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chapter 2 36 - 100 CHAPTER 2 LIMITS AND DERIVATIVES 62(a...

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Unformatted text preview: 100 CHAPTER 2 LIMITS AND DERIVATIVES 62. (a) lim F(1Z) : 0 and lim F(rc) : 0. so lim F(ar) = 0. which is F(0) and hence F is continuous at w = a if z—>0+ m—vO‘ :1:—>O a = 0. Fora > 0. lim F(z) = lim at = a. : F(a). Fora < 0. lim F(a:) : lim(—:c) = —a = F(a). Thus, z—m. m—nz x—>a z—va F is continuous at :c : a; that is. continuous everywhere. (b) Assume that f is continuous on the interval I. Then for a E 1. lim If(m)| = Ilim f(a:)‘ : |f(a)| by I—Nl Theorem 8. (If a is an endpoint of I. use the appropriate one-sided limit.) So |f| is continuous on I. 1 if a: 2 0 (c) No. the converse is false. For example. the function f(a:) : { 1 'f 0 is not continuous at at = 0. but — l m < |f(a:)| = 1 is continuous on R. 63. Define u(t) to be the monks 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 i d)(12) = D. So by the Intermediate Value Theorem. there must be some time to between 0 and 12 such that (u — d)(t0) : 0 <=> u(to) : d(t0). So at time to after 7:00 A.M.. the monk will be at the same place on both days. 2.6 Limits at Infinity; Horizontal Asymptotes ___________—_______‘——————— 1. (a) As :1: becomes large. the values of f(17) approach 5. (b) As 3: becomes large negative. the values of f (2:) approach 3. 2. (a) The graph of a function can intersect a The graph of a function can intersect a horizontal asymptote. vertical asymptote in the sense that it It can even intersect its horizontal asymptote an infinite can meet but not cross it. number of times. No horizontal asymptote One horizontal asymptote Two horizontal asymptotes ...
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