Q

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

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

x = 1 and x =  2.

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Given function is
$f (x) = [x] , 0 < x < 3$
We know that any function is differentiable when both
$\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$       and      $\lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$   are finite and equal
Required condition for function to be differential at x  = 1  is

$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, Left-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^-}\frac{0-1}{h}$
$=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \Rightarrow 1+h<1, \therefore [1+h] =0)$
Right-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^+}\frac{1-1}{h}$
$=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 1+h>1, \therefore [1+h] =1)$
Now, it is clear that
R.H.L. at x= 1  $\neq$  L.H.L. at x= 1  and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x  = 2  is

$\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}$
Now, Left-hand limit of the function at x = 2 is
$\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^-}\frac{1-2}{h}$
$=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \Rightarrow 2+h<2, \therefore [2+h] =1)$
Right-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^+}\frac{2-2}{h}$
$=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 2+h>2, \therefore [2+h] =2)$
Now, it is clear that
R.H.L. at x= 2  $\neq$  L.H.L. at x= 2  and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2

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