontinuous at x = −3, 0, 2,and4. The discontinuities are jump, infinite, infinite, and removable, respectively.
x→2
3. Since the exponential function is continuous, lim
x→1
e x3 −x = e 1−1 = e 0 =1.
x 2 − 9
5. lim
x→−3 x 2 +2x − 3 = lim (x +3)(x − 3)
x→−3 (x +3)(x − 1) = lim x − 3
x→−3 x − 1 = −3 − 3
−3 − 1 = −6
−4 = 3 2
(h − 1) 3 +1 h 3 − 3h 2 +3h − 1 +1 h 3 − 3h 2 +3h
7. lim
= lim
= lim
= lim h 2 − 3h +3 =3
h→0 h
h→0 h
h→0 h
h→0
Another solution: Factor the numerator as a sum of two cubes and then simplify.
(h − 1) 3 +1 (h − 1) 3 +1 3 [(h − 1) + 1] (h − 1) 2 − 1(h − 1) + 1 2
lim
= lim
=lim
h→0 h
h→0 h
h→0 h
9. lim
r→9
= lim
h→0
(h − 1) 2 − h +2 =1− 0+2=3
√ r
(r − 9) 4 = ∞ since (r − 9)4 → 0 as r → 9 and
√ r
> 0 for r 6= 9.
(r − 9)
4
u 4 − 1
11. lim
u→1 u 3 +5u 2 − 6u = lim (u 2 +1)(u 2 − 1)
u→1 u(u 2 +5u − 6) =lim (u 2 +1)(u +1)(u − 1) (u 2 +1)(u +1)
= lim
= 2(2)
u→1 u(u +6)(u − 1) u→1 u(u +6) 1(7) = 4 7
13. Since x is positive, √ x 2 = |x| = x. Thus,
√ √
x2 − 9
lim
x→∞ 2x − 6
= lim
x2 − 9/ √ √
x 2 1 − 9/x
x→∞ (2x − 6)/x = lim
2 1 − 0
x→∞ 2 − 6/x = 2 − 0 = 1 2
15. Let t =sinx. Thenasx → π − , sin x → 0 + ,sot → 0 + . Thus, lim ln(sin x) = lim ln t = −∞.
x→π− t→0 +
√
17. lim x2 +4x +1− x √ √
x2 +4x +1− x x2 +4x +1+x (x 2 +4x +1)− x 2
= lim
· √ = lim √
x→∞
x→∞ 1
x2 +4x +1+x x→∞ x2 +4x +1+x
(4x +1)/x
= lim
x→∞ ( √ x 2 +4x +1+x)/x
divide by x = √
x 2 for x>0
4+1/x
= lim
x→∞ 1+4/x +1/x2 +1 = 4+0
√ = 4 1+0+0+1 2 =2