$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{1 - \cos 2x}}{{{x^2}}},}&{{\rm{ if }}x \ne 0}\\{5,}&{{\rm{ if }}x = 0}\end{array}} \right.$at $x = 0$ }

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{1 - \cos 2x}}{{{x^2}}},}&{{\rm{ if }}x \ne 0}\\{5,}&{{\rm{ if }}x = 0}\end{array}} \right.$ at $x = 0$.

At $x = 0,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{1 - \cos 2x}}{{{x^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 2(0 - h)}}{{{{(0 - h)}^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 2h}}{{{h^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - 1 + 2{{\sin }^2}h}}{{{h^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{(\sin h)}^2}}}{{{{(h)}^2}}}$

$= 2$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 - \cos 2x}}{{{x^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 2(0 + h)}}{{{{(0 + h)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{\sin }^2}h}}{{{h^2}}} = 2$

and $f(0) = 5$

Since, ${\rm{LHL}} = {\rm{RHL}} \ne f(0)$

Therefore we can say that $f(x)$ is not continuous at $x = 0$.