## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{\theta\to0}\frac{\sin\sqrt2\theta}{\sqrt2\theta}=1$$
$$\lim_{\theta\to0}\frac{\sin\sqrt2\theta}{\sqrt2\theta}$$ 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, $$\lim_{\theta\to0}\frac{\sin\sqrt2\theta}{\sqrt2\theta}=1$$