Evaluate the limit: lim (x->0) [tan(2x) - sin(2x)] / x3

Substituting $x=0$ into the expression gives:

$$\frac{\tan(2 \cdot 0) - \sin(2 \cdot 0)}{0^3} = \frac{\tan(0) - \sin(0)}{0} = \frac{0 - 0}{0} = \frac{0}{0}$$

This is an indeterminate form, so we can use L'Hôpital's Rule or Taylor series expansions. We will use Taylor series expansions.

The Taylor series for $\tan(u)$ up to the third order is:

$$\tan(u) = u + \frac{u^3}{3} + O(u^5)$$

The Taylor series for $\sin(u)$ up to the third order is:

$$\sin(u) = u - \frac{u^3}{3!} + O(u^5)$$

For $\tan(2x)$:

$$\tan(2x) = (2x) + \frac{(2x)^3}{3} + O((2x)^5) = 2x + \frac{8x^3}{3} + O(x^5)$$

For $\sin(2x)$:

$$\sin(2x) = (2x) - \frac{(2x)^3}{3!} + O((2x)^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$$

$$\lim_{x \to 0} \frac{2x + \frac{8x^3}{3} - 2x + \frac{4x^3}{3} + O(x^5)}{x^3}$$

Combine the $x^3$ terms:

$$\lim_{x \to 0} \frac{\left(\frac{8}{3} + \frac{4}{3}\right)x^3 + O(x^5)}{x^3}$$ $$\lim_{x \to 0} \frac{\frac{12}{3}x^3 + O(x^5)}{x^3}$$ $$\lim_{x \to 0} \frac{4x^3 + O(x^5)}{x^3}$$

$$\lim_{x \to 0} \left( \frac{4x^3}{x^3} + \frac{O(x^5)}{x^3} \right)$$ $$\lim_{x \to 0} \left( 4 + O(x^2) \right)$$

As $x \to 0$, the term $O(x^2)$ approaches 0.
Therefore, the limit is:

$$4 + 0 = 4$$

The final answer is $\text{4}$.