University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 489: 107

Answer

Proof given below.

Work Step by Step

Let $L=\displaystyle \lim_{n\rightarrow\infty}(n)^{1/n}$. $ L=\displaystyle \lim_{n\rightarrow\infty}(n)^{1/n}\qquad$ ...apply ln(..) to both sides $\displaystyle \ln L=\ln[\lim_{n\rightarrow\infty}(n)^{1/n}]\qquad$ ... ln is continuous, $\displaystyle \ln L=\lim_{n\rightarrow\infty}[\ln(n)^{1/n}]$ $\displaystyle \ln L=\lim_{n\rightarrow\infty}[\frac{1}{n}\ln n]$ $\displaystyle \ln L=\lim_{n\rightarrow\infty}\frac{\ln n}{n}\qquad$ ... $(\displaystyle \frac{\infty}{\infty}$, apply L'Hospital's rule) $\displaystyle \ln L=\lim_{n\rightarrow\infty}\frac{1/n}{1}$ $\displaystyle \ln L=\lim_{n\rightarrow\infty}\frac{1}{n}$ $\ln L=0$ $L=1$
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.