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: 45



Work Step by Step

\begin{align*} \lim _{h\rightarrow 0} \frac{\sin 2h (1-\cos h)}{h^2}&=\lim _{h\rightarrow 0}\frac{\sin 2h }{h} \frac{ 1-\cos h}{h}\\ &=\lim _{h\rightarrow 0}2\frac{\sin 2h }{2h} \frac{ 1-\cos h}{h} \\ &=2 \lim _{2h\rightarrow 0}\frac{\sin 2h }{2h}\lim _{h\rightarrow 0}\frac{ 1-\cos h}{h} \\ &=0. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$ and $\lim _{x\rightarrow 0}\frac{ 1-\cos x}{x}=0. $
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