HWK10 - Valenica, Daniel Homework 10 Due: Nov 6 2007, 3:00...

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Valenica, Daniel – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Cheng 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points IF lim n →∞ a n = 2 , determine the value, iF any, oF lim n →∞ a n +5 . 1. limit = - 3 2. limit = 2 correct 3. limit = 2 5 4. limit doesn’t exist 5. limit = 7 Explanation: To say that lim n →∞ a n = 2 means that a n gets as close as we please to 2 For all su±ciently large n . But then a n +5 gets as close as we please to 2 For all su±ciently large n . Consequently, lim n →∞ a n +5 = 2 . 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 µ 3 6 n + 3 , and iF it does, fnd its limit. 1. the sequence diverges 2. limit = ln 1 3 3. limit = - ln6 4. limit = 0 correct 5. limit = ln 1 2 Explanation: AFter division by n we see that 3 6 n + 3 = 3 n 6 + 3 n , so by properties oF logs, a n = 1 n ln 3 n - 1 n ln µ 6 + 3 n . But by known limits (or use L’Hospital), 1 n ln 3 n , 1 n ln µ 6 + 3 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 ²ind a Formula For the general term a n oF the sequence n 1 , - 4 3 , 16 9 , - 64 27 , ... o assuming that the pattern oF the frst Few terms continues. 1. a n = - 4 3 · n - 1 correct 2. a n = - 5 4 · n - 1
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2 3. a n = - 5 4 · n 4. a n = - 3 4 · n - 1 5. a n = - 3 4 · n 6. a n = - 4 3 · n Explanation: In the sequence n 1 , - 4 3 , 16 9 , - 64 27 , ... o each term is - 4 3 times the preceeding one, i.e. , a n = - 4 3 · a n - 1 . Consequently, a n = - 4 3 · 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, Fnd its limit when a n = 8 n 5 - 2 n 3 + 1 2 n 4 + 5 n 2 + 1 . 1. the sequence diverges correct 2. limit = - 2 5 3. limit = 4 4. limit = 0 5. limit = 1 Explanation: After division by n 4 we see that a n = 8 n - 2 n + 1 n 4 2 + 5 n 2 + 1 n 4 . Now 2 n , 1 n 4 , 5 n 2 , 1 n 4 -→ 0 as n → ∞ ; in particular, the denominator converges and has limit 2 6 = 0. Thus by properties of limits { a n } diverges since the sequence { 8 n } diverges. keywords: 005 (part 1 of 1) 10 points Determine whether the sequence { a n } con- verges or diverges when a n = 16 n 2 8 n + 1 - 2 n 2 + 2 n + 1 , and if it does, Fnd its limit 1. limit = 0 2. limit = 7 8 3. the sequence diverges 4. limit = 7 12 5. limit = 7 4 correct Explanation: After bringing the two terms to a common denominator we see that a n = 16 n 3 + 16 n 2 - (8 n
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HWK10 - Valenica, Daniel Homework 10 Due: Nov 6 2007, 3:00...

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