Question

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

Answer 1

The given function is f(x)=[x],0<x<3 
It is known that a function f is differentiable at a point x=c in its domain if both
limh→0\(-f\frac{(c+h)-f(c)}{h}\) and lim_h→0+\(f\frac{(c+h)-f(c)}{h} \) 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→0-\(f\frac{(1+h)-f(1)}{h}\) =lim_h→0-\(\frac{[1+h]-[1]}{h}\)
=lim_h→0-\(\frac{0-1}{h}\)=lim_h→0-\(\frac{[1+h]-[1]}{h}\)

Consider the right hand limit of f at x=1
lim_h→0+\(f\frac{(1+h)-f(1)}{h}\)=lim_h→0+\(f\frac{[1+h]-[1]}{h}\)
=lim_h→0+\(\frac{1-1}{h}\)
=lim_h→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→0-\(\frac{f(2+h)-f(2)}{h}\) =lim_h→0-\(\frac{[2+h]-[2]}{h}\)
=lim_h→0-\(\frac{1-2}{h}\)=limh→0--1/h =∞

Consider the right hand limit of f at x=1
lim_h→0+\(f\frac{(2+h)-f(2)}{h}\)=lim_h→0+\(f\frac{(2+h)-f(2)}{h}\)
=lim_h→0+\(\frac{2-2}{h}\)
=lim_h→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