Spring07-Test1-solutions

Spring07-Test1-solutions - Math 1205 -- Test 1 14 Feb 2007...

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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 x-values 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.

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Spring07-Test1-solutions - Math 1205 -- Test 1 14 Feb 2007...

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