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Prove that the greatest integer function defined by $\mathbf{f}(\mathbf{x})=[\mathbf{x}], 0 < x < 3$ is not differentiable at $x=1$ and $x=2$.
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(i) At $x=1$ R.H.D. $=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{1-1}{h}=0 \quad \because[1+h]=1$
L.H.D. $=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{0-1}{-\mathrm{h}}=$ not defined $\because[1-\mathrm{h}]=1$
$\therefore \quad \mathrm{f}$ is not differentiable at $\mathrm{x}=1$.
(ii) At $x=3$ R.H.D. $=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$
$=\lim _{h \rightarrow 0} \frac{3-3}{h}=0 ;$ L.H.D. $=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}$
$=\lim _{h \rightarrow 0} \frac{2-3}{h}=\lim _{h \rightarrow 0} \frac{-1}{h}=$ Not defined.
R.H.D. $\neq$ L.H.D. $\Rightarrow$ fis not differentiable at $x=3$.
$=\lim _{h \rightarrow 0} \frac{1-1}{h}=0 \quad \because[1+h]=1$
L.H.D. $=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{0-1}{-\mathrm{h}}=$ not defined $\because[1-\mathrm{h}]=1$
$\therefore \quad \mathrm{f}$ is not differentiable at $\mathrm{x}=1$.
(ii) At $x=3$ R.H.D. $=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$
$=\lim _{h \rightarrow 0} \frac{3-3}{h}=0 ;$ L.H.D. $=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}$
$=\lim _{h \rightarrow 0} \frac{2-3}{h}=\lim _{h \rightarrow 0} \frac{-1}{h}=$ Not defined.
R.H.D. $\neq$ L.H.D. $\Rightarrow$ fis not differentiable at $x=3$.
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