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 366: 39



Work Step by Step

We have $$ \lim _{x \rightarrow 0}\frac{e^x-1}{\sin x}=\frac{0}{0} $$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 0}\frac{e^x-1}{\sin x}= \lim _{x \rightarrow 0}\frac{e^x}{\cos x}=1 $$
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