Answer
$$\lim_{\theta\to\pi/6}\frac{\sin\theta-\frac{1}{2}}{\theta-\frac{\pi}{6}}=\frac{\sqrt3}{2}$$
Work Step by Step
$$A=\lim_{\theta\to\pi/6}\frac{\sin\theta-\frac{1}{2}}{\theta-\frac{\pi}{6}}=\lim_{\theta\to\pi/6}\frac{\sin\theta-\sin\frac{\pi}{6}}{\theta-\frac{\pi}{6}}$$
Let a function $f(x)=\sin x$. According to the definition of derivative, we have
$$f'(x)=\lim_{z\to x}\frac{\sin z-\sin x}{z-x}=(\sin x)'=\cos x$$
So we take $z=\theta$ and $x=\pi/6$ for the limit in this exercise, we have
$$A=\lim_{\theta\to\pi/6}\frac{\sin\theta-\sin\frac{\pi}{6}}{\theta-\frac{\pi}{6}}=\cos\Big(\frac{\pi}{6}\Big)$$
$$A=\frac{\sqrt3}{2}$$
