Yes, the expression $\frac{\sin(3x)}{3x}$ converges to a value as $x$ approaches 0.

To understand this, let's evaluate the limit:

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

We can use the standard limit result:

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

To apply this result, notice that if we let $u = 3x$, then as $x \to 0$, $u \to 0$ as well. The expression can be rewritten in terms of $u$:

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

According to the standard limit:

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

Therefore:

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

This shows that the expression $\frac{\sin(3x)}{3x}$ converges to the value 1 as $x$ approaches 0.

Would you like further details or have any questions?

Here are 5 related questions that expand on this topic:

1. How does the limit $\lim_{x \to 0} \frac{\sin(kx)}{kx}$ change with different values of $k$?
2. What is the significance of the limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ in calculus?
3. How can you apply L'Hopital's rule to evaluate $\lim_{x \to 0} \frac{\sin(3x)}{3x}$?
4. How would you evaluate $\lim_{x \to 0} \frac{\cos(x) - 1}{x}$?
5. What other trigonometric limits are useful to know for calculus?

Tip: When working with limits of trigonometric functions near 0, try to express them in a form involving $\frac{\sin(x)}{x}$ or similar standard limits for easier evaluation.