To solve the limit:

$\lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{2x^3 - 1}$

Step-by-step Solution:

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1. Analyze the numerator and denominator behavior as $x \to 0$:

• Numerator: $\tan(2x) - \sin(2x)$ Using Taylor expansions near $x = 0$: $\tan(2x) \approx 2x + \frac{(2x)^3}{3} + \dots$ $\sin(2x) \approx 2x - \frac{(2x)^3}{6} + \dots$ Subtracting the two: $\tan(2x) - \sin(2x) \approx \frac{(2x)^3}{3} + \frac{(2x)^3}{6} = \frac{(2x)^3}{2}.$

  So, as $x \to 0$, the numerator behaves like $\frac{(2x)^3}{2}$.

• Denominator: $2x^3 - 1$ As $x \to 0$, this approaches $-1$.

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2. Simplify the limit:

Now substitute the approximations: $\lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{2x^3 - 1} \approx \lim_{x \to 0} \frac{\frac{(2x)^3}{2}}{2x^3 - 1}.$

Factor the numerator: $\frac{\frac{(2x)^3}{2}}{2x^3 - 1} = \frac{4x^3 / 2}{2x^3 - 1} = \frac{2x^3}{2x^3 - 1}.$

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3. Evaluate the limit:

As $x \to 0$, the numerator goes to $0$, and the denominator approaches $-1$. Thus: $\lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{2x^3 - 1} = \frac{0}{-1} = 0.$

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Final Answer:

$\boxed{0}$

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Would you like me to explain any part in detail or confirm specific steps?

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Related Questions:

1. What is the Taylor series expansion for $\tan(x)$ around $x = 0$?
2. How do you derive the Taylor series for trigonometric functions?
3. Why does the term $x^3$ dominate as $x \to 0$ in this problem?
4. What would the limit be if the denominator were just $2x^3$?
5. How does $\tan(x)$ compare to $\sin(x)$ for small angles?

Tip:

Always use Taylor expansions when solving limits involving trigonometric functions and small $x$. It simplifies the process significantly!