The given problem is:
$\lim_{x \to 0} \frac{\sin(3x)}{\tan\left(\frac{x}{2}\right)}$

Step-by-step Solution:

1. Expand the trigonometric functions around $x = 0$:

   • The Taylor expansion for $\sin(3x)$ around $x = 0$ is: \sin(3x) \approx 3x \quad \text{(since \sin(x) \sim x$for small$x)}.
   • For $\tan\left(\frac{x}{2}\right)$: \tan\left(\frac{x}{2}\right) \approx \frac{x}{2} \quad \text{(since \tan(x) \sim x$for small$x)}.

2. Simplify the fraction: Substitute the approximations: $\frac{\sin(3x)}{\tan\left(\frac{x}{2}\right)} \approx \frac{3x}{\frac{x}{2}}.$

3. Simplify further: $\frac{3x}{\frac{x}{2}} = \frac{3x \cdot 2}{x} = 6.$

4. Final result: $\lim_{x \to 0} \frac{\sin(3x)}{\tan\left(\frac{x}{2}\right)} = 6.$

Would you like a detailed explanation of the Taylor expansions or other approaches?

Here are 5 related questions to explore further:

1. What are the steps to derive Taylor series expansions for trigonometric functions?
2. How does the small-angle approximation apply to $\sin(x)$ and $\tan(x)$?
3. How do limits involving trigonometric functions differ when higher-order terms are considered?
4. How would this limit change if $\tan(x)$ or $\sin(x)$ were replaced with other trigonometric functions?
5. How can L’Hôpital’s Rule be used to solve this limit instead of Taylor expansions?

Tip: When solving limits with trigonometric functions, always consider whether small-angle approximations or L’Hôpital’s Rule is more efficient!