Question

Show that $f\left(x\right)=\left\{\begin{array}{ll}\frac{x-\left|x\right|}{2},\mathrm{when}& x\ne 0\\ 2,\mathrm{when}& x=0\end{array}\right.$
is discontinuous at x = 0.

Answer 1

The given function can be rewritten as:
$f\left(x\right)=\left\{\begin{array}{c}\frac{x-x}{2},\mathrm{when}x>0\\ \frac{x+x}{2},\mathrm{when}x<0\\ 2,\mathrm{when}x=0\end{array}\right.$
​$\Rightarrow$ $f\left(x\right)=\left\{\begin{array}{c}0,\mathrm{when}x>0\\ x,\mathrm{when}x<0\\ 2,\mathrm{when}x=0\end{array}\right.$

We observe
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$$=\underset{h\to 0}{\lim }\left(-h\right)=0$


(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(h\right)$= $\underset{h\to 0}{\lim }0=0$

And, $f\left(0\right)=2$
∴ ​$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)\ne f\left(0\right)$

Thus, f(x) is discontinuous at x = 0.