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homework.10

# homework.10 - Changed with the DEMO VERSION of CAD-KAS...

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fh2825 – Homework 10 – Radin – (58415) 1 This print-out should have 23 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 →∞ 6 a n b n 5 a n 6 b n when lim n →∞ a n = 6 , lim n →∞ b n = 2 . 1. limit = 12 7 2. limit = 12 7 3. limit = 37 21 4. limit = 37 21 5. limit doesn’t exist Explanation: By properties of limits lim n 2 6 a n b n = 6 lim n →∞ a n lim n →∞ b n = 72 while lim n →∞ (5 a n 6 b n ) = 5 lim n →∞ a n 6 lim n →∞ b n = 42 negationslash = 0 . Thus, by properties of limits again, lim n →∞ 6 a n b n 5 a n 6 b n = 12 7 . 002 10.0 points If lim n → ∞ a n = 6 , determine the value, if any, of lim n → ∞ a n 8 . 1. limit doesn’t exist 2. limit = 6 3. limit = 2 4. limit = 14 5. limit = 3 4 Explanation: To say that lim n → ∞ a n = 6 means that a n gets as close as we please to 6 for all sufficiently large n . But then a n 8 gets as close as we please to 6 for all sufficiently large n . Consequently, lim n → ∞ a n 8 = 6 . 003 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 5 4 , 25 16 , 125 64 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 4 5 parenrightBig n 2. a n = parenleftBig 6 5 parenrightBig n 1 3. a n = parenleftBig 5 4 parenrightBig n 4. a n = parenleftBig 6 5 parenrightBig n Changed with the DEMO VERSION of CAD-KAS PDF-Editor (http://www.cadkas.com). Changed with the DEMO VERSION of CAD-KAS PDF-Editor (http://www.cadkas.com). Changed with the DEMO VERSION of CAD-KAS PDF-Editor (http://www.cadkas.com).

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fh2825 – Homework 10 – Radin – (58415) 2 5. a n = parenleftBig 5 4 parenrightBig n 1 6. a n = parenleftBig 4 5 parenrightBig n 1 Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 5 4 , 25 16 , 125 64 , . . . bracerightBig have the property that a n = ra n 1 = parenleftBig 5 4 parenrightBig a n 1 . Thus a n = ra n 1 = r 2 a n 2 = . . . = r n 1 a 1 = parenleftBig 5 4 parenrightBig n 1 a 1 . Consequently, a n = parenleftBig 5 4 parenrightBig n 1 since a 1 = 1. keywords: sequence, common ratio 004 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 5 n 5 2 n 3 + 4 5 n 4 + 3 n 2 + 5 . 1. limit = 2 3 2. limit = 1 3. the sequence diverges 4. limit = 0 5. limit = 4 5 Explanation: After division by n 4 we see that a n = 5 n 2 n + 4 n 4 5 + 3 n 2 + 5 n 4 . Now 2 n , 4 n 4 , 3 n 2 , 5 n 4 −→ 0 as n → ∞ ; in particular, the denominator converges and has limit 5 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 5 n } diverges. 005 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 8 n 2 8 n + 5 n 2 + 6 n + 1 , and if it does, find its limit 1. limit = 3 16 2. limit = 0 3. the sequence diverges 4. limit = 3 8 5. limit = 1 8 Explanation: After bringing the two terms to a common denominator we see that a n = 8 n 3 + 8 n 2 (8 n + 5) ( n 2 + 6 ) (8 n + 5) ( n + 1) = 3 n 2 48 n 30
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homework.10 - Changed with the DEMO VERSION of CAD-KAS...

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