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Unformatted text preview: fh2825 Homework 10 Radin (58415) 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 6 a n b n 5 a n 6 b n when lim n a n = 6 , lim n b n = 2 . 1. limit = 12 7 2. limit = 12 7 3. limit = 37 21 4. limit = 37 21 5. limit doesnt exist Explanation: By properties of limits lim n 2 6 a n b n = 6 lim n a n lim n b n = 72 while lim n (5 a n 6 b n ) = 5 lim n a n 6 lim n b n = 42 negationslash = 0 . Thus, by properties of limits again, lim n 6 a n b n 5 a n 6 b n = 12 7 . 002 10.0 points If lim n a n = 6 , determine the value, if any, of lim n a n 8 . 1. limit doesnt exist 2. limit = 6 3. limit = 2 4. limit = 14 5. limit = 3 4 Explanation: To say that lim n a n = 6 means that a n gets as close as we please to 6 for all sufficiently large n . But then a n 8 gets as close as we please to 6 for all sufficiently large n . Consequently, lim n a n 8 = 6 . 003 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 5 4 , 25 16 , 125 64 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 4 5 parenrightBig n 2. a n = parenleftBig 6 5 parenrightBig n 1 3. a n = parenleftBig 5 4 parenrightBig n 4. a n = parenleftBig 6 5 parenrightBig n Changed with the DEMO VERSION of CADKAS PDFEditor (http://www.cadkas.com). Changed with the DEMO VERSION of CADKAS PDFEditor (http://www.cadkas.com). fh2825 Homework 10 Radin (58415) 2 5. a n = parenleftBig 5 4 parenrightBig n 1 6. a n = parenleftBig 4 5 parenrightBig n 1 Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 5 4 , 25 16 , 125 64 , . . . bracerightBig have the property that a n = ra n 1 = parenleftBig 5 4 parenrightBig a n 1 . Thus a n = ra n 1 = r 2 a n 2 = . . . = r n 1 a 1 = parenleftBig 5 4 parenrightBig n 1 a 1 . Consequently, a n = parenleftBig 5 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 = 5 n 5 2 n 3 + 4 5 n 4 + 3 n 2 + 5 . 1. limit = 2 3 2. limit = 1 3. the sequence diverges 4. limit = 0 5. limit = 4 5 Explanation: After division by n 4 we see that a n = 5 n 2 n + 4 n 4 5 + 3 n 2 + 5 n 4 . Now 2 n , 4 n 4 , 3 n 2 , 5 n 4 as n ; in particular, the denominator converges and has limit 5 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 5 n } diverges....
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This note was uploaded on 04/30/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

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