Question

If a function $f(x)$ is defined as
$f(x)=\{\begin{array}{l}\frac{x}{\sqrt{x^2}},x\ne 0\\ 0,x=0\end{array}$ 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 $:f(x)=\{\begin{array}{ll}\frac{x}{\sqrt{x^2}}& ,x\ne 0\\ 0& ,x=0\end{array}.$
$\therefore f(x)=\{\begin{array}{ll}\frac{x}{|x|}& ,x\ne 0\\ 0& ,x=0\end{array}.$
$\therefore f(0)=0$
$R.H.L=\underset{x\to 0^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }\frac{0+h}{|0+h|}=\underset{h\to 0}{\lim }\frac{n}{h}=1$
$L.H.L.=\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }\frac{(0-h)}{|(0-h)|}=\underset{h\to 0}{\lim }\frac{-h}{h}=-1$
$R.H.L\ne L.H.L$
i.e. $\underset{x\to 0^{+}}{\lim }f(x)\ne \underset{x\to 0^{-}}{\lim }f(x)$
$\therefore f(x)$ is discontinuous at $x=0$