University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 58



Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\ln f(x)=\dfrac{\ln (e^x+x)}{x} $ Now, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $e^{\lim\limits_{x \to 0} \frac{e^x+1/e^x+x}{1}}=e^{\lim\limits_{x \to 0} \frac{e^0+1}{e^0+0}}$ Thus, $e^{\lim\limits_{x \to 0} \frac{e^0+1}{e^0+0}}=e^{2}$
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