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.

Discover how to solve the limit lim x -> 0 (tan(2x) - 2x)/x^3 using Taylor series expansion and trigonometric function properties. This problem focuses on advanced calculus concepts, including Taylor series and higher-order terms, making it suitable for college-level students.

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