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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 doesnt 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 doesnt 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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 Spring '10
 ZHENG

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