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