Question

$f\left(x\right)=\{\begin{array}{cc}{x}^{2}sin\frac{1}{x},& ifx\ne 0\\ 0,& ifx=0\end{array}$

Answer 1

We have, $f\left(x\right)=\{\begin{array}{cc}{x}^{2}sin\frac{1}{x},& ifx\ne 0\\ 0,& ifx=0\end{array}$

For differentiability at x=0,

$L{f}^{\prime }\left(0\right)=\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{-}}{lim}\frac{x^{2}sin\frac{1}{x}-0}{x-0}=\underset{h\to 0}{lim}\frac{(0-h{)}^{2}sin\left(\frac{1}{0-h}\right)}{0-h}=\underset{h\to 0}{lim}\frac{h^{2}sin\left(\frac{-1}{h}\right)}{-h}=\underset{h\to 0}{lim}+hsin\left(\frac{1}{h}\right)[\because sin(-\theta )=-sin\theta ]$
$=0\times$ [an oscillating number between -1 and 1 ]=0

Rf'(0)$=\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{lim}\frac{x^{2}sin\frac{1}{x}-0}{x-0}=\underset{h\to 0}{lim}\frac{(0-h{)}^{2}sin\left(\frac{1}{0+h}\right)}{0+h}=\underset{h\to 0}{lim}\frac{h^{2}sin\left(\frac{1}{h}\right)}{h}$
$=0\times$ [an oscillating number between -1 and 1 ]=0
$\because L{f}^{\prime }\left(0\right)=R{f}^{\prime }\left(0\right)$
So, f(x) is differentiable at x=0