Prove that the greatest integer function defined by

The given function $f$ is $f(x)=[x], 0

It is known that a function f is differentiable at a point x = c in its domain if both
$\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \rightarrow 0^{\circ}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1

$\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[1+h]-[1]}{h}$

$=\lim _{h \rightarrow 0^{-}} \frac{0-1}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$\

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \rightarrow 0^{0}} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[1+h]-[1]}{h}$

$=\lim _{h \rightarrow 0^{+}} \frac{1-1}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at

x = 1

To check the differentiability of the given function at x = 2, consider the left hand limit

of f at x = 2

$\lim _{h \rightarrow 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \rightarrow 0^{-}} \frac{1-2}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \rightarrow 0^{+}} \frac{2-2}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2