1. **State the problem:** Find the limit $$\lim_{x \to 0} 4x \sin\left(\frac{1}{3x}\right)$$. 2. **Recall the limit property:** The sine function is bounded between -1 and 1, i.e., $$-1 \leq \sin(\theta) \leq 1$$ for any real number $$\theta$$. 3. **Apply the squeeze theorem:** Since $$\sin\left(\frac{1}{3x}\right)$$ oscillates between -1 and 1, multiply these inequalities by $$4x$$: $$-4x \leq 4x \sin\left(\frac{1}{3x}\right) \leq 4x$$ 4. **Evaluate the limits of the bounding functions as $$x \to 0$$:** $$\lim_{x \to 0} -4x = 0$$ $$\lim_{x \to 0} 4x = 0$$ 5. **By the squeeze theorem, the original limit is:** $$\lim_{x \to 0} 4x \sin\left(\frac{1}{3x}\right) = 0$$ **Final answer:** $$0$$