Question
Which one of the following is correct in respect of the function $$f\left( x \right) = \frac{{{x^2}}}{{\left| x \right|}}$$ for $$x \ne 0$$ and $$f\left( 0 \right) = 0?$$
A.
$$f\left( x \right)$$ is discontinuous every where
B.
$$f\left( x \right)$$ is continuous every where
C.
$$f\left( x \right)$$ is continuous at $$x = 0$$ only
D.
$$f\left( x \right)$$ is discontinuous at $$x = 0$$ only
Answer :
$$f\left( x \right)$$ is continuous every where
Solution :
\[\begin{array}{l}
f\left( x \right) = \left\{ \begin{array}{l}
\frac{{{x^2}}}{x},\,\,x \ne 0\\
\,\,\,0,\,\,\,x = 0
\end{array} \right.\\
= \left\{ \begin{array}{l}
\frac{{{x^2}}}{x} = x,\,\,\,\,\,\,\,\,\,\,\,x > 0\\
\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\\
\frac{{{x^2}}}{{ - x}} = - x,\,\,\,\,\,x < 0
\end{array} \right.
\end{array}\]
$$\eqalign{
& {\text{Now, }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0 \cr
& \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( x \right) = 0{\text{ and }}f\left( 0 \right) = 0 \cr} $$
So, $$f\left( x \right)$$ is continuous at $$x = 0$$
Also, $$f\left( x \right)$$ is continuous for all other values of $$x.$$
Hence, $$f\left( x \right)$$ is continuous everywhere.