Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 367: 46



Work Step by Step

To find the limit $$ \lim _{x \rightarrow \infty}x^{1/x^2} $$ we have to find $$\ln \lim _{x \rightarrow \infty}x^{1/x^2}=\lim _{x \rightarrow \infty}\ln x^{1/x^2}\\ =\lim _{x \rightarrow \infty}\frac{\ln x}{x^2}=\frac{\infty}{\infty}$$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow \infty}\frac{\ln x}{x^2}=\lim _{x \rightarrow \infty}\frac{1}{2x^2}=0. $$ Now, $$\ln \lim _{x \rightarrow \infty}x^{1/x^2}=0\Longrightarrow \lim _{x \rightarrow \infty}x^{1/x^2}=e^0=1. $$
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