Question

Show that $f\left(x\right)=\left\{\begin{array}{ll}\frac{\left|x-a\right|}{x-a},\mathrm{when}& x\ne a\\ 1,\mathrm{when}& x=a\end{array}\right.$
is discontinuous at x = a.

Answer 1

The given function can be rewritten as:
$f\left(x\right)=\left\{\begin{array}{c}\frac{x-a}{x-a},\mathrm{when}x>a\\ \frac{a-x}{x-a},\mathrm{when}x<a\\ 1,\mathrm{when}x=a\end{array}\right.$
​$\Rightarrow$ $f\left(x\right)=\left\{\begin{array}{c}1,\mathrm{when}x>a\\ -1,\mathrm{when}x<a\\ 1,\mathrm{when}x=a\end{array}\right.$

$\Rightarrow$ $f\left(x\right)=\left\{\begin{array}{l}1,\mathrm{when}x\ge a\\ -1,\mathrm{when}x<a\end{array}\right.$

We observe
(LHL at x = a) = $\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(a-h\right)$$=\underset{h\to 0}{\lim }\left(-1\right)=-1$


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


∴ ​$\underset{x\to a^{-}}{\lim }f\left(x\right)\ne \underset{x\to a^{+}}{\lim }f\left(x\right)$

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