f(x)= $L{t}_{n\to \infty }\left({\left(\sin x\right)}^{2n}\right)$ then set of points of discontinuity of f(x) is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}\to \infty }{\lim }{(\sin \mathrm{x})}^{2\mathrm{n}} \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}\to \infty }{\lim }{(\sin \mathrm{x})}^{2\mathrm{n}}=\left\{\begin{array}{ll}0 \mathrm{if}& |\sin \mathrm{x}|<1\\ 1 \mathrm{if}& |\sin \mathrm{x}|=1\end{array}\right. \therefore \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{discontinuous} \mathrm{for} \mathrm{the} \mathrm{values} \mathrm{of} \mathrm{x} \\ \mathrm{for} \mathrm{which} |\sin \mathrm{x}|=1\Rightarrow \mathrm{x}=(2\mathrm{n}+1) \frac{\mathrm{\pi }}{2}\forall \mathrm{n}\in \mathrm{z} \end{gathered}$$