To evaluate the limit $\lim_{x \to 0} \frac{\sin(3x)}{x}$, we can use the standard trigonometric limit:

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

We will manipulate the expression to use this standard limit.

Rewrite the expression by introducing a factor of 3 in the denominator:

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

Now, notice that the limit inside the product can be recognized as a form of the standard trigonometric limit:

$\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1.$

Thus, we can evaluate the original limit as follows:

$\lim_{x \to 0} \frac{\sin(3x)}{x} = \left( \lim_{x \to 0} \frac{\sin(3x)}{3x} \right) \cdot 3 = 1 \cdot 3 = 3.$

Therefore, the limit is:

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

Would you like further details or have any questions?

Here are 5 related questions that expand on this topic:

1. How would you evaluate $\lim_{x \to 0} \frac{\sin(5x)}{x}$?
2. What is the limit $\lim_{x \to 0} \frac{\sin(x)}{\sin(2x)}$?
3. How can L'Hopital's rule be applied to solve limits involving trigonometric functions?
4. Why is the standard limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ important in calculus?
5. How would you find $\lim_{x \to 0} \frac{\tan(2x)}{x}$?

Tip: Whenever evaluating limits involving trigonometric functions, see if you can rewrite them to use standard limits like $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$.