exam1practicesolutions - Math 2250 Exam #1 Practice Problem...

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Unformatted text preview: Math 2250 Exam #1 Practice Problem Solutions 1. Find the vertical asymptotes (if any) of the functions g ( x ) = 1 + 2 x , h ( x ) = 4 x 4- x 2 What are the domains of g and h ? Answer: The function g has a vertical asymptote at x = 0. The function h has vertical asymptotes when 4- x 2 = 0, so theyre at x =- 2 and x = 2. 2. Evaluate (a) lim x 2 x 2- 4 x 2- 5 x + 6 (b) lim x - 2 | x + 2 | x + 2 (c) lim x 4 x 3 + 2 x- 4 4 x 2- 5 x + 6 x 3 (a) We can factor the numerator as x 2- 4 = ( x + 2)( x- 2) and the denominator as x 2- 5 x + 6 = ( x- 2)( x- 3) . Therefore, lim x 2 x 2- 4 x 2- 5 x + 6 = lim x 2 ( x + 2)( x- 2) ( x- 2)( x- 3) = lim x 2 x + 2 x- 3 = 4- 1 =- 4 . (b) When x <- 2, the quantity x + 2 is negative, so | x + 2 | =- ( x + 2) . Hence, lim x - 2- | x + 2 | x + 2 = lim x - 2-- ( x + 2) x + 2 =- 1 . On the other hand, when x >- 2, the quantity x + 2 is positive, so | x + 2 | = x + 2 . Therefore, lim x - 2 + | x + 2 | x + 2 = lim x - 2 + x + 2 x + 2 = 1 . Since the limits from the left and right dont agree, lim x - 2 | x + 2 | x + 2 does not exist....
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

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exam1practicesolutions - Math 2250 Exam #1 Practice Problem...

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