MATH
Prelim1Solutions

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Math 1110 Prelim 1 Answers Please find below abbreviated answers to the questions on the first prelim. On several of the questions, more explanation would be necessary to receive full credit. 1. ( 20 points ) Compute the following limits. (a) ( 5 points ) lim x 1 (3 x 3 - 2 x + 5) (b) ( 5 points ) lim x →∞ 3 x 2 + 10 x - 1 2 x 2 - 25 x + 120 (c) ( 5 points ) lim x →- 2 1 2 + 1 x x 2 - 4 (d) ( 5 points ) lim x 0 6 x 2 cos x sin x sin 2 x Solution (a) lim x 1 (3 x 3 - 2 x + 5) = 3 · 1 3 - 2 · 1 + 5 = 6 (b) ( Please note that in order to have received full credit on this question, you would have had to do polynomial division. ) Since the degree of the numerator and the degree of the denominator are the same, the limit is just the ratio of the leading coefficients. That is, lim x →∞ 3 x 2 + 10 x - 1 2 x 2 - 25 x + 120 = 3 2 (c) lim x →- 2 1 2 + 1 x x 2 - 4 = lim x →- 2 2+ x 2 x ( x + 2)( x - 2) = lim x →- 2 1 2 x ( x - 2) = 1 16 (d) lim x 0 6 x 2 cos x sin x sin 2 x = lim x 0 3 cos x x sin x 2 x sin 2 x = 3 2. ( 15 points ) Let f ( x ) be the function defined below.

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• Fall '06
• MARTIN,C.
• Calculus, Limits, lim, Continuous function, lim |f, lim x→0 sin

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