University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Practice Exercises - Page 552: 10

Answer

Converges to $0$.

Work Step by Step

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=\dfrac{\ln (2n^3+1)}{n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[\dfrac{\ln (2n^3+1)}{n}]$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{\infty}{\infty}$ Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, so take the help of L-Hospital's rule. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{\frac{6n^2}{(2n^3+1)}}{1}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{6n^2}{(2n^3+1)}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{6}{(2n^3/n^2+1/n^2)}=\frac{6}{\infty}$ $\lim\limits_{n \to \infty}a_n=0$ Therefore, the sequence converges to $0$.
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