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2n+1 In Section 8, Example 2, if you follow the same procedure for the result is Зп-5 N = + ??? 9e 3 What integer should replace ????

2n+1 In Section 8, Example 2, if you follow the same procedure for the result is Зп-5 N = + ??? 9e 3 What integer should replace ????

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.1: Infinite Sequences And Summation Notation
Problem 72E
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What number is ???

2n+1 In Section 8, Example 2, if you follow the same procedure for the result is 3n-5' ??? N 9€ What integer should replace ????
Transcribed Image Text:2n+1 In Section 8, Example 2, if you follow the same procedure for the result is 3n-5' ??? N 9€ What integer should replace ????
Prove lim n+ = . Example 2 Зп+1 7n-4 40 2. Sequences Discussion. For each e > 0, we need to to decide how big n must be to guarantee n - 2< e. Thus we want 7n-4 21n + 7 – 21n + 12 7(4n – 4) 19 < E. 7(7n – 4) or Since 7n – 4 > 0, we can drop the absolute value and manipulate the inequality further to "solve" for n: 19 < 7n – 4 7€ 19 + 4 < 7n 7€ 4 < n. 19 or or 49€ Our steps are reversible, so we will put N = could have chosen N to be any number larger than 19 49€ +. Incidentally, we 19 49€ Formal Proof 19 Let e > 0 and let N +. Then n > N implies n > 49€ 19 49€ 19 7e 19 hence 7e ) 19 hence 7n + 4, hence 7n – 4 > < €, and hence | 7(7n-4) Зп+1 7n-4 - < €. This proves lim Зп+1 7n-4
Transcribed Image Text:Prove lim n+ = . Example 2 Зп+1 7n-4 40 2. Sequences Discussion. For each e > 0, we need to to decide how big n must be to guarantee n - 2< e. Thus we want 7n-4 21n + 7 – 21n + 12 7(4n – 4) 19 < E. 7(7n – 4) or Since 7n – 4 > 0, we can drop the absolute value and manipulate the inequality further to "solve" for n: 19 < 7n – 4 7€ 19 + 4 < 7n 7€ 4 < n. 19 or or 49€ Our steps are reversible, so we will put N = could have chosen N to be any number larger than 19 49€ +. Incidentally, we 19 49€ Formal Proof 19 Let e > 0 and let N +. Then n > N implies n > 49€ 19 49€ 19 7e 19 hence 7e ) 19 hence 7n + 4, hence 7n – 4 > < €, and hence | 7(7n-4) Зп+1 7n-4 - < €. This proves lim Зп+1 7n-4
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