Solução_Calculo_Stewart_6e

F.
46 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
39. No matter how many times we zoom in toward the origin, the graphs of f(x) =sin(π/x) appear to consist of almost-vertical
lines. This indicates more and more frequent oscillations as x → 0.
41. There appear to be vertical asymptotes of the curve y =tan(2sinx) at x ≈ ±0.90
and x ≈ ±2.24. Tofind the exact equations of these asymptotes, we note that the
graph of the tangent function has vertical asymptotes at x = π + πn. Thus,we
2
must have 2sinx = π + πn, or equivalently, sin x = π + π n.Since
2 4 2
−1 ≤ sin x ≤ 1,wemusthavesin x = ± π and so x = ± 4 sin−1 π (corresponding
4
to x ≈ ±0.90). Just as 150 ◦ is the reference angle for 30 ◦ , π − sin −1 π 4
is the
reference angle for sin −1 π .Sox = ±
4
π − sin −1 π 4 are also equations of
vertical asymptotes (corresponding to x ≈ ±2.24).
2.3 Calculating Limits Using the Limit Laws
1. (a) lim
x→2
[f(x)+5g(x)] = lim
x→2
f(x) + lim
x→2
[5g(x)] [Limit Law 1]
=lim
x→2
f(x)+5lim
x→2
g(x) [Limit Law 3]
=4+5(−2) = −6
3
(b) lim [g(x)] 3 = lim g(x) [Limit Law 6]
x→2 x→2
=(−2) 3 = −8
(c) lim
x→2
f(x)=
lim
x→2
f(x) [Limit Law 11]
= √ 4=2