Business Calc Homework w answers_Part_10

# Business Calc Homework w answers_Part_10 - 46 Section 2.1...

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48. continued (b) lim x 2 1 1 f ( x ) 5 0; lim x 2 1 2 f ( x ) 5 0 (c) Yes. The limit is 0. 49. (a) [ 2 2 p , 2 p ] by [ 2 2, 2] (b) ( 2 2 p , 0) < (0, 2 p ) (c) c 5 2 p (d) c 5 2 2 p 50. (a) [ 2 p , p ] by [ 2 3, 3] (b) 1 2 p , } p 2 } 2 < 1 } p 2 } , p 2 (c) c 5 p (d) c 5 2 p 51. (a) [ 2 2, 4] by [ 2 1, 3] (b) (0, 1) < (1, 2) (c) c 5 2 (d) c 5 0 52. (a) [ 2 4.7, 4.7] by [ 2 3.1, 3.1] (b) ( 2‘ , 2 1) < ( 2 1, 1) < (1, ) (c) None (d) None 53. [ 2 4.7, 4.7] by [ 2 3.1, 3.1] lim x 0 ( x sin x ) 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 ) x ) and h ( x ) 5 ) x ) . ) x sin x ) 5 ) x ) ? ) sin x ) # ) x ) ? 1 5 ) x ) 2 ) x ) # x sin x # ) x ) Because lim x 0 ( 2 ) x ) ) 5 lim x 0 ) x ) 5 0, the Sandwich Theorem gives lim x 0 ( x sin x ) 5 0. 54. [ 2 4.7, 4.7] by [ 2 5, 5] lim x 0 ( x 2 sin x ) 5 0 Confirm using the Sandwich Theorem, with g ( x ) = 2 x 2 and h ( x ) 5 x 2 . ) x 2 sin x ) 5 ) x 2 ) ? ) sin x ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 sin x # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem gives lim x 0 ( x 2 sin x ) 5 0 55. [ 2 0.5, 0.5] by [ 2 0.25, 0.25] lim x 0 1 x 2 sin } x 1 2 } 2 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 x 2 and h ( x ) 5 x 2 . ) x 2 sin } x 1 2 } ) 5 ) x 2 ) ? ) sin } x 1 2 } ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 sin } x 1 2 } # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem give lim x 0 1 x 2 sin } x 1 2 } 2 5 0. 56. [ 2 0.5, 0.5] by [ 2 0.25, 0.25] lim x 0 1 x 2 cos } x 1 2 } 2 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 x 2 and h ( x ) 5 x 2 . ) x 2 cos } x 1 2 } ) 5 ) x 2 ) ? ) cos } x 1 2 } ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 cos } x 1 2 } # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem give lim x 0 1 x 2 cos } x 1 2 } 2 5 0. 57. (a) In three seconds, the ball falls 4.9(3) 2 5 44.1 m, so its average speed is } 44 3 .1 } 5 14.7 m/sec. 46 Section 2.1

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(b) The average speed over the interval from time t 5 3 to time 3 1 h is } D D y t } 5 5 } 4.9(6 h h 1 h 2 ) } 5 29.4 1 4.9 h Since lim h →0 (29.4 1 4.9 h ) 5 29.4, the instantaneous speed is 29.4 m/sec. 58. (a) y 5 gt 2 20 5 g (4 2 ) g 5 } 2 1 0 6 } 5 } 5 4 } or 1.25 (b) Average speed 5 } 2 4 0 } 5 5 m/sec (c) If the rock had not been stopped, its average speed over the interval from time t 5 4 to time t 5 4 1 h is } D D y t } 5 5 } 1.25(8 h h 1 h 2 ) } 5 10 1 1.25 h Since lim h →0 (10 1 1.25 h ) 5 10, the instantaneous speed is 10 m/sec.
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