$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$ is equal to
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$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$ is equal to

(A) 3

(B) -3

(C) 6

(D) 0

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Best answer

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).

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