To solve the limit:

$\lim_{x \to 0} \frac{\sin(8x)}{\tan(2x)}$

We can use the small-angle approximation for sine and tangent, which states that as $x \to 0$,

$\sin(x) \approx x \quad \text{and} \quad \tan(x) \approx x.$

Let's break it down step-by-step:

1. The given limit is:

$\lim_{x \to 0} \frac{\sin(8x)}{\tan(2x)}.$

2. Applying the small-angle approximations:

$\sin(8x) \approx 8x \quad \text{and} \quad \tan(2x) \approx 2x \quad \text{as} \quad x \to 0.$

3. Substituting these approximations into the original expression:

$\lim_{x \to 0} \frac{8x}{2x}.$

4. Simplifying the expression:

$\lim_{x \to 0} \frac{8x}{2x} = \lim_{x \to 0} 4 = 4.$

Thus, the value of the limit is:

$\boxed{4}.$

Would you like more details on any of these steps or have further questions?

Here are five related questions to expand on this topic:

1. What happens if you substitute other values of $x$ in the limit equation?
2. How does L'Hopital's Rule apply to solving limits like this?
3. Can the small-angle approximations be used for larger values of $x$?
4. What is the geometric interpretation of the small-angle approximation for sine and tangent?
5. How does the limit change if you replace the functions with their hyperbolic counterparts, like $\sinh(x)$ and $\tanh(x)$?

Tip: When solving limits involving trigonometric functions near zero, small-angle approximations are a powerful tool for simplifying expressions.