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

Toolbox:
• Suppose f is a real function and c is a point in its domain the derivative of f at c is defined by
• $\lim_{h \to 0}\large\frac{f(c+h)-f(c)}{h}$
• $f'(x)=\lim_{h \to 0}\large\frac{f(x+h)-f(x)}{h}$
Step 1:
At $x=1$
Substitute $x=1$
RHD=$\lim_{h \to 0}\large\frac{f(1+h)-f(1)}{h}$
$\qquad=\lim_{h \to 0}\large\frac{1-1}{h}$$=0 Step 2: LHD=\lim_{h \to 0}\large\frac{f(1-h)-f(1)}{-h} \qquad=\lim_{h \to 0}\large\frac{0-1}{h}=not defined. Therefore f is not differentiable at x=1. Step 3: At x=3 RHD=\lim_{h \to 0}\large\frac{f(3+h)-f(1)}{h} \qquad=\lim_{h \to 0}\large\frac{3-3}{h}$$=0$
Step 4:
LHD=$\lim_{h \to 0}\large\frac{f(3-h)-f(3)}{-h}$
$\qquad=\lim_{h \to 0}\large\frac{2-3}{h}$
$\Rightarrow \lim_{h\to 0}\large\frac{-1}{h}$=not defined.
$\Rightarrow$ RHD $\neq$LHD
Therefore f is not differentiable at $x=3$
hello madam in this question in the  limit application in i.e how the value of the value of  f(3+h) will becomes 3 in step 3 and also in step 4 how f(3-h) will become 2 and limit application in i.e how the value of  f(1+h) will becomes 1 in step 1 and also in step 2 how the value of f(1-h) will become not defined value can u pls explain in detail