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# solution_pdf10 - Horstman(mdh995 HW10 Radin(58505 This...

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Horstman (mdh995) – HW10 – Radin – (58505) 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 →∞ 2 a n b n 6 a n b n when lim n →∞ a n = 6 , lim n →∞ b n = 2 . 1. limit doesn’t exist 2. limit = 13 19 3. limit = 12 19 correct 4. limit = 12 19 5. limit = 13 19 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 →∞ (6 a n b n ) = 6 lim n →∞ a n lim n →∞ b n = 38 negationslash = 0 . Thus, by properties of limits again, lim n →∞ 2 a n b n 6 a n b n = 12 19 . 002 10.0 points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 4 6 n + 4 parenrightbigg , and if it does, find its limit. 1. the sequence diverges 2. limit = ln 2 5 3. limit = ln 2 3 4. limit = ln 6 5. limit = 0 correct Explanation: After division by n we see that 4 6 n + 4 = 4 n 6 + 4 n , so by properties of logs, a n = 1 n ln 4 n 1 n ln parenleftbigg 6 + 4 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 4 n , 1 n ln parenleftbigg 6 + 4 n parenrightbigg −→ 0 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 = 3 n 5 n 3 + 4 7 n 4 + 4 n 2 + 2 . 1. the sequence diverges correct 2. limit = 1 4

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Horstman (mdh995) – HW10 – Radin – (58505) 2 3. limit = 0 4. limit = 2 5. limit = 3 7 Explanation: After division by n 4 we see that a n = 3 n 1 n + 4 n 4 7 + 4 n 2 + 2 n 4 . Now 1 n , 4 n 4 , 4 n 2 , 2 n 4 −→ 0 as n → ∞ ; in particular, the denominator converges and has limit 7 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 3 n } diverges. 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 8 n 2 8 n + 4 n 2 + 2 n + 1 , and if it does, find its limit 1. the sequence diverges 2. limit = 1 2 correct 3. limit = 1 4 4. limit = 1 6 5. limit = 0 Explanation: After bringing the two terms to a common denominator we see that a n = 8 n 3 + 8 n 2 (8 n + 4) ( n 2 + 2 ) (8 n + 4) ( n + 1) = 4 n 2 16 n 8 8 n 2 + 12 n + 4 .
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