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.

Explore the convergence of the expression (sin(3x))/3x as x approaches 0. Learn how to evaluate the limit using trigonometric identities and theorems. Suitable for advanced high school students and calculus enthusiasts.

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