SOLUTION —
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{4 \sin x \cos x+\cos x}{4 \sin x \cos x-3 \cos x}$
By L'Hospital's Rule
$\begin{array}{l}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\cos x(4 \sin x+1)}{\cos x(4 \sin x-3)} \\=\frac{4 \sin \frac{\pi}{6}+1}{4 \sin \frac{\pi}{6}-3}=\frac{2+1}{2-3}=-3\end{array}$
So, The correct option will be (B).