The problem asks for the limit of the following trigonometric function as $x \to 0$:

$\lim_{{x \to 0}} \frac{{\sin(3x)}}{x}$

Solution:

To solve this, we can use the standard limit property:

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

where $k$ is a constant. In this case, $k = 3$.

Therefore, applying this property to the given expression:

$\lim_{{x \to 0}} \frac{{\sin(3x)}}{x} = 3$

Thus, the solution to the limit is:

$\boxed{3}$

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

Here are some related questions you might find useful:

1. How can we generalize this limit rule for other trigonometric functions?
2. What would happen if the limit involved a different coefficient, like $\sin(5x)$?
3. How is the limit $\lim_{x \to 0} \frac{\sin(x)}{x}$ derived?
4. What other types of limits are common in trigonometry?
5. How would this change if the limit was $\lim_{x \to 0} \frac{\cos(3x)}{x}$?

Tip: When working with limits involving trigonometric functions, remember key standard limits like $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, as these often simplify the problem.