$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{|x - 4|}}{{2(x - 4)}},}&{{\rm{ if }}x \ne 4}\\{0,}&{{\rm{ if }}x = 4}\end{array}} \right.$at $x = 4$ }

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{|x - 4|}}{{2(x - 4)}},}&{{\rm{ if }}x \ne 4}\\{0,}&{{\rm{ if }}x = 4}\end{array}} \right.$at $x = 4$.
At $x = 4,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {4^ - }} \frac{{|x - 4|}}{{2(x - 4)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{|4 - h - 4|}}{{2[(4 - h) - 4]}} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 - h|}}{{(8 - 2h - 8)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{ - 2h}} = \frac{{ - 1}}{2}\quad$ and $f(4) = 0 \ne {\rm{LHL}}$

Therefore we can say that $f(x)$ is discontinuous at $x = 4$.