Solução_Calculo_Stewart_6e

F.
64 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
45. From the graph, it appears y =1is a horizontal asymptote.
3x 3 +500x 2
3x 3 +500x 2
lim
x→∞ x 3 + 500x 2 +100x +2000 = lim
x 3
3+(500/x)
= lim
x→∞ x 3 +500x 2 + 100x + 2000 x→∞ 1+(500/x) + (100/x 2 )+(2000/x 3 )
x 3
3+0
=
=3, soy =3is a horizontal asymptote.
1+0+0+0
The discrepancy can be explained by the choice of the viewing window. Try
[−100,000, 100,000] by [−1, 4] to get a graph that lends credibility to our
calculation that y =3is a horizontal asymptote.
47. Let’s look for a rational function.
(1) lim f(x) =0 ⇒ degree of numerator < degree of denominator
x→±∞
(2) lim f(x) =−∞ ⇒ there is a factor of x 2 in the denominator (not just x, since that would produce a sign
x→0
change at x =0), and the function is negative near x =0.
(3) lim f(x) =∞ and lim f(x) =−∞ ⇒ vertical asymptote at x =3;thereisafactorof(x − 3) in the
x→3− x→3 +
denominator.
(4) f(2) = 0 ⇒ 2 is an x-intercept; there is at least one factor of (x − 2) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us
f(x) =
2 − x as one possibility.
x 2 (x − 3)
49. y = f(x) =x 4 − x 6 = x 4 (1 − x 2 )=x 4 (1 + x)(1 − x). They-intercept is
f(0) = 0. Thex-intercepts are 0, −1,and1 [found by solving f(x) =0for x].
Since x 4 > 0 for x 6= 0, f doesn’t change sign at x =0. The function does change
sign at x = −1 and x =1.Asx → ±∞, f(x) =x 4 (1 − x 2 ) approaches −∞
because x 4 →∞and (1 − x 2 ) →−∞.
51. y = f(x) =(3− x)(1 + x) 2 (1 − x) 4 .They-intercept is f(0) = 3(1) 2 (1) 4 =3.
The x-intercepts are 3, −1,and1. There is a sign change at 3,butnotat−1 and 1.
When x is large positive, 3 − x is negative and the other factors are positive, so
lim f(x) =−∞. Whenx is large negative, 3 − x is positive, so
x→∞
lim f(x) =∞.
x→−∞
53. (a) Since −1 ≤ sin x ≤ 1 for all x, − 1 x ≤ sin x
x
≤ 1 for x>0. Asx →∞, −1/x → 0 and 1/x → 0, so by the Squeeze
x
sin x
Theorem, (sin x)/x → 0. Thus, lim
x→∞ x =0.