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: 46



Work Step by Step

Using the fact that $$\sin^2x=\frac{1}{2}(1-\cos 2x)$$ we have \begin{align*} \lim _{t\rightarrow 0} \frac{ 1-\cos 2t}{\sin^23t}&=\lim _{t\rightarrow 0} \frac{2\sin^2t}{\sin^23t}\\ &=\lim _{t\rightarrow 0} \frac{2}{9}\frac{\sin^2t}{t^2} \frac{ 9t^2}{\sin^23t}\\ &=\frac{2}{9}\lim _{t\rightarrow 0} \frac{\sin^2t}{t^2} \lim _{3t\rightarrow 0}\frac{ (3t)^2}{\sin^23t}\\ &=\frac{2}{9}. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$.
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