1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\tan(6x)}{\sin(2x)}.$$\n\n2. **Recall important formulas and rules:**\n- As $x \to 0$, $\tan x \approx x$ and $\sin x \approx x$.\n- The limit $$\lim_{x \to 0} \frac{\sin(ax)}{x} = a$$ and similarly for tangent.\n\n3. **Apply the approximations:**\n$$\lim_{x \to 0} \frac{\tan(6x)}{\sin(2x)} \approx \lim_{x \to 0} \frac{6x}{2x}.$$\n\n4. **Simplify the fraction:**\n$$\frac{6x}{2x} = \frac{\cancel{6} \cancel{x}}{\cancel{2} \cancel{x}} = \frac{6}{2} = 3.$$\n\n5. **Conclusion:**\nTherefore, $$\lim_{x \to 0} \frac{\tan(6x)}{\sin(2x)} = 3.$$