Fall06-Test1-solutions

Fall06-Test1-solutions - Math 1205 Test 1 20 Sep 2006 f Hx...

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Math 1205 -- Test 1 20 Sep 2006 NAME: Explain all work! Use only methods developed in class. - 6 - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 6 x - 4 - 3 - 2 - 1 1 2 3 4 5 f H x L 1 . [16] 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 - 4 lim x ض f H x L 4 lim x Ø- 3 f H x L 2 lim x Ø H - 1 L - f H x L - 4 lim x Ø H - 1 L + f H x L - 2 lim x Ø- 1 f H x L DNE left right lim x Ø 1 - f H x L H DNE L lim x Ø 1 + f H x L H DNE L lim x Ø 1 f H x L H DNE L lim x Ø 2 - f H x L 2 lim x Ø 2 + f H x L 2 lim x Ø 2 f H x L 2
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( 1 continued) b) Using the definition of asymptotes, explain why x = 1 is a vertical asymptote, but x = - 1 is not. Vert. Asymps( lim x Ø a ± f H x L = ± ): x = 1 satisfies the definition; f H x L is finite near x = - 1 c) Give the x - values for which f is not continuous. Explain the kind of discontinuity of each. At x = - 1 left and right limits exist but are unequal(jump discontinuity); at x = 1 the limit does not exist; at x = 2 left and right limits exist and are equal but are not equal to f H 2 L (removable discontinu- ity). f is continuous at x = - 3 2 . [8] Can f H x L approach a limit as x Ø c even if f H c L is undefined? If so, give an example. Yes, f H x L = sin H x L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x with c=0 is an example. lim x Ø 0 f H x L = 1 , but f is undefined at x = 0 . 3 . [8] If we know that 1 - x ÅÅÅÅÅ 2 § log H x + 1 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ x § 1 + x § for » x » § .5 , then show how to use the Squeeze Theo- rem to find lim x Ø 0 log H x + 1 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ x . The Squeeze Theorem tells us that if g 1 H x L § f H x L § g 2 H x L near a and lim x Ø a g 1 H x L = b = lim x Ø a g 2 H x L , then lim x Ø a f H x L = b . Let g 1 H x L = 1 - x ÅÅÅÅÅ 2 and g 2 H x L = 1 + » x » , a=0, we see that since lim x Ø 0 g 1 H x L = lim x Ø 0 I 1 - x ÅÅÅÅÅ 2 M = 1 = lim x Ø 0 H 1 + » x »L = lim x Ø 0 g 2 H x L , then lim x Ø 0 f H x L = 1 .
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