Homework 10 - Su, Yung Homework 10 Due: Nov 6 2007, 3:00 am...

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Su, Yung – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Shinko Harper 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 = 7 , determine the value, iF any, oF lim n →∞ a n +8 . 1. limit = 7 correct 2. limit doesn’t exist 3. limit = - 1 4. limit = 15 5. limit = 7 8 Explanation: To say that lim n →∞ a n = 7 means that a n gets as close as we please to 7 For all su±ciently large n . But then a n +8 gets as close as we please to 7 For all su±ciently large n . Consequently, lim n →∞ a n +8 = 7 . 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 4 2 n + 3 , and iF it does, fnd its limit. 1. limit = - ln2 2. limit = ln 4 5 3. the sequence diverges 4. limit = 0 correct 5. limit = ln2 Explanation: AFter division by n we see that 4 2 n + 3 = 4 n 2 + 3 n , so by properties oF logs, a n = 1 n ln 4 n - 1 n ln µ 2 + 3 n . But by known limits (or use L’Hospital), 1 n ln 4 n , 1 n ln µ 2 + 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 , - 5 3 , 25 9 , - 125 27 , ... o assuming that the pattern oF the frst Few terms continues. 1. a n = - 3 2 · n 2. a n = - 5 3 · n
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2 3. a n = - 3 2 · n - 1 4. a n = - 5 3 · n - 1 correct 5. a n = - 3 5 · n - 1 6. a n = - 3 5 · n Explanation: In the sequence n 1 , - 5 3 , 25 9 , - 125 27 , ... o each term is - 5 3 times the preceeding one, i.e. , a n = - 5 3 · a n - 1 . Consequently, a n = - 5 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 = 6 n 5 - 3 n 3 + 5 5 n 4 + 4 n 2 + 4 . 1. the sequence diverges correct 2. limit = 6 5 3. limit = 5 4 4. limit = - 3 4 5. limit = 0 Explanation: After division by n 4 we see that a n = 6 n - 3 n + 5 n 4 5 + 4 n 2 + 4 n 4 . Now 3 n , 5 n 4 , 4 n 2 , 4 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 = 7 n 2 7 n + 2 - n 2 + 3 n + 1 , and if it does, Fnd its limit 1. the sequence diverges 2. limit = 5 14 3. limit = 0 4. limit = 5 21 5. limit = 5 7 correct Explanation: After bringing the two terms to a common denominator we see that a n = 7 n 3 + 7 n 2 - (7 n
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Homework 10 - Su, Yung Homework 10 Due: Nov 6 2007, 3:00 am...

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