Show that the function f(x)={ `x^2sin(1/x) , when x!=0 and 0 when x=0` } is differentiable at x=0.

Explanation

To show that the function \( f(x) = \begin{cases} x^2 \sin(1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \) is differentiable at \( x = 0 \), we need to calculate the left-hand derivative and the right-hand derivative at that point and check if they are equal.

### Step 1: Define the derivatives
The left-hand derivative (LHD) at \( x = 0 \) is defined as:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} \]
The right-hand derivative (RHD) at \( x = 0 \) is defined as:
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