Prove that the greatest integer function defined by $f(x) = [x],0 < x < 3$ is not differentiable at x $=$ 1 and x $=$ 2.

At x $=$ 1:
$Rf'(1) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to 0} \cfrac{{[1 + h] - [1]}}{h} = 0$
[ $[1 + h = 1\,and\,[1] = 1]$

and $Lf'(1) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(1 - h) - f(1)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \cfrac{{[1 - h] - [1]}}{{ - h}} = \infty$

Thus $Rf'(1) \ne Lf'(1)$

Hence f(x)$= [x]$ is not differentiable at x $=$ 1
At x $=$ 2 :

$Rf'(2) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(2 + h) - f(2)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \cfrac{{[2 + h] - [2]}}{h} = 0$
$Lf'(2) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(2 - h) - f(2)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \cfrac{{[2 - h] - [2]}}{{ - h}} = \cfrac{{1 - 2}}{0} = \infty$

therefore, $Rf'(2) \ne Lf'(2)$

Hence we can say that $f(x) = [x]$ is not differentiable at x $=$ 2.