Calculus: One and Several Variables

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Unformatted text preview: 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 → 2 x − 5 x 2 x = lim x → (2 − 5 x ) = 2 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. 36. does not exist 37. 1 38. 3 39. 16 40. 41. does not exist 42. 2 43. 4 44. 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. 51. 2 52. . 167 53....
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This note was uploaded on 04/30/2011 for the course MATH 1431 taught by Professor Any during the Spring '08 term at University of Houston.

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ch02[1] - P1 PBU/OVY P2 PBU/OVY QC PBU/OVY T1 PBU...

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