$L=\underset{x\to a}{lim} \frac{|2\sin x-1|}{2\sin x-1}.\text{ Then }$

$L=\underset{x\to a}{lim} \frac{|2\sin x-1|}{2\sin x-1}$

For $a=\pi /6$ $\text{ L.H.L. }=\underset{x\to \frac{\pi }{6}}{lim} \frac{1-2\sin x}{2\sin x-1}=-1$

$\text{ R.H.L. }=\underset{x\to \frac{{\pi }^{+}}{6}}{lim} \frac{2\sin x-1}{2\sin x-1}=-1$

Hence, the limit does not exist

$\text{ For }a=\pi ,\underset{x\to \pi }{lim} \frac{1-2\sin x}{2\sin x-1}=-1$

(as in neighborhood of$\mathrm{\pi }$, sinx is less than 1/2)

$\text{ For }a=\pi ,\underset{x\to \pi /2}{lim} \frac{2\sin x-1}{2\sin x-1}=1$

(as in neighborhood of $\mathrm{\pi }/2$, sinx approaches 1).