Answer
$$\lim_{x\to0}\frac{\cos2x-1}{\sin x}=0$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\cos2x-1}{\sin x}$$
Recall the identity: $\cos2x=1-2\sin^2x$
$$A=\lim_{x\to0}\frac{1-2\sin^2x-1}{\sin x}=\lim_{x\to0}\frac{-2\sin^2x}{\sin x}=\lim_{x\to0}(-2\sin x)$$
$$A=-2\sin0=-2\times0=0$$
