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: 41


The limit does not exist.

Work Step by Step

We have $$ \lim _{x \rightarrow 0}\frac{e^{2x}-1-x}{ x^2}=\frac{0}{0} $$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 0}\frac{e^{2x}-1-x}{ x^2}=\lim _{x \rightarrow 0}\frac{2e^{2x}-1}{ 2x}$$ by checking the one-sidded limits we have $$ \lim _{x \rightarrow 0^+}\frac{2e^{2x}-1}{ 2x}=\infty \neq \lim _{x \rightarrow 0^-}\frac{2e^{2x}-1}{ 2x}=-\infty $$ hence the limit does not exist.
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