University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises - Page 111: 24


$$\lim_{x\to0}\frac{\cos2x-1}{\sin x}=0$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\cos2x-1}{\sin x}$$ Recall the identity: $\cos2x=1-2\sin^2x$ $$A=\lim_{x\to0}\frac{1-2\sin^2x-1}{\sin x}=\lim_{x\to0}\frac{-2\sin^2x}{\sin x}=\lim_{x\to0}(-2\sin x)$$ $$A=-2\sin0=-2\times0=0$$
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