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HW 10 - Kim Jin Homework 10 Due Nov 6 2007 3:00 am Inst...

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Kim, Jin – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Diane Radin 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points If lim n → ∞ a n = 9 , determine the value, if any, of lim n → ∞ a n - 2 . 1. limit = 7 2. limit = 11 3. limit = 9 2 4. limit doesn’t exist 5. limit = 9 correct Explanation: To say that lim n → ∞ a n = 9 means that a n gets as close as we please to 9 for all sufficiently large n . But then a n - 2 gets as close as we please to 9 for all sufficiently large n . Consequently, lim n → ∞ a n - 2 = 9 . keywords: sequence, limit, properties of limits 002 (part 1 of 1) 10 points Determine if the sequence { a n } converges when a n = 1 n ln 2 6 n + 5 , and if it does, find its limit. 1. limit = 0 correct 2. limit = ln 2 11 3. limit = - ln 6 4. the sequence diverges 5. limit = ln 1 3 Explanation: After division by n we see that 2 6 n + 5 = 2 n 6 + 5 n , so by properties of logs, a n = 1 n ln 2 n - 1 n ln 6 + 5 n . But by known limits (or use L’Hospital), 1 n ln 2 n , 1 n ln 6 + 5 n -→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . keywords: limit, sequence, log function, 003 (part 1 of 1) 10 points Find a formula for the general term a n of the sequence n 1 , - 3 5 , 9 25 , - 27 125 , . . . o assuming that the pattern of the first few terms continues. 1. a n = - 5 3 · n 2. a n = - 5 3 · n - 1

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Kim, Jin – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Diane Radin 2 3. a n = - 3 5 · n - 1 correct 4. a n = - 2 3 · n 5. a n = - 2 3 · n - 1 6. a n = - 3 5 · n Explanation: In the sequence n 1 , - 3 5 , 9 25 , - 27 125 , . . . o each term is - 3 5 times the preceeding one, i.e. , a n = - 3 5 · a n - 1 . Consequently, a n = - 3 5 · n - 1 since a 1 = 1. keywords: sequence, exponential 004 (part 1 of 1) 10 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 6 n 5 - n 3 + 4 5 n 4 + 2 n 2 + 2 . 1. limit = - 1 2 2. the sequence diverges correct 3. limit = 0 4. limit = 6 5 5. limit = 2 Explanation: After division by n 4 we see that a n = 6 n - 1 n + 4 n 4 5 + 2 n 2 + 2 n 4 . Now 1 n , 4 n 4 , 2 n 2 , 2 n 4 -→ 0 as n → ∞ ; in particular, the denominator converges and has limit 5 6 = 0. Thus by properties of limits { a n } diverges since the sequence { 6 n } diverges. keywords: 005 (part 1 of 1) 10 points Determine whether the sequence { a n } con- verges or diverges when a n = 20 n 2 5 n + 2 - 4 n 2 + 8 n + 1 , and if it does, find its limit 1. limit = 0 2. limit = 12 5 correct 3. limit = 4 5 4. limit = 6 5 5. the sequence diverges Explanation: After bringing the two terms to a common denominator we see that a n = 20 n 3 + 20 n 2 - (5 n + 2) ( 4 n 2 + 8 ) (5 n + 2) ( n + 1) = 12 n 2 - 40 n - 16 5 n 2 + 7 n + 2 .
Kim, Jin – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Diane Radin 3 Thus a n = 12 - 40 n - 16 n 2 5 + 7 n + 2 n 2 .

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