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Q. $\displaystyle \lim_{x \to \pi/6}$ $\frac{2\,sin^{2}\,x+sin\,x-1}{2\,sin^{2}\,x-3\,sin\,x+1}$

Limits and Derivatives

Solution:

We have, $\displaystyle \lim_{x \to \frac{\pi}{6}}$ $\frac{2\,sin^{2}\,x+sin\,x-1}{2\,sin^{2}\,x-3\,sin\,x+1}$ $( \frac{0}{0}$ form$)$
$=\displaystyle \lim_{x \to \frac{\pi}{6}}$$\frac{\left(sin\,x+1\right)\left(2\,sin x-1\right)}{\left(sin\,x-1\right)\left(2\,sin\,x-1\right)}$
$=\displaystyle \lim_{x \to \frac{\pi}{6}}$$\frac{\left(sin\,x+1\right)}{\left(sin\,x-1\right)}$
$=\frac{ \displaystyle \lim_{x \to \pi/6}(sin\,x+1)}{ \displaystyle \lim_{x \to \pi/6}(sin\,x-1)}$
$=\frac{\left(\frac{1}{2}+1\right)}{\left(\frac{1}{2}-1\right)}=-3$