Question

If a function $f ( x )$ is defined as
$f(x)=\begin{cases} \frac{x}{\sqrt{x^{2}}}, x \neq 0 \\ 0, x=0 \end{cases}$ then :

• A. $f ( x )$ is continuous at $x =0$ but not differentiable at $x =0$
• B. $f ( x )$ is continuous as well as diffcrcntiable at $x =0$
• C. $f ( x )$ is discontinuous at $x =0$
• D. None of these.

Answer 1

Answer: (C)

Given $: f(x)=\begin{cases}\frac{x}{\sqrt{x^{2}}} & , x \neq 0 \\ 0 & , x=0\end{cases}.$
$\therefore f(x)=\begin{cases}\frac{x}{|x|} & , x \neq 0 \\ 0 & , x=0\end{cases}.$
$\therefore f (0)=0$
$R.H.L =\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{0+h}{|0+h|}=\displaystyle\lim _{h \rightarrow 0} \frac{n}{h}=1$
$L.H.L. =\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{(0-h)}{|(0-h)|}=\displaystyle\lim _{h \rightarrow 0} \frac{-h}{h}=-1$
$R.H.L \neq L.H.L$
i.e. $\displaystyle\lim _{x \rightarrow 0^{+}} f(x) \neq \displaystyle\lim _{x \rightarrow 0^{-}} f(x)$
$\therefore f ( x )$ is discontinuous at $x =0$