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Limits of sequences Evaluate the limit of the | Ch 8 - 8RE

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 8RE Chapter 8

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 8RE

Problem 8RE

Limits of sequences Evaluate the limit of the sequence or state that it does not exist.

Step-by-Step Solution:
Step 1 of 3

STEP _ BY_ STEP SOLUTION Step-1 n Given sequence is ; a = snn( 6 ) In the given sequence ‘n’ has no fixed value , In this cases we can use mathematicalinduction method . Because sin( n ) is depends upon n. 6We know that , sin(x) lies between -1 to 1. That is , sin function is oscillating between -1 to 1.Now , we have to check the values of a n Let , a = sin( n ) n 6 If n =1 , then the value of sin( (1)) = sin( ) = 1 , since sin( ) = 1 . 6 6 2 6 2 If n =2 , then the value of sin( (2)) = sin( ) = 3 , since sin( ) = 3 . 6 3 2 3 2 (3) If n =3 , then the value of sin( 6 ) = sin( )2= 1 , since sin( ) = 1.2 (4) 2 3 If n =4 , then the value of sin( 6 ) = sin( 3 ) = sin ( ) =3sin( ) = 3 2 , 3since sin( ) 3 2 . (5) 1 1 If n =5 , then the value of sin( 6 ) = sin( 6) = sin( )6= 2 , since sin( )6= 2. If n =6 , then the value of sin( (6)) = sin( ) = 0 , since sin( ) = 0. 6 (7) If n =7 , then the value of sin( ) = sin( + ) = -sin( )= - 1 , 6 6 6 2 since sin( + x ) = sin(x) and sin( ) = 1. 6 2 (8) 4 3 If n =8 , then the value of sin( 6 ) = sin( 3 ) = sin( + 3 ) = -sin( )3 - 2 , 3 since sin( + x ) = sin(x) and sin( ) = 3 2 . If n =9 , then the value of sin( (9)) = sin( 3 ) = sin( + ) = -sin( )= - 1 , 6 2 2 2\n since sin( + x ) = sin(x) and sin( ) = 1. If n =10 , then the value of sin( (10)) = sin( 5 ) = sin(2 ) = - sin( )= - 3 , 6 3 3 3 2 since sin(2 x ) = sin(x) and sin( ) = 3 . 3 2 Clearly observe these values , the sequence oscillate among the values+ 1 + 3 + 2 , 2 , 1 , 0. Therefore , a isnoscillate between -1 to1 . Therefore, lim a = lnm sin( n ) has no limit . n n 6

Step 2 of 3

Chapter 8, Problem 8RE is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Limits of sequences Evaluate the limit of the | Ch 8 - 8RE