homework.10 - fh2825 Homework 10 Radin (58415) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 doesnt 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 doesnt 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). 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 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....
View Full Document

This note was uploaded on 04/30/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

Page1 / 10

homework.10 - fh2825 Homework 10 Radin (58415) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online