Question

Find if the given below function
is continuous or discontinuous
at the indicated point
$f\left(x\right)=\{\begin{array}{cc}\frac{|x-4|}{2(x-4)},& \text{if}x\ne 4\\ 0,& \text{if}x=4\end{array}\text{at}x=4$

Answer 1

Use the concept of continuity of a function at a point

Given that
$f\left(x\right)=\{\begin{array}{cc}\frac{|x-4|}{2(x-4)},& \text{if}x\ne 4\\ 0,& \text{if}x=4\end{array}\text{at}x=4\dots \left(1\right)$

At $x=4,L.H.L=\underset{x\to 4^{-}}{\text{lim}}f\left(x\right)$
$=\underset{h\to 0^{+}}{\text{lim}}f(4-h)$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{|(4-h)-4|}{2\left[\right(4-h)-4]}$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{|-h|}{-2h}=\underset{h\to 0^{+}}{\text{lim}}\frac{-(-h)}{-2h}$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{h}{-2h}=\frac{-1}{2}\dots \left(2\right)$
{$\because |-h|=-(-h)=h\text{as}h>0$}

At $x=4,R.H.L=\underset{x\to 4^{+}}{\text{lim}}f\left(x\right)$
$=\underset{h\to 0^{+}}{\text{lim}}f(4+h)$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{|(4+h)-4|}{2\left[\right(4+h)-4]}$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{|h|}{-2h}=\underset{h\to 0^{+}}{\text{lim}}\frac{h}{-2h}=\frac{-1}{2}$

$\therefore L.H.L=R.H.L\dots \left(3\right)$

So, limit exists. Now check for continuity,
Given that $f\left(4\right)=0\dots \left(4\right)$
Since, $LHL=RHL\ne f\left(4\right)$
[Using (3) and (4)]
So, $f\left(x\right)$ is discontinuous at $x=4$