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Unformatted text preview: Math 1205  Test 1 14 Feb 2007 NAME: ANSWER KEY 4 3 2 1 1 2 3 4 x 4 3 2 1 1 2 f H x L 1 . a) Let f H x L be defined by the graph given above. Determine the following limits, writing + or  as appropriate. If a limit does not exist, explain why. lim x  f H x L 1 lim x 3 f H x L 1 lim x 4 f H x L 1 lim x 1 f H x L 2 lim x 1 + f H x L 1 lim x 1 f H x L DNE left and right limits are not equal lim x  1 f H x L DNE lim x  1 f H x L DNE lim x 2 f H x L 1 b) Using the definition of asymptotes, explain why x =  1 is a vertical asymptote, but x = 3 is not. f(x) has a vertical asymptote at x=a if and only if the left or right (or both) limit of f(x) as x a is . That is the case for x=1, but not for x=3. c) From the graph of f , state the xvalues at which f is discontinuous and explain why. f does not satisfy lim x a f H x L =f(a) at x=3, 4 the left and right limits of f are not equal at x=1 the values of f decrease without bound at x=1 2 . If we know that x cos H x L sin H x L x for 0 x p 2 , find lim x + sin H x L x . Support your answer....
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This test prep was uploaded on 04/01/2008 for the course MATH 1205 taught by Professor Fbhinkelmann during the Fall '08 term at Virginia Tech.
 Fall '08
 FBHinkelmann
 Calculus, Limits

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