Answer
$$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}=1$$
Work Step by Step
$$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}$$
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: $$\lim_{h\to0}\frac{\sin(\sin h)}{\sin h}=1$$