Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 49



Work Step by Step

Since $|x|=-x $ when $ x<0$, then we have \begin{align*} \lim _{x \rightarrow 0^-} \frac{\sin x}{|x|}&= \lim _{x \rightarrow 0} \frac{\sin x}{-x}\\ &= (-1) \lim _{x \rightarrow 0} \frac{\sin x}{x}\\ &=-1. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$ .
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