The problem asks you to determine the value of the following limit:

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

Solution:

To solve this limit, we can use the fact that both $\tan(x)$ and $\sin(x)$ can be approximated by their Taylor series expansions around $x = 0$:

$\tan(x) \approx x + \frac{x^3}{3} + \mathcal{O}(x^5)$

$\sin(x) \approx x - \frac{x^3}{6} + \mathcal{O}(x^5)$

Let's substitute these into the limit expression:

$\lim_{x \to 0} \frac{\left( x + \frac{x^3}{3} + 2x \right)}{\left( x - \frac{x^3}{6} + 2x \right)}$

This simplifies to:

$\lim_{x \to 0} \frac{3x + \frac{x^3}{3}}{3x - \frac{x^3}{6}}$

As $x \to 0$, the $x^3$ terms become negligible, so the expression simplifies further to:

$\lim_{x \to 0} \frac{3x}{3x} = \lim_{x \to 0} 1 = 1$

Therefore, the value of the limit is:

$\boxed{1}$

Would you like more details or have any questions?

Related Questions:

1. How do you find the Taylor series of $\tan(x)$ and $\sin(x)$?
2. What is L'Hopital's rule, and how could it be applied here?
3. What are the typical forms where Taylor series expansion is useful in limits?
4. How can you use the small-angle approximation in trigonometric limits?
5. How do higher-order terms in a Taylor expansion affect the calculation of limits?

Tip: When calculating limits involving trigonometric functions, approximating the functions using Taylor series can simplify the problem significantly.