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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.
 Fall '08
 CHESTKOFSKY
 Calculus, Asymptotes

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