1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\sin 2x}{3x}$$. 2. **Recall the standard limit:** We know that $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. 3. **Rewrite the expression:** $$\lim_{x \to 0} \frac{\sin 2x}{3x} = \lim_{x \to 0} \frac{\sin 2x}{2x} \cdot \frac{2x}{3x}$$ 4. **Simplify the fraction:** $$\lim_{x \to 0} \frac{\sin 2x}{2x} \cdot \frac{\cancel{2x}}{\cancel{3x}} = \lim_{x \to 0} \frac{\sin 2x}{2x} \cdot \frac{2}{3}$$ 5. **Evaluate the limits separately:** - $$\lim_{x \to 0} \frac{\sin 2x}{2x} = 1$$ (by the standard limit rule) - $$\frac{2}{3}$$ is constant 6. **Multiply the results:** $$1 \times \frac{2}{3} = \frac{2}{3} \approx 0.67$$ **Final answer:** $$0.67$$