Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 1

Answer

$\frac{1}{2}$

Work Step by Step

A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant. $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{2+n^{3}}{1+2n^{3}}$ Divide numerator and denominator by $n^{3}$. $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{\frac{2+n^{3}}{n^{3}}}{\frac{1+2n^{3}}{n^{3}}}$ $=\lim\limits_{n \to \infty}\frac{\frac{2}{n^{3}}+1}{\frac{1}{n^{3}}+2}$ $=\frac{0+1}{0+2}$ $=\frac{1}{2}$ Hence, the given sequence converges to $\frac{1}{2}$.
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