Pre-Calc Homework Solutions 334

Pre-Calc Homework Solutions 334 - 334 Section 8.3 f(x xa...

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Unformatted text preview: 334 Section 8.3 f (x) xa g(x) 44. (a) f o(g) as xa if lim 0 3. If p 1 0 1, then 1 Suppose f and g are both positive in some open interval containing a. Then f O(g) as xa if there is a f (x) g(x) dx xp lim c0 c dx xp lim c0 x positive integer M for which close to a. M for x sufficiently lim c0 p 1 1 p 1 c 1 c p 1 p 1 because ( p 1) 0. (b) From Section 5.5, we know that ES ES h 4 b a 4 h M 180 (4) 4. If 0 1 0 p 1, then 1 where M is a bound for the absolute value of f [a, b]. Thus, (b M a) 180 on 1 dx xp lim c0 c dx xp p 1 int (b M a) 180 lim c0 x as h0, so ES O(h 4). Thus as h0, ES0. b 12 a 2 h M lim c0 1 p 1 c 1 c p 1 p 1 1 1 p (c) From Section 5.6, we know that ET ET h2 Quick Review 8.3 3 where M is a bound for the absolute value of f on [a, b]. Thus h0, so ET 45. (a) lim x 1. 0 dx x 3 x dx x 2 3 ln x 3 0 ln 6 1 ln 3 1 ln 2 2 ln 2 1 ln 2 2 (b 2 a) M 12 int (b a) M 12 1 as 2. 1 1 O(h ). Thus as h0, ET0. lim x 1 1 4 1 ln x 2 2 dx x 2 2 1 1 0 f(x) g(x) f (x) g(x) f (x) x g(x) lim 3. dx x 2 4 1 Thus f grows faster than g as x by definition. (b) lim x f(x) g(x) lim x f (x) g(x) f (x) lim x g(x) 1 x 2 tan 1 4 2 1 x tan 1 C 2 2 1 3 x 4 dx x 3 C L 4. dx x4 Thus f grows at the same rate as g as x by definition. 46. (a) lim x C 3 f ( x) g( x) lim x f (x) g(x) 5. 9 x 2 0 for 3 x The domain is ( 3, 3). 6. x 1 0 for x 1 The domain is (1, ). 7. L 1 cos x x2 Thus f ( x) grows faster than g( x) by definition. f ( x) (b) lim x g( x) f (x) lim g(x) x cos x cos x x2 1, so cos x 1 x2 1. Thus f ( x) grows at the same rate as g(x) by definition. 8. x 2 1 x2 1 1 x 2 so 1 x x2 1 x2 x for x 1 s Section 8.3 Improper Integrals (pp. 433444) 1 9. lim f (x) x g(x) 4e x x x 3e lim 5 7 4e x x x 3e lim lim x 4 3 4 3 Exploration 1 1. Because 1 Investigating 0 dx xp 0. Thus f and g grow at the same rate as x . 1 has an infinite discontinuity at x xp 1 10. lim x f (x) g(x) lim x 2x x 2x x 2 1 3 1 3 1 x 3 x 2. 0 dx x lim c0 c dx x lim ln x c0 1 c lim ( ln c) c0 lim x lim x 2 1 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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