Spring07-Test1-solutions

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

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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 Ø-¶ f H x L - 1 lim x Ø 3 f H x L - 1 lim x Ø 4 f H x L 1 lim x Ø 1 - f H x L 2 lim x Ø 1 + f H x L - 1 lim x Ø 1 f H x L DNE - left and right limits are not equal lim x Ø- 1 - f H x L DNE - ¶ lim x Ø- 1 f H x L DNE - ¶ lim x Ø 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 Ø 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 Ø 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
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2 . If we know that x cos H x L § sin H x L § x for 0 § x § p ÅÅÅÅ 2 , find lim x Ø 0 + sin H x L ÅÅÅÅÅÅÅÅÅÅÅÅÅ x . Support your answer. The given inequality can be rewritten as cos H x L § sin H x L ÅÅÅÅÅÅÅÅÅÅÅÅ x § 1 by dividing through by x. Since lim x
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