ma1012_hw9_201012

ma1012_hw9_201012 - Mathematics I: Homework 9. Limit...

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Unformatted text preview: Mathematics I: Homework 9. Limit Laws. Thomas’ Calculus 11th Edition, Chapter 2, Section 2: 1, 7, 13, 17, 23, 31, 35, 39, 43, 47, 49. Total: 14 Answers 1.  ­9 7. 5/8 13. 16 47. È7 b h c È7 lim h hÄ! Section 2.2 Calculating Limits Using the Limit Laws œ lim ŠÈ7 b h c È7‹ ŠÈ7 b h b È7‹ h ŠÈ7 b h b È7‹ " 77 17. 3/2 23. 3/2 h hÄ! œ lim h Ä ! h ŠÈ7 b h b È7‹ (7 b h) c 7 œ lim31. 4 È h Ä ! h Š 7 bh bÈ 7 ‹ œ lim h Ä ! È 7 bh b È 7 œ " #È 7 48. lim È3(0 b h) b 1 c È3(0) b 1 lim h hÄ! ŠÈ7 b h c È7‹ ŠÈ7 b h b È7‹ (7 b h) c 7 3h œ lim 3 œ lim 7 œ limÈ b h b 1 7 1 h ŠÈ7 b h b Èh ‹È3h b 1 b "‹ Ä ! ÄŠ! 7È3hb Èb‹ h hÄ! h Š hÄ! h 35. 3/2 ŠÈ3h b 1 c ‹ ŠÈ3h b Using Section 2.2 Calculating "Limits 1 b "‹ the Limit b 1) c " (3h Laws œ lim œ lim h ŠÈ3h b 1b "‹ hÄ! h Ä ! h ŠÈ3h b 1 b 1 ‹ 39.  ­10,  ­20,  ­1, 5/7 43. 2 œ 3 # 77 È 7‹ " " œ lim È7bhbÈ7 œ #È7 hÄ! 49. lim È47. 2x# œ È5 c 2(0)# œ È5 and lim È5 c x# œ È5 c (0)# œ È5; by the sandwich theorem, 5c È3(0) b 1 1 b "‹ # a œ 50. xlim! b 2 b 1 x b #œ 2 c 0 œ 2 and xlim! 2 cos x œ 2(1) œ 2; by the sandwich theorem, xlim! g(x) œ 2 lim Ä 3 c œ 3 Ä Ä h Ä ! È3h 1 x sin x 2c2 cos x limŠÈ3h bœcÈ5 È3h b 1 b "‹ f(x) 1 "‹ Š x œ limÄ ! 49. œ lim È hÄ! hŠ 3h b 1b "‹ xÄ! xÄ! h Ä ! h ŠÈ3h b 1 b 1 ‹ (3h b 1) c " 51. (a) 1c x ‹ 1 œ 1; c 0 œ 1 and # 6 È5 c 2(0)# œ Èxlim! Šlim È5 œ 1 # œ È5 c (0)xlim!È5; by by the sandwich theorem, xlim! 5Ä and the sandwich theorem, cx 6 œ Ä Ä xÄ! (b) For x Á 0, y œ (x sin x)/(2 c 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0. c 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ! xÄ! x sin x œ1 (x sin x)/(2 c 2 cos x) he other two graphslimthe " c 24 ‹ œ lim 52. (a) in Š # x xÄ! xÄ! e graphs converge as x Ä 0. 1ccos x " lim œ #. x xÄ! ‹œ1c 0 6 œ 1 and lim 1 œ 1; by the sandwich theorem, lim xÄ! 1 # x Ä ! 2c2 cos x " # œ1 " c lim x Ä ! #4 x œ c0œ " # and lim xÄ! # œ " ; by the sandwich theorem, # (b) For all x Á 0, the graph of f(x) œ (1 c cos x)/x# lies between the line y œ " and the parabola # yœ 4‹ " # x c x# /24, and the graphs converge as x Ä 0. œ " # œ lim 1 xÄ! # c lim x Ä ! #4 c0œ " # and lim " xÄ! # œ " ; by the sandwich theorem, # œ ". # the graph of f(x) œ (1 c cos x)/x# " he line y œ3. and the parabola at those points c where lim x% œ lim x# . Thus, c% œ c# Ê c# a1 c c# b œ 0 5 # xlimc f(x) exists Ä xÄc xÄc , and the graphs converge as x Ä 0. Ê c œ 0, 1, or c1. Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. fpm@itesm.mx xÄ! xÄ! x Ä c1 xÄ1 54. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the ...
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