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Answer: Limits of sequences Evaluate the limit of the

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

Solution for problem 5RE 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 5RE

Problem 5RE

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-1L’Hopital’s Rule for zero over zero ; Suppose that lim f(x) = 0 ,limg (x) = 0 , and that functions ‘f’ and ‘g’ are differentiable on an xa xa 1open interval ‘I Containing a. Assume also that, g (x) = / 0, in I , if x / a then ; 1 lim f(x) = lim f (x). xa g(x) xa g (x) so long as the limit is finite, +, or -. Similar results hold for x and x.L’Hopital’s Rule for infinity over infinity;suppose that functions ‘f’ and ‘g’ are differentiable for all x larger than some fixed number. If lim f(x) = ,limg (x) = then ; xa xa 1 lim f(x) = lim f (x). xa g(x) xa g (x) so long as the limit is finite, +, or -. Similar results hold for x and x. Step-2 n 1/n Given sequence is ; a n n = n It is an exponential form , here the base and power contains ‘n ‘ . So , in this casestake ‘ln’ on both sides .Then the given sequence becomes; Let , f(x) = n 1/n ln( f(x) ) = ln( n 1/n )\n 1 m = n ln(n) , since ln(a ) = m ln(a). lim 1 ln(n) = lim ln(n. n n n n ln(n)is a rational function , let P(n) = ln(n) , Q(n) = n n As , n tends to infinity , P(n) also tending to infinity , and as , n tends to infinity , Q(n) alsotends to infinity. So , in this cases we can use L’Hopital’s Rule 1 P(x) P (x) That is , xa Q(x) = xa Q (x) . ln(n) d(ln(n)) Therefore , lim = lim d(n) n n n dn 1 = lim n , since d ln(x) = 1/x. n 1 dx = lim n. n = 0 , since as n , then 1 . n 1/n Therefore , lim ln(n ) = 0. n lim n1/n = e 0 = 1, since ln(a ) = b ,that implies a = e b n Therefore, lim a = lim n1/n = 1 n n n

Step 2 of 3

Chapter 8, Problem 5RE is Solved
Step 3 of 3

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

Since the solution to 5RE from 8 chapter was answered, more than 371 students have viewed the full step-by-step answer. The answer to “Limits of sequences Evaluate the limit of the sequence or state that it does not exist.” is broken down into a number of easy to follow steps, and 16 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 5RE from chapter: 8 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This full solution covers the following key subjects: evaluate, exist, Limit, Limits, sequence. This expansive textbook survival guide covers 85 chapters, and 5218 solutions.

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Answer: Limits of sequences Evaluate the limit of the