Question

lf a function $f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{c}\frac{x}{\sqrt{x^2}},x\ne 0\\ 0x=0\end{array}$ then

• A. f(x) is continuous at x=0 but not differentiable at x=0
• B. f(x) is continuous as well as differentiable at x=0
• C. f(x) is discontinuous at x=0
• D. f(x) is continuous everywhere

Answer 1

Answer: (C)

f(x) is discontinuous at x=0

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

$f\left(x\right)=\frac{x}{\pm x}x\ne 0$

if $x>0f\left(x\right)=\frac{x}{x}=1$

if $x<0f\left(x\right)=\frac{x}{-x}=-1$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=1$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=-1$

LHL$\ne$RHL $\Rightarrow$Discontinuous at $x=0$