Answer
$$\lim_{\theta\to0}\frac{1-\cos\theta}{\sin2\theta}=0$$
Work Step by Step
$$A=\lim_{\theta\to0}\frac{1-\cos\theta}{\sin2\theta}=\lim_{\theta\to0}\frac{1-\cos\theta}{2\sin\theta\cos\theta}$$
Multiply both numerator and denominator by $(1+\cos\theta)$:
$$A=\lim_{\theta\to0}\frac{(1-\cos\theta)(1+\cos\theta)}{2\sin\theta\cos\theta(1+\cos\theta)}$$
$$A=\lim_{\theta\to0}\frac{1-\cos^2\theta}{2\sin\theta\cos\theta(1+\cos\theta)}$$
Recall that $\sin^2\theta=1-\cos^2\theta$
$$A=\lim_{\theta\to0}\frac{\sin^2\theta}{2\sin\theta\cos\theta(1+\cos\theta)}$$
$$A=\lim_{\theta\to0}\frac{\sin\theta}{2\cos\theta(1+\cos\theta)}$$
$$A=\frac{\sin0}{2\cos0(1+\cos0)}$$
$$A=\frac{0}{2\times1(1+1)}=0$$
