Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 746: 89


$\lim\limits_{n\to\infty} (a_nb_n)=0$

Work Step by Step

Need to prove that $\lim\limits_{n\to\infty} (a_nb_n)=0$ When a composition of a function is continuous , then the limit also becomes continuous. Let us consider that $\lim\limits_{n\to\infty} a_n=L$ and the function $f(a_n)$ is continuous at the limit $L$. The property of absolute value: $-|a_n| \leq a_n \leq |a_n|$ for all the values of $n$, thus $\lim\limits_{n\to\infty} |a_n|=0$ According to the limits laws of sequences: $\lim\limits_{n\to\infty} |a_n|=0$; $\lim\limits_{n\to\infty} -|a_n|=-\lim\limits_{n\to\infty} |a_n|=0$ Apply the squeeze theorem for sequence. $\lim\limits_{n\to\infty} a_n=0$. Thus, If $\lim\limits_{n\to\infty} |a_n|=0$, then $\lim\limits_{n\to\infty} a_n=0$ Need to consider a small number $\epsilon \gt 0$ , then $a_n \to 0$ This means that there exists a number $N$ such that the value $|a_n| \lt \dfrac{\epsilon }{M}$ for the all values of $n \geq N$ This gives: $|a_nb_n|=|a_n||b_n| \lt \dfrac{\epsilon }{M} (M)=\epsilon$ So, this is verified that $\lim\limits_{n\to\infty} (a_nb_n)=0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.