Answer
$$0$$
Work Step by Step
\begin{align*}
\lim _{\theta \rightarrow 0} \frac{ \cos 2\theta -\cos \theta }{ \theta}&= \lim _{\theta \rightarrow 0} \frac{ 1-2\sin^2\theta -\cos \theta }{ \theta}\\
&=\lim _{\theta \rightarrow 0} \left( \frac{ 1 -\cos \theta }{ \theta} - \frac{ 2\sin^2\theta }{ \theta}\right)\\
&= \lim _{\theta \rightarrow 0} \frac{ 1 -\cos \theta }{ \theta} -\lim _{\theta \rightarrow 0} \frac{ 2\sin^2\theta }{ \theta} \\
&= \lim _{\theta \rightarrow 0} \frac{ 1 -\cos \theta }{ \theta} -2\lim _{\theta \rightarrow 0} \frac{ \sin \theta }{ \theta} \lim _{\theta \rightarrow 0} \sin\theta \\
&=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. $
