Let $\text{⨍ (x)=[x]}$ and $\mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}0,& \mathrm{x}\in \mathrm{I}\\ {\mathrm{x}}^2,& \mathrm{x}\in \mathrm{R}-\mathrm{I}\end{array}\right.$

$$\begin{gathered} \begin{array}{r}\underset{x\to 1^{+}}{lim} f\left(x\right)=1,\underset{x\to 1^{-}}{lim}f\left(x\right)=0,f\left(1\right)=1\\ LHL\ne RHL\\ \end{array} \\ \text{ So limit does not exist so f is discontinous at x=}1 \\ \begin{array}{r}g\left(x\right)=0\text{ at }x\in Z\\ g\left(x\right)=x^2\text{ at }x\in R-Z\\ \underset{x\to 1^{-}}{lim} g\left(x\right)=1\\ \underset{x\to 1^{-}}{lim} g\left(x\right)=1\\ g\left(1\right)=0\end{array} \\ \text{ limit exists but function is discontinuous as value at }1\text{ is not same as limit } \\ \mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{g}\left(\left[\mathrm{x}\right]\right)=0\text{ for all }\mathrm{x} \\ \text{ It is continuous at all }\mathrm{x}\text{ being a consatant function } \end{gathered}$$