To solve the limit $\lim_{x \to 0} \frac{\sin(6x)}{\sin(2x)}$, let's approach it step-by-step.

Step 1: Rewrite the expression using known limits

We can use the fact that $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$ for any constant $k$.

Now, let's rewrite the given limit as follows:

$\lim_{x \to 0} \frac{\sin(6x)}{\sin(2x)} = \lim_{x \to 0} \frac{\sin(6x) / x}{\sin(2x) / x}$

Step 2: Apply the limits

Now apply the known limits for sine at small values of $x$: $\lim_{x \to 0} \frac{\sin(6x) / x}{\sin(2x) / x} = \frac{6}{2} = 3$

Thus, the limit evaluates to:

$\boxed{3}$

Would you like more details or have any other questions?

Here are five related questions:

1. What is the limit of $\lim_{x \to 0} \frac{\sin(4x)}{\sin(2x)}$?
2. How does the small angle approximation for sine apply to this problem?
3. How can L'Hopital's Rule be applied to compute similar limits?
4. What is the general form of limits involving $\frac{\sin(kx)}{\sin(mx)}$?
5. How would you evaluate $\lim_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)}$?

Tip: When working with trigonometric limits, small angle approximations like $\sin(x) \approx x$ near zero are often very helpful.