To solve the limit:

$\lim_{x \to 0} \frac{\tan(2x) - 2x}{x^3}$

Step 1: Use the Taylor expansion of $\tan(2x)$ around $x = 0$.

The Taylor series expansion of $\tan(2x)$ around 0 is:

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

Substitute this expansion into the expression:

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

So the expression becomes:

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

Step 2: Take the limit as $x \to 0$.

As $x \to 0$, the higher-order terms $O(x^2)$ go to zero. Thus, the limit is:

$\lim_{x \to 0} \frac{\tan(2x) - 2x}{x^3} = \frac{8}{3}$

Final Answer:

$\lim_{x \to 0} \frac{\tan(2x) - 2x}{x^3} = \frac{8}{3}$

Would you like more details or have any questions? Here are 5 related questions:

1. How is the Taylor series expansion derived?
2. What are common Taylor expansions used in limits?
3. Can you explain the higher-order term notation $O(x^5)$?
4. How does L'Hopital's rule apply to this type of limit?
5. What happens if we replace $\tan(2x)$ with $\sin(2x)$ in this limit?

Tip: For limits involving trigonometric functions near zero, using Taylor series expansions can greatly simplify the solution.