HW10 - momin (rrm497) Homework 10 cheng (58520) 1 This...

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Unformatted text preview: momin (rrm497) Homework 10 cheng (58520) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Compute the value of lim n 5 a n b n 3 a n b n when lim n a n = 6 , lim n b n = 2 . 1. limit = 31 10 2. limit = 3 3. limit = 3 correct 4. limit doesnt exist 5. limit = 31 10 Explanation: By properties of limits lim n 2 5 a n b n = 5 lim n a n lim n b n = 60 while lim n (3 a n b n ) = 3 lim n a n lim n b n = 20 negationslash = 0 . Thus, by properties of limits again, lim n 5 a n b n 3 a n b n = 3 . 002 10.0 points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 2 5 n + 2 parenrightbigg , and if it does, find its limit. 1. limit = ln5 2. limit = ln 2 7 3. limit = ln 2 5 4. limit = 0 correct 5. the sequence diverges Explanation: After division by n we see that 2 5 n + 2 = 2 n 5 + 2 n , so by properties of logs, a n = 1 n ln 2 n 1 n ln parenleftbigg 5 + 2 n parenrightbigg . But by known limits (or use LHospital), 1 n ln 2 n , 1 n ln parenleftbigg 5 + 2 n parenrightbigg as n . Consequently, the sequence { a n } converges and has limit = 0 . 003 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 5 n 5 4 n 3 + 2 8 n 4 + n 2 + 1 . 1. limit = 4 2. limit = 2 3. the sequence diverges correct momin (rrm497) Homework 10 cheng (58520) 2 4. limit = 5 8 5. limit = 0 Explanation: After division by n 4 we see that a n = 5 n 4 n + 2 n 4 8 + 1 n 2 + 1 n 4 . Now 4 n , 2 n 4 , 1 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 { 5 n } diverges. 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 5 n 2 5 n + 3 n 2 + 2 n + 1 , and if it does, find its limit 1. limit = 2 15 2. the sequence diverges 3. limit = 1 5 4. limit = 2 5 correct 5. limit = 0 Explanation: After bringing the two terms to a common denominator we see that a n = 5 n 3 + 5 n 2 (5 n + 3) ( n 2 + 2 ) (5 n + 3) ( n + 1) = 2 n 2 10 n 6 5 n 2 + 8 n + 3 ....
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HW10 - momin (rrm497) Homework 10 cheng (58520) 1 This...

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