Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 15



Work Step by Step

Given: $\lim\limits_{t\to0}\dfrac{e^{2t}-1}{\sin t}$ Here, $\lim_{t\to0}(e^{2t}-1)=e^{2\times0}-1=1-1=0$ and $\lim_{t\to0}\sin t=\sin0=0.$ This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{t\to0}\dfrac{\dfrac{d}{dt}(e^{2t}-1)}{\dfrac{d}{dt}(\sin t)}=\lim\limits_{t\to0}\dfrac{2e^{2t}}{\cos t}\\=\dfrac{2e^{2\times0}}{\cos 0}\\=2$
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