Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 54



Work Step by Step

Here, we have $\ln f(x)=\frac{\ln (\ln x)}{x-e}$ and $f(x)=e^{\frac{\ln (\ln x)}{x-e}}$ But $e^{\lim\limits_{x \to e^{+}} ( \frac{\ln(\ln x)}{x})}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $e^{\lim\limits_{x \to e^{+}} ( \frac{1/x ln x}{1})}=e^{ \frac{1/e ln e}{1})}=e^{1/e}$
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