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 78: 57


See the proof below.

Work Step by Step

Since $ m\neq 0$, then we have \begin{align*} \lim _{x \rightarrow 0}\frac{\cos mx-1}{x^2}&= \lim _{x \rightarrow 0}m^2 \frac{\cos m x-1}{(mx)^2}\\ &= - m^2\lim _{mx \rightarrow 0} \frac{1-\cos mx}{(mx)^2}\\ &=- \frac{m^2}{2}.\\ \end{align*} Where we used the fact that $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}$.
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