Solução_Calculo_Stewart_6e

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F. (c) lim x→−∞ √ x2 + x +1+x = lim x→−∞ = lim x→−∞ = √ x2 + x +1+x √ x 2 + x +1− x √ = lim x2 + x +1− x x→−∞ (x +1)(1/x) √ x2 + x +1− x (1/x) = 1+0 − √ 1+0+0− 1 = − 1 2 lim x→−∞ 1+(1/x) − 1+(1/x)+(1/x 2 ) − 1 Note that for x
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.
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