<s>[INST] 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.
 [/INST] As this is in the form of $\frac{0}{0}$,so apply L Hospital rule
$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$
$=\lim_{x \to \frac{\pi}{6}} \frac{4 \sin x\cos x+\cos x}{4 \sin x\cos x-3 \cos x}=\frac{\sqrt3+\frac{\sqrt3}{2}}{\sqrt3-\frac{3\sqrt3}{2}}=-3$
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