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

Differentiable at $x = 0$ and $f'(0) = 0$.

The correct answer is Option (2) → Differentiable at $x = 0$ and $f'(0) = 0$. ##

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

For differentiability at $x = 0$,

$Lf'(0) = \lim\limits_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}$

$= \lim\limits_{x \to 0^-} \frac{x^2 \sin \frac{1}{x} - 0}{x - 0}$

Put $x = 0 - h$,

$= \lim\limits_{h \to 0} \frac{(0 - h)^2 \sin \left( \frac{1}{0 - h} \right)}{-h} = \lim\limits_{h \to 0} \frac{h^2 \sin \left( -\frac{1}{h} \right)}{-h}$

$= \lim\limits_{h \to 0} h \sin \left( \frac{1}{h} \right) \quad [∵ \sin(-\theta) = -\sin\theta]$

$= 0 \times [\text{an oscillating number between } -1 \text{ and } 1] = 0$

$Rf'(0) = \lim\limits_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^+} \frac{x^2 \sin \frac{1}{x} - 0}{x - 0}$

Put $x = 0 + h$,

$= \lim\limits_{h \to 0} \frac{(0 + h)^2 \sin \left( \frac{1}{0 + h} \right)}{0 + h} = \lim\limits_{h \to 0} \frac{h^2 \sin (1/h)}{h} = \lim\limits_{h \to 0} h \sin (1/h)$

$= 0 \times [\text{an oscillating number between } -1 \text{ and } 1] = 0$

$∵Lf'(0) = Rf'(0)$

So, $f(x)$ is differentiable at $x = 0$.