Answer
$$0$$
Work Step by Step
\begin{align*}
\lim _{h\rightarrow 0} \frac{\sin 2h (1-\cos h)}{h^2}&=\lim _{h\rightarrow 0}\frac{\sin 2h }{h} \frac{ 1-\cos h}{h}\\
&=\lim _{h\rightarrow 0}2\frac{\sin 2h }{2h} \frac{ 1-\cos h}{h} \\
&=2 \lim _{2h\rightarrow 0}\frac{\sin 2h }{2h}\lim _{h\rightarrow 0}\frac{ 1-\cos h}{h} \\
&=0.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$ and $\lim _{x\rightarrow 0}\frac{ 1-\cos x}{x}=0. $
