1. **State the problem:** We want to find the limit $$\lim_{x \to \infty} \frac{\sin x}{x^2}$$. 2. **Recall the properties:** The sine function oscillates between -1 and 1 for all real numbers, so $$-1 \leq \sin x \leq 1$$. 3. **Apply the squeeze theorem:** Since $$-1 \leq \sin x \leq 1$$, dividing all parts by $$x^2$$ (which is positive for large $$x$$) gives: $$\frac{-1}{x^2} \leq \frac{\sin x}{x^2} \leq \frac{1}{x^2}$$. 4. **Evaluate the limits of the bounding functions:** $$\lim_{x \to \infty} \frac{-1}{x^2} = 0$$ and $$\lim_{x \to \infty} \frac{1}{x^2} = 0$$. 5. **Conclude by squeeze theorem:** Since both bounding limits are 0, the original limit is also 0: $$\lim_{x \to \infty} \frac{\sin x}{x^2} = 0$$. **Final answer:** 0