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: 47


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Work Step by Step

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