The problem requires finding the limit:

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

Step-by-Step Solution

1. Analyze the Denominator: For small values of $x$, we can use the Taylor series expansion for $\tan(x)$ around $x = 0$: $\tan(x) = x + \frac{x^3}{3} + O(x^5)$ Therefore, $x - \tan(x) = x - \left(x + \frac{x^3}{3} + O(x^5)\right) = -\frac{x^3}{3} + O(x^5)$

2. Rewrite the Limit: Substitute the approximation for $x - \tan(x)$: $\lim_{{x \to 0}} \frac{x^3}{x - \tan(x)} = \lim_{{x \to 0}} \frac{x^3}{-\frac{x^3}{3} + O(x^5)}$

3. Simplify the Expression: As $x \to 0$, the $O(x^5)$ term becomes negligible compared to $-\frac{x^3}{3}$, so we approximate: $\lim_{{x \to 0}} \frac{x^3}{-\frac{x^3}{3}} = \lim_{{x \to 0}} -3 = -3$

Final Answer

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

Would you like further clarification on any step?

Related Questions:

1. How can we use Taylor series to approximate other trigonometric functions around zero?
2. Why does the higher-order term $O(x^5)$ become negligible in this problem?
3. How can L'Hôpital's Rule be applied to evaluate limits involving indeterminate forms?
4. What is the general process for finding limits of ratios that go to indeterminate forms?
5. How would the solution change if the power in the numerator was different?

Tip:

Taylor series expansions are useful for approximating complex functions near specific points, especially for calculating limits and simplifying expressions.