The limit of $f(x) = \frac{\sin(5x)}{x}$ as $x$ approaches 0 can be evaluated using standard trigonometric limit properties.

We know that:

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

Let's manipulate the given function $\frac{\sin(5x)}{x}$. We can rewrite it as:

$\frac{\sin(5x)}{x} = \frac{\sin(5x)}{5x} \cdot 5$

Now, using the fact that:

$\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1$

We multiply by 5:

$\lim_{x \to 0} \frac{\sin(5x)}{x} = 5$

Thus, the limit is:

$\boxed{5}$

Would you like more details on this process or have any other questions?

Here are 5 related questions:

1. What is the limit of $\frac{\sin(kx)}{x}$ as $x \to 0$ for any constant $k$?
2. Can you use L'Hopital's Rule to solve this limit? How?
3. How does the small-angle approximation for sine function help in evaluating limits?
4. What would the limit be if we had $\frac{\sin(5x)}{x^2}$ as $x \to 0$?
5. What happens to the limit when you replace sine with another function like cosine?

Tip: Whenever dealing with trigonometric limits, especially involving sine and cosine, the small-angle approximation is very useful for simplifying expressions.