Question

A function is defined as follows:$F\left(x\right)=$ $\{\begin{array}{cc}-\frac{x}{\sqrt{x^2}},& x\ne 0\\ 0,& x=0\end{array}$
Which one of the following is correct in respect of the above function?

• A. $f\left(x\right)$ is continuous at $x=0$ but not differentiable at $x=0$
• B. $f\left(x\right)$ is continuous as well as differentiable at $x=0$
• C. $f\left(x\right)$ is discontinuous at $x=0$
• D. None of the above

Answer 1

Answer: (C)

$f\left(x\right)$ is discontinuous at $x=0$

Given function is $-\frac{x}{\sqrt{x^2}}$ for $x\ne 0$

which is equal to $-\frac{x}{\left|x\right|}$

means for $x<0-\frac{x}{\left|x\right|}=-\frac{x}{-x}=1$

$x>0-\frac{x}{\left|x\right|}=\frac{x}{x}=-1$

at $x=0$, given function=0

$\underset{x\to 0+}{lim}\frac{x}{\left|x\right|}=-1$

$\underset{x\to 0-}{lim}\frac{x}{\left|x\right|}=1$

LHL(at x=0) and RHL(x=0) and value of function at x=0 are not equal so the function is discontinious.