The given limit expression is:

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

Step-by-Step Solution:

1. Expand $\tan(x)$ and $\sin(x)$ using Taylor series expansions around $x = 0$:

   • The Taylor series for $\tan(x)$ around 0 is: $\tan(x) = x + \frac{x^3}{3} + O(x^5)$
   • The Taylor series for $\sin(x)$ around 0 is: $\sin(x) = x - \frac{x^3}{6} + O(x^5)$

2. Subtract $\tan(x)$ and $\sin(x)$:

   = \frac{x^3}{3} + \frac{x^3}{6} = \frac{2x^3}{6} + \frac{x^3}{6} = \frac{x^3}{2}$$

3. Substitute the result into the original limit: $\lim_{{x \to 0}} \frac{\tan(x) - \sin(x)}{x^3} = \lim_{{x \to 0}} \frac{\frac{x^3}{2}}{x^3} = \frac{1}{2}$

Thus, the value of the limit is:

$\boxed{\frac{1}{2}}$

Would you like further details on the steps?

Here are 5 related questions to expand on this concept:

1. How are Taylor series used to approximate functions near a point?
2. What is the Taylor series expansion of $\cos(x)$ around $x = 0$?
3. How do you handle limits involving higher powers of $x$?
4. Can you find the limit $\lim_{{x \to 0}} \frac{\sin(x) - x}{x^3}$?
5. What is L'Hopital's Rule, and could it be applied to solve this limit?

Tip: Taylor series expansion is very useful when working with limits, especially for small values of $x$ where the higher order terms can be ignored for simplification.