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Unformatted text preview: lawrence (cdl678) – Homework 2 – Odell – (56280) 1 This printout should have 37 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 →∞ 2 a n b n 3 a n − 2 b n when lim n →∞ a n = 6 , lim n →∞ b n = − 2 . 1. limit = 12 11 2. limit = 13 11 3. limit = − 12 11 correct 4. limit = − 13 11 5. limit doesn’t exist Explanation: By properties of limits lim n → 2 2 a n b n = 2 lim n →∞ a n lim n →∞ b n = − 24 while lim n →∞ (3 a n − 2 b n ) = 3 lim n →∞ a n − 2 lim n →∞ b n = 22 negationslash = 0 . Thus, by properties of limits again, lim n →∞ 2 a n b n 3 a n − 2 b n = − 12 11 . 002 10.0 points If lim n →∞ a n = 7 , determine the value, if any, of lim n →∞ a n − 9 . 1. limit doesn’t exist 2. limit = 16 3. limit = − 2 4. limit = 7 9 5. limit = 7 correct Explanation: To say that lim n →∞ a n = 7 means that a n gets as close as we please to 7 for all sufficiently large n . But then a n − 9 gets as close as we please to 7 for all sufficiently large n . Consequently, lim n →∞ a n − 9 = 7 . 003 10.0 points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 6 5 n + 4 parenrightbigg , and if it does, find its limit. 1. limit = ln 6 5 2. limit = 0 correct 3. limit = − ln5 4. limit = ln 2 3 5. the sequence diverges Explanation: lawrence (cdl678) – Homework 2 – Odell – (56280) 2 After division by n we see that 6 5 n + 4 = 6 n 5 + 4 n , so by properties of logs, a n = 1 n ln 6 n − 1 n ln parenleftbigg 5 + 4 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 6 n , 1 n ln parenleftbigg 5 + 4 n parenrightbigg −→ as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 004 10.0 points Determine if the sequence { a n } n converges when a n = 7 n 3 n − 1 and if it does, find its limit when 1. converges with limit = 3 7 2. converges with limit = 7 2 3. converges with limit = 0 4. converges with limit = 7 3 correct 5. the sequence diverges Explanation: After division by n we see that a n = 7 3 − 1 n −→ 7 3 as n → ∞ . Thus { a n } n converges and has limit = 7 3 . 005 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 7 n + ( − 1) n 4 n + 3 . 1. converges with limit = 2 2. converges with limit = 1 3. sequence does not converge 4. converges with limit = 7 4 correct 5. converges with limit = 3 2 Explanation: After division by n we see that a n = 7 + ( − 1) n n 4 + 3 n . But ( − 1) n n , 3 n −→ as n → ∞ , so a n → 7 4 as n → ∞ . Conse quently, the sequence converges and has limit = 7 4 ....
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This note was uploaded on 04/09/2010 for the course M 56280 taught by Professor Odell during the Spring '10 term at University of Texas.
 Spring '10
 odell

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