Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-02 JWDD027-Salas-v1 November 25, 2006 15:53 SECTION 2.1 33 CHAPTER 2 SECTION 2.1 1. (a) 2 (b) 1 (c) does not exist (d) 3 2. (a) 4 (b) 4 (c) 4 (d) 2 3. (a) does not exist (b) 3 (c) does not exist (d) 3 4. (a) 1 (b) does not exist (c) does not exist (d) 1 5. (a) does not exist (b) does not exist (c) does not exist (d) 1 6. (a) 1 (b) 2 (c) does not exist (d) 2 7. (a) 2 (b) 2 (c) 2 (d) 1 8. (a) 2 (b) 2 (c) 2 (d) 2 9. (a) 0 (b) 0 (c) 0 (d) 0 10. (a) does not exist (b) does not exist (c) does not exist (d) 1 11. c = 0 , 6 12. c = 3 13. 1 14. 3 15. 12 16. 5 17. 1 18. does not exist 19. 3 2 20. does not exist 21. does not exist 22. does not exist 23. lim x 3 2 x 6 x 3 = lim x 3 2 = 2 24. lim x 3 x 2 6 x + 9 x 3 = lim x 3 ( x 3) 2 x 3 = lim x 3 ( x 3) = 0 25. lim x 3 x 3 x 2 6 x + 9 = lim x 3 x 3 ( x 3) 2 = lim x 3 1 x 3 ; does not exist 26. lim x 2 x 2 3 x + 2 x 2 = lim x 2 ( x 1)( x 2) x 2 = lim x 2 ( x 1) = 1 27. lim x 2 x 2 x 2 3 x + 2 = lim x 2 x 2 ( x 1)( x 2) = lim x 2 1 x 1 = 1 28. does not exist 29. does not exist 30. lim x 1 x + 1 x = 2 31. lim x 0 2 x 5 x 2 x = lim x 0 (2 5 x ) = 2
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-02 JWDD027-Salas-v1 November 25, 2006 15:53 34 SECTION 2.1 32. lim x 3 x 3 6 2 x = lim x 3 x 3 2(3 x ) = lim x 3 1 2 = 1 2 33. lim x 1 x 2 1 x 1 = lim x 1 ( x 1)( x + 1) x 1 = lim x 1 ( x + 1) = 2 34. lim x 1 x 3 1 x 1 = lim x 1 ( x 1)( x 2 + x + 1) ( x 1) = lim x 1 x 2 + x + 1 = 3 35. 0 36. does not exist 37. 1 38. 3 39. 16 40. 0 41. does not exist 42. 2 43. 4 44. 0 45. does not exist 46. 2 47. lim x 1 x 2 + 1 2 x 1 = lim x 1 ( x 2 + 1 2)( x 2 + 1 + 2) ( x 1)( x 2 + 1 + 2) = lim x 1 x 2 1 ( x 1)( x 2 + 1 + 2) = lim x 1 x + 1 x 2 + 1 + 2 = 2 2 2 = 1 2 48. lim x 5 x 2 + 5 30 x 5 = lim x 5 ( x 2 + 5 30)( x 2 + 5 + 30) ( x 5)( x 2 + 5 + 30) = lim x 5 x 2 25 ( x 5)( x 2 + 5 + 30) = lim x 5 x + 5 x 2 + 5 + 30 = 10 2 30 = 5 30 49. lim x 1 x 2 1 2 x + 2 2 = lim x 1 x 2 1 2 x + 2 2 2 x + 2 + 2 2 x + 2 + 2 = lim x 1 ( x 1)( x + 1) ( 2 x + 2 + 2 ) 2 x + 2 4 = lim x 1 ( x 1)( x + 1) ( 2 x + 2 + 2 ) 2( x 1) = lim x 1 ( x + 1) ( 2 x + 2 + 2 ) 2 = 2 · 4 2 = 4 50. 0 51. 2 52. 0 . 167 53. 3 2 54. 1 12 55. ( i ) 5 ( ii ) does not exist 56. ( i ) 0 ( ii ) 2 3 57. ( i ) 5 4 ( ii ) 0 58. ( i ) 17 32 ( ii ) 1 2 59. c = 1 60. c = 5 2 61. c = 2
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-02 JWDD027-Salas-v1 November 25, 2006 15:53 SECTION 2.2 35 62. c = 3 63. lim x 0 f ( x ) = 1; lim x 0 g ( x ) = 0 64. lim x 0 f ( x ) = 1; lim x 0 g ( x ) = 0 SECTION 2.2 1. 1 2 2. lim x 0 x 2 (1 + x ) 2 x = lim x 0 1 2 x (1 + x ) = 0 3. lim x 0 x (1 + x ) 2 x 2 = lim x 0 1 + x 2 x ; does not exist 4. 4 3 5. lim x 1 x 4 1 x 1 = lim x 1 ( x 3 + x 2 + x + 1) = 4 6. does not exist 7. does not exist 8. lim x 1 x 2 1 x 2 2 x + 1 = lim x 1 ( x 1)( x + 1) ( x 1) 2 = lim x 1 x + 1 x 1 ; does not exist 9. 1 10. does not exist 11. does not exist 12. 1 13. 0 14. 0
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