To solve the given limit:

$\lim_{x \to 0} \frac{2x - \sin(2x)}{x \sin(2x)}$

Step 1: Simplify the numerator

The numerator $2x - \sin(2x)$ can be approximated using the Taylor expansion of $\sin(2x)$ around $x = 0$: $\sin(2x) \approx 2x - \frac{(2x)^3}{6} + \dots = 2x - \frac{8x^3}{6} + \dots$

Thus, $2x - \sin(2x) = \left(2x - 2x + \frac{8x^3}{6}\right) = \frac{4x^3}{3}$

Step 2: Simplify the denominator

The denominator $x \sin(2x)$ can also be approximated. Using $\sin(2x) \approx 2x$, we have: $x \sin(2x) \approx x \cdot 2x = 2x^2$

Step 3: Simplify the fraction

Substituting the approximations into the limit: $\frac{2x - \sin(2x)}{x \sin(2x)} \approx \frac{\frac{4x^3}{3}}{2x^2}$

Simplify: $\frac{\frac{4x^3}{3}}{2x^2} = \frac{4x}{6} = \frac{2x}{3}$

Step 4: Take the limit as $x \to 0$

As $x \to 0$, the expression $\frac{2x}{3} \to 0$. Thus: $\lim_{x \to 0} \frac{2x - \sin(2x)}{x \sin(2x)} = 0$

Final Answer:

$\boxed{0}$

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Would you like a detailed breakdown of any step or additional examples?

Related Questions:

1. How do Taylor series help in solving limits?
2. What is the small-angle approximation for trigonometric functions?
3. Can L'Hôpital's Rule be used to solve this limit? How?
4. What happens if we substitute $\sin(2x)$ directly without expansions?
5. Are there cases where higher-order terms in Taylor expansion are necessary?

Tip:

When dealing with limits involving trigonometric functions, the Taylor series expansion is often very effective, especially near zero!