Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 48


converges to $1$

Work Step by Step

Given: $a_n=\sqrt[n] n$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\sqrt[n] n$ Let $x=\lim\limits_{n \to \infty }\sqrt[n] n=n^{1/n}$ Take natural logarithmic to both sides. $lnx=\lim\limits_{n \to \infty }n^{1/n}$ $lnx=\lim\limits_{n \to \infty}\frac{lnn}{n}$ Use L-Hospital's Rule. $lnx=\lim\limits_{n \to \infty}\frac{1/n}{1}$ $lnx=0$ Raise the power to base $e$ $e^{lnx}=e^{0}$ $x=1$ Hence, the sequence converges to $1$.
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