Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.2 Sequences - 8.2 Exercises - Page 616: 24

Answer

$\lim\limits_{n \to \infty} a_n=0$

Work Step by Step

Definition of a sequence defined by a function: Let us consider a function $f(x)$ whose limit $\lim\limits_{x \to \infty} f(x)$ exists then the sequence $a_n=f(n)$ will converge to the same limit. That is, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$ Here, in this problem we have $a_n=\dfrac{\ln (1/n)}{n}$ and $f(x)=\dfrac{\ln (1/x)}{x}$ Next, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{\ln (1/x)}{x}$ or, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{\ln (1)-\ln x}{x}=\lim\limits_{x \to \infty} \dfrac{-\ln x}{x}$ In order to compute the limit, we will apply L-Hospital's Rule. $\lim\limits_{n \to \infty} a_n=-\lim\limits_{x \to \infty} \dfrac{1/x}{1}$ or, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{1}{x}$ Since, $n \gt 0$ for all the values of $n$ and $\lim\limits_{n \to \infty} e^n=\infty$ and $\lim\limits_{n \to \infty} e^{-n}=0$ Thus. $\lim\limits_{n \to \infty} a_n=0$
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