Question

Consider $f\left(x\right)=\left\{\begin{array}{c}\frac{x^2}{|x|},x\ne 0\\ f\left(x\right)=0,x=0\end{array}\right\}$

Then which of the following is true?

• A. $f\left(x\right)$ is discontinuous everywhere
• B. $f\left(x\right)$ is continuous everywhere
• C. $f\left(x\right)$ is not defined
• D. $f\left(x\right)$ is differentiable at x= 0

Answer 1

The correct option is B $f\left(x\right)$ is continuous everywhere

$f\left(x\right)=\frac{x^2}{|x|},x\ne 0$ and $f\left(x\right)=0,x=0$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{x^2}{|x|}=\underset{x\to 0^{+}}{lim}\frac{x^2}{x}=\underset{x\to 0^{+}}{lim}x=0\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{x^2}{|x|}=\underset{x\to 0^{-}}{lim}\frac{x^2}{-x}=\underset{x\to 0^{-}}{lim}-x=0$
$\therefore$ limit value = function value
Hence, $f\left(x\right)$ is continuous everywhere but is not differentiable.