University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 85: 21



Work Step by Step

$$\lim_{\theta\to0}\frac{\sin\sqrt2\theta}{\sqrt2\theta}$$ Take $u=\sqrt2\theta$. Then as $\theta\to0$, $u\to(\sqrt2\times0)=0$ Thus, $$\lim_{u\to0}\frac{\sin u}{u}$$ Apply Theorem 7, we have: $$\lim_{u\to0}\frac{\sin u}{u}=1$$ Therefore, $$\lim_{\theta\to0}\frac{\sin\sqrt2\theta}{\sqrt2\theta}=1$$
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