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exam1practicesolutions

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

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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 they’re 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 don’t agree, lim x →- 2 | x + 2 | x + 2 does not exist. (c) Dividing numerator and denominator by x 3 , we get that lim x →∞ 4 x 3 + 2 x - 4 4 x 2 - 5 x + 6 x 3 = lim x →∞ 1 x 3 ( 4 x 3 + 2 x - 4 ) 1 x 3 (4 x 2 - 5 x + 6 x 3 ) = lim
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