06Fe3p1 - 1 lim n →∞ 1 2 n = 2 lim n →∞(17 n = 1 3 lim n →∞ 5 n 3 6 n 3 17 n 3 9 n 2 4 = 4 lim n →∞ n 3 e n = 5 lim n →∞ 3 √

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Prof. Girardi Math 142 Fall 2006 11.21.06 Exam 3 - Part 1 Warning: there are 5 problems: 2 problems on the front and 3 problems on the back. Turn your paper over. There are 5 problems. Each problem is worth 2 points. 10 NAME: please check the box of your section Section 005 (WF 8:00 am) or Section 006 (WF 9:05 am) INSTRUCTIONS : Indicate your reasoning. Put answers in box and show work below the box.
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Unformatted text preview: 1. lim n →∞ 1 2 n = 2. lim n →∞ (17) n = 1 3. lim n →∞ 5 n 3 + 6 n + 3 17 n 3 + 9 n 2 + 4 = 4. lim n →∞ n 3 e n = 5. lim n →∞ 3 √ n 2 + 5 6 √ 64 n 4 + 17 n = hint: 3 √ n 2 + 5 6 √ 64 n 4 + 17 n = ( n 2 + 5) 1 3 (64 n 4 + 17 n ) 1 6 = ( n 2 + 5) 2 6 (64 n 4 + 17 n ) 1 6 = [( n 2 + 5) 2 ] 1 6 (64 n 4 + 17 n ) 1 6 2...
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This note was uploaded on 12/13/2011 for the course MATH 142 taught by Professor Kustin during the Fall '11 term at South Carolina.

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06Fe3p1 - 1 lim n →∞ 1 2 n = 2 lim n →∞(17 n = 1 3 lim n →∞ 5 n 3 6 n 3 17 n 3 9 n 2 4 = 4 lim n →∞ n 3 e n = 5 lim n →∞ 3 √

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