Answer
$$0$$
Work Step by Step
We evaluate the limit:
\begin{align*}
\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin x}&=\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin x} \cdot \frac{\cos x+1}{\cos x+1}\\
&=\lim _{x \rightarrow 0} \frac{-\sin ^{2} x}{\sin x(\cos x+1)}\\
&=-\lim _{x \rightarrow 0} \frac{\sin x}{\cos x+1}\\
&=-\frac{0}{1+1}=0
\end{align*}
