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Unformatted text preview: tgo72 Homework 10 Gompf (58370) 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 3 a n 6 b n when lim n a n = 6 , lim n b n = 2 . 1. limit = 5 3 2. limit = 8 5 correct 3. limit = 8 5 4. limit doesnt exist 5. limit = 5 3 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 (3 a n 6 b n ) = 3 lim n a n 6 lim n b n = 30 negationslash = 0 . Thus, by properties of limits again, lim n 4 a n b n 3 a n 6 b n = 8 5 . 002 10.0 points If lim n a n = 3 , determine the value, if any, of lim n a n +2 . 1. limit = 5 2. limit = 1 3. limit = 3 correct 4. limit doesnt exist 5. limit = 3 2 Explanation: To say that lim n a n = 3 means that a n gets as close as we please to 3 for all sufficiently large n . But then a n +2 gets as close as we please to 3 for all sufficiently large n . Consequently, lim n a n +2 = 3 . 003 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 3 4 , 9 16 , 27 64 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 4 3 parenrightBig n 1 2. a n = parenleftBig 4 5 parenrightBig n 3. a n = parenleftBig 4 3 parenrightBig n 4. a n = parenleftBig 4 5 parenrightBig n 1 tgo72 Homework 10 Gompf (58370) 2 5. a n = parenleftBig 3 4 parenrightBig n 1 correct 6. a n = parenleftBig 3 4 parenrightBig n Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 3 4 , 9 16 , 27 64 , . . . bracerightBig have the property that a n = ra n 1 = parenleftBig 3 4 parenrightBig a n 1 . Thus a n = ra n 1 = r 2 a n 2 = . . . = r n 1 a 1 = parenleftBig 3 4 parenrightBig n 1 a 1 . Consequently, a n = parenleftBig 3 4 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 4 n 3 + 4 8 n 4 + 3 n 2 + 1 . 1. limit = 4 2. limit = 0 3. the sequence diverges correct 4. limit = 1 4 5. limit = 4 3 Explanation: After division by n 4 we see that a n = 2 n 4 n + 4 n 4 8 + 3 n 2 + 1 n 4 . Now 4 n , 4 n 4 , 3 n 2 , 1 n 4 as n ; in particular, the denominator converges and has limit 8 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 2 n } diverges....
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 Spring '08
 RAdin
 Calculus

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