The value of

Given,

$f(x)=\left\{\begin{array}{c}x^{2} \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0\end{array}\right.$ at $x=0$

For differentiability we know that:

$\mathrm{L} f^{\prime}(c)=\mathrm{R} f^{\prime}(c)$

$\therefore \mathrm{L} f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$

$=\lim _{h \rightarrow 0} \frac{(0-h)^{2} \sin \frac{1}{(0-h)}-0}{-h}=\frac{h^{2} \cdot \sin \left(-\frac{1}{h}\right)}{-h}$

$=h \cdot \sin \left(\frac{1}{h}\right)=0 \times\left[-1 \leq \sin \left(\frac{1}{h}\right) \leq 1\right]$

$=0$

$R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{(0+h)^{2} \sin \left(\frac{1}{0+h}\right)-0}{h}$

$=\lim _{h \rightarrow 0} \frac{h^{2} \sin \left(\frac{1}{h}\right)}{h}=\lim _{h \rightarrow 0} h \cdot \sin \left(\frac{1}{h}\right)$

$=0 \times\left[-1 \leq \sin \left(\frac{1}{h}\right) \leq 1\right]=0$

Hence, $L f^{\prime}(0)=R f^{\prime}(0)=0$

Therefore, f(x) is differentiable at x = 0