Evaluate the following limits:


Evaluate the following limits:

$\lim _{x \rightarrow \pi / 6} \frac{\left(2 \sin ^{2} x+\sin x-1\right)}{\left(2 \sin ^{2} x-3 \sin x+1\right)}$



To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is $\frac{0}{0}$

Formula used: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ or we can used $L$ hospital Rule,

So, by using the rule, Differentiate numerator and denominator

$\lim _{x \rightarrow \frac{\pi}{6}} \frac{4 \sin x \cos x+\cos x}{4 \sin x \cos x-3 \cos x}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{4 \sin x+1}{4 \sin x-3}=\frac{2+1}{2-3}=-3$

Therefore, $\lim _{x \rightarrow 0} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}=-3$


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