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MIT18_014F10_pr_ex3_sols

MIT18_014F10_pr_ex3_sols - Practice Exam 3 Solutions...

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0 Practice Exam 3 Solutions t 3 + t 1+ t 2 dt Problem 1. Evaluate Solution The problem can be simplified as t 3 + t = t ( t 2 + 1). Then by substitution with u = t 2 + 1 and thus du = 2 t dt we have t t 2 + 1 dt = 1 u 1 / 2 du = 1 u 3 / 2 + C = 1 ( t 2 + 1) 3 / 2 + C. 2 3 3 2 Problem 2. Evaluate 5 x 3 x 9 dx 3 Solution We begin by making the substitution x 2 9 = u . Then 2 xdx = du and x 2 = u + 9. Substituting in, we get u 3 / 2 16 16 16 du 1 1 2 18 1 ( u +9) u 3 / 2 +9 u 1 / 2 du = 5 / 2 16 5 / 2 +3 16 3 / 2 + = u u = . · 2 2 2 5 3 5 0 0 Problem 3: Suppose that lim x a + g ( x ) = B = 0 where B is finite and lim x a + h ( x ) = 0, but h ( x ) = 0 in a neighborhood of a . Prove that g ( x ) h ( x ) lim x a + = . Solution Let M R + and set = 1 / (2 M ) > 0. By hypothesis, there exist δ 1 , δ 2 such that | g ( x ) B | < | B | / 2 if 0 < x a < δ 1 and | h ( x ) | < | B | if 0 < x a < δ 2 . Choose δ = min { δ 1 , δ 2 } . Then for 0 < x a < δ , | g ( x ) | > | B | / 2 and | h ( x ) | 1 > 1 / ( B � ). Thus | | | g ( x ) | > | B | = 1 = M.
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