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Unformatted text preview: nav277 – Homework 10 – Odell – (58340) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Compute the value of lim n →∞ 4 a n b n 2 a n − 6 b n when lim n →∞ a n = 6 , lim n →∞ b n = − 2 . 1. limit = − 2 correct 2. limit = − 25 12 3. limit = 25 12 4. limit = 2 5. limit doesn’t exist Explanation: By properties of limits lim n → 2 4 a n b n = 4 lim n →∞ a n lim n →∞ b n = − 48 while lim n →∞ (2 a n − 6 b n ) = 2 lim n →∞ a n − 6 lim n →∞ b n = 24 negationslash = 0 . Thus, by properties of limits again, lim n →∞ 4 a n b n 2 a n − 6 b n = − 2 . 002 10.0 points If lim n →∞ a n = 2 , determine the value, if any, of lim n →∞ a n − 8 . 1. limit doesn’t exist 2. limit = − 6 3. limit = 10 4. limit = 2 correct 5. limit = 1 4 Explanation: To say that lim n →∞ a n = 2 means that a n gets as close as we please to 2 for all sufficiently large n . But then a n − 8 gets as close as we please to 2 for all sufficiently large n . Consequently, lim n →∞ a n − 8 = 2 . 003 10.0 points Find a formula for the general term a n of the sequence { a n } ∞ n =1 = braceleftBig 1 , − 4 3 , 16 9 , − 64 27 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = − parenleftBig 4 3 parenrightBig n 2. a n = − parenleftBig 3 4 parenrightBig n 3. a n = − parenleftBig 5 4 parenrightBig n 4. a n = parenleftBig − 4 3 parenrightBig n − 1 correct nav277 – Homework 10 – Odell – (58340) 2 5. a n = parenleftBig − 5 4 parenrightBig n − 1 6. a n = parenleftBig − 3 4 parenrightBig n − 1 Explanation: By inspection, consecutive terms a n − 1 and a n in the sequence { a n } ∞ n =1 = braceleftBig 1 , − 4 3 , 16 9 , − 64 27 , . . . bracerightBig have the property that a n = ra n − 1 = parenleftBig − 4 3 parenrightBig a n − 1 . Thus a n = ra n − 1 = r 2 a n − 2 = . . . = r n − 1 a 1 = parenleftBig − 4 3 parenrightBig n − 1 a 1 . Consequently, a n = parenleftBig − 4 3 parenrightBig n − 1 since a 1 = 1. keywords: sequence, common ratio 004 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 2 n 5 − 5 n 3 + 5 7 n 4 + 3 n 2 + 2 . 1. limit = 2 7 2. limit = 0 3. limit = − 5 3 4. the sequence diverges correct 5. limit = 5 2 Explanation: After division by n 4 we see that a n = 2 n − 5 n + 5 n 4 7 + 3 n 2 + 2 n 4 . Now 5 n , 5 n 4 , 3 n 2 , 2 n 4 −→ as n → ∞ ; in particular, the denominator converges and has limit 7 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 2 n } diverges....
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This note was uploaded on 04/15/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.
 Spring '10
 ZHENG

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