I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having trouble getting to it. Any ideas would be greatly appreciated.

Notice, $$\lim_{x\to \pi/6}\frac{2\sin^2 x+\sin x-1}{2\sin^2 x-3\sin x+1}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-(1-2\sin^2 x)}{2-3\sin x-(1-2\sin^2 x)}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-\cos 2x}{2-3\sin x-\cos 2x}$$
Apply L'hospital's rule for $\frac00$ form 
$$=\lim_{x\to \pi/6}\frac{\cos x+2\sin 2x}{-3\cos x+2\sin2x}$$
$$=\frac{\frac{\sqrt 3}{2}+\sqrt 3}{-\frac{3\sqrt 3}{2}+\sqrt 3}=\color{red}{-3}$$
