1. **State the problem:** We want to find the limit as $x$ approaches 0 of the function $$\frac{x^2 \sin\left(\frac{1}{x^2}\right)}{x}.$$\n\n2. **Simplify the expression:** Notice that the denominator is $x$, so we can simplify the fraction:\n$$\frac{x^2 \sin\left(\frac{1}{x^2}\right)}{x} = x \sin\left(\frac{1}{x^2}\right).$$\n\n3. **Recall the properties of sine:** The sine function is bounded between -1 and 1, so:\n$$-1 \leq \sin\left(\frac{1}{x^2}\right) \leq 1.$$\n\n4. **Use the Squeeze Theorem:** Multiply the inequality by $x$ (which approaches 0):\n$$-x \leq x \sin\left(\frac{1}{x^2}\right) \leq x.$$\n\n5. **Evaluate the limits of the bounds:** As $x \to 0$, both $-x$ and $x$ approach 0:\n$$\lim_{x \to 0} -x = 0, \quad \lim_{x \to 0} x = 0.$$\n\n6. **Conclude the limit:** By the Squeeze Theorem,\n$$\lim_{x \to 0} x \sin\left(\frac{1}{x^2}\right) = 0.$$\n\n**Final answer:** $$\boxed{0}.$$