Question

Let $f\left(x\right)=\left[x\right]$ and $g\left(x\right)=$$\{\begin{array}{c}0,x\in Z\\ x^2,x\in R-Z\end{array}$
$\left(\right[.]$ represents greatest integer function$).$ Then

• A. $f\left(x\right)$ is not continuous at $x=1.$
• B. $gof$ is continuous for all $x.$
• C. $fog$ is continuous for all $x.$
• D. $\underset{x\to 1}{lim}g\left(x\right)$ exists but $g\left(x\right)$ is not continuous at $x=1.$

Answer 1

Answer: (A), (B), (D)

$\underset{x\to 1}{lim}g\left(x\right)$ exists but $g\left(x\right)$ is not continuous at $x=1.$

Since, $\underset{x\to 1^{-}}{lim}g\left(x\right)=\underset{x\to 1^{+}}{lim}g\left(x\right)=1$ and $g\left(1\right)=0,$
so, $g\left(x\right)$ is not continuous at $x=1$ but $\underset{x\to 1}{lim}g\left(x\right)$ exists.
We have $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}[1-h]=0$
and $\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}[1+h]=1$
So, $\underset{x\to 1}{lim}f\left(x\right)$ does not exist and hence $f\left(x\right)$ is not continuous at $x=1$
We have $gof\left(x\right)=g\left(f\right(x\left)\right)=g\left(\right[x\left]\right)=0,\forall x\in R$
So, $gof$ is continuous for all $x.$
We have $fog\left(x\right)=f\left(g\right(x\left)\right)$$=\{\begin{array}{c}f\left(0\right),x\in Z\\ f\left(x^2\right),x\in R-Z\end{array}$
$=$$\{\begin{array}{c}0,x\in Z\\ \left[x^2\right],x\in R-Z\end{array}$
which is clearly not continuous for all $x$.