Show that $f(x) = |x - 5|$ is continuous but not differentiable at $x = 5$

We have, $f(x) = |x - 5|$
therefore,$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{ - (x - 5),}&{{\rm{ if }}x < 5}\\{x - 5,}&{{\rm{ if }}x \ge 5}\end{array}} \right.$

For continuity at $x = 5$,

${\rm{LHL}} = \mathop {\lim }\limits_{x \to {5^ - }} ( - x + 5)$
$= \mathop {\lim }\limits_{h \to 0} [ - (5 - h) + 5] = \mathop {\lim }\limits_{h \to 0} h = 0$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {5^ + }} (x - 5)$
$= \mathop {\lim }\limits_{h \to 0} (5 + h - 5) = \mathop {\lim }\limits_{h \to 0} h = 0$

therefore,$f(5) = 5 - 5 = 0$

$\Rightarrow$ ${\rm{LHL}} = {\rm{RHL}} = f(5)$

Hence, $f(x)$ is continuous at $x = 5$.

Now, $L{f^\prime }(5) = \mathop {\lim }\limits_{x \to {5^ - }} \frac{{f(x) - f(5)}}{{x - 5}}$
$= \mathop {\lim }\limits_{x \to {5^ - }} \frac{{ - x + 5 - 0}}{{x - 5}} = - 1$

$R{f^\prime }(5) = \mathop {\lim }\limits_{x \to {5^ + }} \frac{{f(x) - f(5)}}{{x - 5}}$

$= \mathop {\lim }\limits_{x \to {5^ + }} \frac{{x - 5 - 0}}{{x - 5}} = 1$

$\Rightarrow L{f^\prime }(5) \ne R{f^\prime }(5)$

Therefore we can say that $f(x) = |x - 5|$ is not differentiable at $x = 5$.