Let $f(x)=\underset{x→0}{\lim}\frac{x}{1+(2\sin x)^2n}$, then f is discontinuous at

$π/6$

$f(x)=\left\{\begin{matrix}x&if&|\sin x|<\frac{1}{2}i.e.-\frac{π}{6}+nπ<x<\frac{π}{6}+nπ\\\frac{x}{2}&if&|\sin x|=\frac{1}{2}i.e.x=±\frac{π}{6}+nπ\\0&if&|\sin x|>\frac{1}{2}i.e.\frac{π}{6}+nπ<x<\frac{5π}{6}+nπ\end{matrix}\right.$

Thus points of discontinuities of f are of the form $x=±\frac{π}{6}+nπ$

In particular, f is discontinuous at $x=\frac{π}{6}$