Math_161_HW_Even_Solns_Ch._10

# Math_161_HW_Even_Solns_Ch._10 - 10 LIMITS AND THE...

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EXERCISE 10-1 303 10 LIMITS AND THE DERIVATIVE EXERCISE 10-1 2. (A) lim x 1 f ( ) = 2 (B) lim 1 + ( ) = 2 (C) lim 1 ( ) = 2 (D) (1) = 2 4. (A) lim 4 ( ) = 4 (B) lim 4 + ( ) = 4 (C) lim 4 ( ) = 4 (D) (4) does not exist (E) Yes, define (4) = 4 6. (A) lim 2 g ( ) = 2 (B) lim 2 + ( ) = 2 (C) lim 2 ( ) = 2 (D) (2) = 2 8. (A) lim 4 ( ) = 0 (B) lim 4 + ( ) = 0 (C) lim 4 ( ) = 0 (D) (4) = 0 10. (A) lim →− 2 + ( ) = 3 (B) lim 2 ( ) = -3 (C) Since lim 2 + ( ) ± lim 2 ( ), lim 2 ( ) does not exist. (D) (-2) = -3 (E) No, lim 2 ( ) does not exist. 12. (A) lim 2 + ( ) = -3 (B) lim 2 ( ) = 3 (C) lim 2 ( ) does not exist since lim 2 + ( ) ± lim 2 ( ) (D) (2) = 3 (E) No, lim 2 ( ) does not exist. 14. 4 12 as 3; thus lim 3 4 = 12 16. - 3 5 - 3 = 2 as 5; thus lim 5 ( - 3) = 2 18. ( + 3) (-1)(-1 + 3) = -2 as -1; thus lim 1 ( + 3) = -2 20. - 2 4 - 2 = 2 as x 4; thus lim 4 2 = 2 4 = 1 2 22. 16 7 16 7(0) = 16 = 4 as 0; thus lim 0 16 7 = 4 24. lim 1 2 ( ) = 2 lim 1 ( ) = 2(4) = 8 26. lim 1 [ ( ) - 3 ( )] = lim 1 ( ) - 3 lim 1 ( ) = 4 - 3(-5) = 19 28. lim 1 3 ( ) 1 4 ( ) = lim 1 [3 ( )] lim 1 [1 4 ( )] = 3 lim 1 ( ) 1 4 lim 1 ( ) = 3 ( 5) 1 4(4) = - 8 15

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304 CHAPTER 10 LIMITS AND THE DERIVATIVE 30. lim x 1 [ f ( ) - 7 ] g ( ) = lim 1 [ ( ) 7 ] · lim 1 ( ) = lim 1 ( ) 7 lim 1 · (4) = {-5 - 7} · (4) = -48 32. lim 1 + 2 ( ) 3 = lim 1 [2 + 2 ( )] 3 = 2 lim 1 + 2 lim 1 ( ) 3 = 10 3 = -2 34. lim 1 [2 - ( )] 3 = [2 - lim 1 ( )] 3 = [2 - 4] 3 = -8 36. 38. 40. ( ) = 2 + if 0 2 if > 0 (A) lim 0 + ( ) = lim 0 + (2 - ) = 2 (B) lim 0 ( ) = lim 0 (2 + ) = 2 (C) lim 0 ( ) = 2 since lim 0 + ( ) = lim 0 ( ) = 2 (D) (0) = 2 + 0 = 2 42. ( ) = + 3i f <− 2 + 2 if >− 2 (A) lim →− 2 + ( ) = lim 2 + + 2 = 0 (B) lim 2 ( ) = lim 2 ( + 3) = 1 (C) lim 2 ( ) does not exist since lim 2 + ( ) ± lim 2 ( ) (D) (-2) does not exist; is not defined at = -2. 44. ( ) = + 3 if < 0 3 if > 0 (A) lim 3 ( ) = lim 3 + 3 does not exist since = -3 is a non-removable zero of the denominator.
EXERCISE 10-1 305 (B) lim x 0 f ( ) = lim 0 + 3 = lim 0 + + 3 = 0 (C) lim 3 ( ) does not exist, since lim 3 + ( ) does not exist. 46. ( ) = 3 3 = 3 ( 3) =− 1i f < 3 3 3 = f > 3 (Note : Observe that for < 3, | - 3| = 3 - = -( - 3) and for > 3, | - 3| = - 3) (A) lim 3 + ( ) = lim 3 + 1 = 1 (B) lim 3 ( ) = lim 3 (-1) = -1 (C) lim 3 ( ) does not exist, since lim 3 + ( ) ± lim 3 ( ) (D) (3) does not exist; is not defined at = 3. 48. ( ) = + 3 2 + 3 = + 3 ( + (A) lim →− 3 + 3 ( + = lim 3 1 = - 1 3 (B) lim 0 ( ) = lim 0 1 does not exist. 50. ( ) = 2 + 6 + 3 = ( + 3)( 2) ( + (A) lim 3 ( ) = lim 3 ( + ( + = lim 3 ( - 2) = -5 (B) lim 0 ( ) = lim 0 2 + 6 + 3 = 6 3 = -2 (C) lim 2 ( ) = lim 2 2 + 6 + 3 = 0 5 = 0 52. ( ) = 2 1 ( + 1) 2 = ( 1)( + ( + 2 (A) lim 1 ( ) = lim 1 ( + ( + 2 = lim 1 1 + 1 does not exist since lim 1 ( - 1) = -2 but lim 1 ( + 1) = 0. (B) lim 0 ( ) = lim 0 2 1 ( + 2 = 1 1 = -1 (C) lim 1 ( ) = lim 1 2 1 ( + 2 = 0 4 = 0 54. ( ) = 3 2 + 2 1 2 + 3 + 2 = (3 + ( + 2) ( + (A) lim 3 ( ) = lim 3 3 2 + 2 1 2 + 3 + 2 = 20 2 = 10

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306 CHAPTER 10 LIMITS AND THE DERIVATIVE (B) lim x →− 1 f ( ) = lim 1 (3 1)( + 1) ( + 2) ( + = lim 1 3 1 + 2 = 4 1 = -4 (C) lim 2 ( ) = lim 2 3 2 + 2 1 2 + 3 + 2 = 15 12 = 5 4 56. ( ) = 5 - 1 lim h 0 (2 + ) (2) = lim 0 5(2 + ) 1 (1 0 = lim 0 10 + 5 1 9 = lim 0 5 = lim 0 5 = 5 58. ( ) = 2 - 2 lim 0 + ) = lim 0 + ) 2 2 (4 2) = lim 0 4 + 4 + 2 2 2 = lim 0 4 + 2 = lim 0 (4 + ) = 4 60. ( ) = 1 + lim 0 + ) = lim 0 1 + + 1 + () = lim 0 + = lim 0 + · + + + + = lim 0 2 + 2 + + = lim 0 + + = lim 0 1 + + = 1 62. ( ) = 2 + | - 2| lim 0 + ) = lim 0 2 + 2 + 2
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## This note was uploaded on 05/11/2010 for the course MATH mth161 taught by Professor L during the Spring '10 term at St. Martins.

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Math_161_HW_Even_Solns_Ch._10 - 10 LIMITS AND THE...

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