To find the limit of $\frac{\sin(5x)}{x}$ as $x \to 0$, we can use the standard limit:

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

We will rewrite the expression to match the standard form. Let's consider:

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

We can manipulate this by multiplying and dividing by 5:

$\lim_{x \to 0} \frac{\sin(5x)}{x} = \lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \cdot 5 \right).$

Now, we can separate the constant 5 from the limit:

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

Next, we recognize that if $u = 5x$, as $x \to 0$, $u \to 0$ as well. Therefore:

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

Substitute this result back:

$= 5 \cdot 1 = 5.$

Thus, the limit is:

$\boxed{5}.$

Would you like more details on this solution, or have any questions?

Here are five related questions to explore further:

1. What is the limit of $\frac{\sin(3x)}{x}$ as $x \to 0$?
2. How do you find the limit of $\frac{\tan(x)}{x}$ as $x \to 0$?
3. Why is the limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ fundamental in calculus?
4. How would the limit change if the function were $\frac{\sin(x)}{x^2}$ as $x \to 0$?
5. What is the Taylor series expansion of $\sin(x)$, and how does it relate to limits?

Tip: Remember that many trigonometric limits can be solved using small-angle approximations or Taylor series expansions!