Answer
$$0$$
Work Step by Step
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\sin x}
&=\lim _{x \rightarrow 0}\left(\frac{\cos 2 x-1}{\sin x} \cdot \frac{\cos 2 x+1}{\cos 2 x+1}\right)\\
&=\lim _{x \rightarrow 0} \frac{\cos ^{2} 2 x-1}{\sin x(\cos 2 x+1)}\\
&=\lim _{x \rightarrow 0} \frac{-\sin ^{2} 2 x}{\sin x(\cos 2 x+1)} \\
&=\lim _{x \rightarrow 0} \frac{-4 \sin x \cos ^{2} x}{\cos 2 x+1}\\
&=\frac{-4(0)(1)^{2}}{1+1}\\
&=0
\end{aligned}
