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

Explanation

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

### Step 1: Define the left-hand derivative at \( x = 0 \)

The left-hand derivative of \( f \) at \( x = 0 \) is given by the limit:

\[
f'_{-}(0) = \lim_{h \to 0^{-}} \frac{f(0 + h) - f(0)}{h}
...