Pre-Calc Homework Solutions 329

Pre-Calc Homework Solutions 329 - Section 8.2 53. (a) f (x)...

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53. (a) f ( x ) 5 e x ln (1 1 1/ x ) 1 1 } 1 x } . 0 when x ,2 1 or x . 0 Domain: ( 2‘ , 2 1) < (0, ) (b) The form is 0 2 1 , so lim x 2 1 2 f ( x ) 5‘ (c) lim x →- x ln 1 1 1 } 1 x } 2 5 lim x →- 5 lim x →- 5 lim x →- 5 1 lim x →- f ( x ) 5 lim x →- e x ln (1 1 1/ x ) 5 e 54. (a) Because the difference in the numerator is so small compared to the values being subtracted, any calculator or computer with limited precision will give the incorrect result that 1 2 cos x 6 is 0 for even moderately small values of x . For example, at x 5 0.1, cos x 6 < 0.9999999999995 (13 places), so on a 10-place calculator, cos x 6 5 1 and 1 2 cos x 6 5 0. (b) Same reason as in part (a) applies. (c) lim x 0 } 1 2 x c 1 o 2 s x 6 }5 lim x 0 } 6 x 1 5 2 s x in 11 x 6 } 5 lim x 0 } si 2 n x x 6 6 } 5 lim x 0 } 6 x 5 12 co x s 5 x 6 } 5 lim x 0 } cos 2 x 6 } 5 } 1 2 } (d) The graph and/or table on a grapher show the value of the function to be 0 for x -values moderately close to 0,
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