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G 0 (0) is the only negative value. The slope at x = 4 is smaller than the slope at x = 2 and both are smaller than the slope at x = −2. Thus, g 0 (0) < 0 < g 0 (4) < g0 (2) < g 0 (−2). 19. We begin by drawing a curve through the origin with a slope of 3 to satisfy f (0) = 0 and f 0 (0) = 3. Since f 0 (1) = 0, we will round off our figure so that there is a horizontal tangent directly over x = 1. Last, we make sure that the curve has a slope of −1 as we pass over x = 2. Two of the many possibilities are shown.

Then x > N = max(1, ln M) ⇒ ex > max(e, M) ≥ M , so lim ex = ∞. x→∞ 71. Suppose that lim f (x) = L. Then for every ε > 0 there is a corresponding positive number N such that |f (x) − L| < ε x→∞ whenever x > N . If t = 1/x, then x > N ⇔ 0 < 1/x < 1/N ⇔ 0 < t < 1/N. Thus, for every ε > 0 there is a corresponding δ > 0 (namely 1/N) such that |f(1/t) − L| < ε whenever 0 < t < δ. This proves that lim f (1/t) = L = lim f(x). t→0+ x→∞ Now suppose that lim f (x) = L. Then for every ε > 0 there is a corresponding negative number N such that x→−∞ |f (x) − L| < ε whenever x < N .

5|. Note that 2x2 + x + 1 lim g(x) = x→∞ 3 2 and lim f(x) = 0. 05. 804, so we choose N = 15 (or any larger number). 63. 5 whenever x ≤ N. We graph the three parts of this x+1 inequality on the same screen, and see that the inequality holds for x ≤ −6. So we choose N = −6 (or any smaller number). 9 whenever x ≤ N. From the x+1 graph, it seems that this inequality holds for x ≤ −22. So we choose N = −22 (or any smaller number). ¤ 65 66 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 65. 0001 = 10 000 ⇔ x > 100 (x > 0) √ √ (b) If ε > 0 is given, then 1/x2 < ε ⇔ x2 > 1/ε ⇔ x > 1/ ε.

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Calculus, 6ed., solutions manual by Stewart J.

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