Let $f(x)=\lim\limits_{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^n-(2 \cos x)^{2 n}}, n \in Z$. Then,

all the above

We have, $f(x)=\lim\limits_{n \rightarrow \infty} \frac{\left(\sin ^2 x\right)^n}{\left(\frac{3}{4}\right)^n-\left(\cos ^2 x\right)^n}$

In the neighbourhood of $x=n \pi \pm \frac{\pi}{6}$, we have $f(x)=\frac{0}{0-0}$, which does not exist.

Hence, f(x) is discontinuous at $x=n \pi \pm \pi / 6, n \in Z$.

Also, we have

$f\left(\frac{\pi}{3}\right)=\lim\limits_{n \rightarrow \infty} \frac{\left(\frac{3}{4}\right)^n}{\left(\frac{3}{4}\right)^n-\left(\frac{1}{4}\right)^n}=\lim\limits_{n \rightarrow \infty} \frac{1}{1-\left(\frac{1}{3}\right)^n}=1$

and, $f(0)=\lim\limits_{n \rightarrow \infty} \frac{0}{\left(\frac{3}{4}\right)^n-1}=0$

Hence, option (a), (b), (c) are correct.