Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Review - Exercises - Page 359: 7



Work Step by Step

$\lim\limits_{x \to 0}$$\frac{e^x-1}{tanx}$ After plugging in 0 for x in the numerator and denominator, we get $\frac{0}{0}$, an indeterminate form. This is one of the forms where we can use L'Hopital's Rule. We take the derivative of the top and the bottom and get: $\lim\limits_{x \to 0}$$\frac{e^x}{(sec^2x)}$ We then plug in 0 again, and we get: $\frac{e^0}{(sec0)^2}$ = $\frac{1}{1}$ = 1
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