1. **State the problem:** We need to find the limit $$\lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{x^3}.$$\n\n2. **Recall the formulas and expansions:** For small $x$, use the Taylor series expansions:\n- $\tan(2x) = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$\n- $\sin(2x) = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$\n\n3. **Substitute expansions into the numerator:**\n$$\tan(2x) - \sin(2x) = \left(2x + \frac{8x^3}{3}\right) - \left(2x - \frac{4x^3}{3}\right) + O(x^5) = 2x + \frac{8x^3}{3} - 2x + \frac{4x^3}{3} + O(x^5) = \frac{12x^3}{3} + O(x^5) = 4x^3 + O(x^5).$$\n\n4. **Divide by $x^3$:**\n$$\frac{\tan(2x) - \sin(2x)}{x^3} = \frac{4x^3 + O(x^5)}{x^3} = 4 + O(x^2).$$\n\n5. **Take the limit as $x \to 0$:**\n$$\lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{x^3} = 4.$$\n\n**Final answer:** $$4.$$