$f\left(x\right)=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\left\{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\left(\pi x\right)\right\}}^n+\left[x\right]$

$=\left\{\begin{array}{lll}\left[x\right]& :& {\cos }^2\left(\pi x\right)\in \left[0,1\right)\\ 1+\left[x\right]& :& {\cos }^2\left(\pi x\right)=1\end{array}\right.$
$=\left\{\begin{array}{lll}\left[x\right]& :& x\notin \mathrm{{\rm I}}\\ 1+\left[x\right]& :& x\in \mathrm{{\rm I}}\end{array}\right.$
For $x=1$,
$f\left(1^{-}\right)=\left[1^{-}\right]=0, f\left(1^{+}\right)=\left[1^{+}\right]=1, f\left(1\right)=1+\left[1\right]=2$
$\because f\left(1^{+}\right)\ne f\left(1^{-}\right)\Rightarrow f\left(x\right)$ is discontinuous at $x=1$
For $x=\frac{3}{2}$,

$f\left({\frac{3}{2}}^{+}\right)=\left[{\frac{3}{2}}^{+}\right]=1, f\left({\frac{3}{2}}^{-}\right)=\left[{\frac{3}{2}}^{-}\right]=1, f\left(\frac{3}{2}\right)=\left[\frac{3}{2}\right]=1$
$\because f\left({\frac{3}{2}}^{-}\right)=f\left({\frac{3}{2}}^{+}\right)=f\left(\frac{3}{2}\right)$
$\Rightarrow f\left(x\right)$ is continuous at $x=\frac{3}{2}$