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

$$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}=1$$
$$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}$$ Let's take $\theta=\sin h$ Then as $h\to0$, $\theta\to(\sin0)=0$ Therefore, $$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}=\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ Apply Theorem 7 here: $$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}=1$$