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Unformatted text preview: ASSIGNMENT 2  SOLUTIONS Problem 1. a) Both F 1 ( x ) and F 2 ( x ) have x = 0 as the possible vertical asymptote. We check by computing onesided limits: lim x → + F 1 ( x ) = lim x → + √ x 2 + 1 + 1 x = ∞ , lim x → F 1 ( x ) = lim x → √ x 2 + 1 + 1 x =∞ Either one of the above limits suffices to conclude that x = 0 is, indeed, a vertical asymptote for F 1 ( x ). But it’s a good idea to compute them both whenever possible, firstly because it’s good practice, and secondly because we’ll have to do that latter on when we’ll sketch graphs. Both lim x → + F 2 ( x ) and lim x → F 2 ( x ) are of the form . If we use the conjugate, we get: √ x 2 + 1 1 x = ( √ x 2 + 1 1)( √ x 2 + 1 + 1) x ( √ x 2 + 1 + 1) = x 2 x ( √ x 2 + 1 + 1) = x √ x 2 + 1 + 1 ( x Ó = 0) We now see that lim x → + F 2 ( x ) = lim x → + √ x 2 + 1 1 x = lim x → + x √ x 2 + 1 + 1 = 0 and, similarly, lim x → F 2 ( x ) = 0. Therefore x = 0 is not a vertical asymptote for...
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This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.
 Spring '10
 MARYAMNAMAZI
 Calculus, Logic, Limits

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