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(cos n) Find the limit of the sequence an 47

(cos n) Find the limit of the sequence an 47

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:** Find the limit of the sequence \( a_n = \frac{\cos n}{4^n} \). **Solution:** To determine the limit of the sequence as \( n \) approaches infinity, observe the following: 1. The function \(\cos n\) oscillates between -1 and 1 for all integer values of \(n\). 2. The denominator \(4^n\) grows exponentially as \(n\) increases. 3. Since the numerator \(\cos n\) remains bounded while the denominator increases without bound, the terms of the sequence \(a_n\) approach zero. Therefore, the limit of the sequence \( a_n = \frac{\cos n}{4^n} \) as \( n \rightarrow \infty \) is 0.
Transcribed Image Text:**Problem Statement:** Find the limit of the sequence \( a_n = \frac{\cos n}{4^n} \). **Solution:** To determine the limit of the sequence as \( n \) approaches infinity, observe the following: 1. The function \(\cos n\) oscillates between -1 and 1 for all integer values of \(n\). 2. The denominator \(4^n\) grows exponentially as \(n\) increases. 3. Since the numerator \(\cos n\) remains bounded while the denominator increases without bound, the terms of the sequence \(a_n\) approach zero. Therefore, the limit of the sequence \( a_n = \frac{\cos n}{4^n} \) as \( n \rightarrow \infty \) is 0.
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