Solução_Calculo_Stewart_6e

F.
60 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
lim
x→0 (−x4 )=0and lim x 4 =0, the Squeeze Theorem gives us lim(x 4 sin(1/x)) = 0,whichequalsf(0). Thus,f is
x→0 x→0
continuous at 0 and, hence, on (−∞, ∞).
65. Define u(t) to be the monk’s distance from the monastery, as a function of time, on the first day, and define d(t) to be his
distance from the monastery, as a function of time, on the second day. Let D be the distance from the monastery to the top of
the mountain. From the given information we know that u(0) = 0, u(12) = D, d(0) = D and d(12) = 0. Now consider the
function u − d, which is clearly continuous. We calculate that (u − d)(0) = −D and (u − d)(12) = D.Sobythe
Intermediate Value Theorem, there must be some time t 0 between 0 and 12 such that (u − d)(t 0)=0 ⇔ u(t 0)=d(t 0).
So at time t 0 after 7:00 AM, the monk will be at the same place on both days.
2.6 Limits at Infinity; Horizontal Asymptotes
1. (a) As x becomes large, the values of f(x) approach 5.
(b) As x becomes large negative, the values of f(x) approach 3.
3. (a) lim f(x) =∞ (b) lim f(x) =∞ (c) lim f(x) =−∞
x→2 − +
x→−1
(d) lim f(x) =1 (e) lim f(x) =2 (f ) Vertical: x = −1, x =2; Horizontal: y =1, y =2
x→∞ x→−∞
x→−1
5. f(0) = 0, f(1) = 1,
lim f(x) =0,
x→∞
f is odd
7. lim f(x) =−∞, lim f(x) =∞, 9. f(0) = 3, lim f(x) =4,
x→2 x→∞ x→0− lim f(x) =0, lim f(x) =∞, lim f(x) =2,
x→−∞ x→0 +
x→0 +
lim f(x) =−∞ lim f(x) =−∞, lim f(x) =−∞,
x→0 − x→−∞ x→4− lim f(x) =∞, lim
x→4 +
f(x) =3
x→∞
11. If f(x) =x 2 /2 x , then a calculator gives f(0) = 0, f(1) = 0.5, f(2) = 1, f(3) = 1.125, f(4) = 1, f(5) = 0.78125,
f(6) = 0.5625, f(7) = 0.3828125, f(8) = 0.25, f(9) = 0.158203125, f(10) = 0.09765625, f(20) ≈ 0.00038147,
f(50) ≈ 2.2204 × 10 −12 , f(100) ≈ 7.8886 × 10 −27 .
It appears that lim x 2 /2 x =0.
x→∞