$f(x)= \begin{cases}x^{2} \sin \frac{1}{x} & \text {, if } x \neq 0 \\ 0 & \text {, if } x=0\end{cases}$
at $x=0$
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
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