1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{\sqrt[3]{27x - 81}}{x^2 - 2x - 3}$$. 2. **Analyze the expression:** The numerator is the cube root of $$27x - 81$$ and the denominator is $$x^2 - 2x - 3$$. 3. **Simplify inside the cube root:** $$27x - 81 = 27(x - 3)$$. 4. **Factor the denominator:** $$x^2 - 2x - 3 = (x - 3)(x + 1)$$. 5. **Rewrite the limit:** $$\lim_{x \to 3} \frac{\sqrt[3]{27(x - 3)}}{(x - 3)(x + 1)} = \lim_{x \to 3} \frac{\sqrt[3]{27} \sqrt[3]{x - 3}}{(x - 3)(x + 1)} = \lim_{x \to 3} \frac{3 \sqrt[3]{x - 3}}{(x - 3)(x + 1)}$$. 6. **Simplify the expression by canceling terms:** We can write: $$\frac{3 \sqrt[3]{x - 3}}{(x - 3)(x + 1)} = \frac{3 (x - 3)^{1/3}}{(x - 3)(x + 1)} = \frac{3}{(x + 1)} \cdot \frac{(x - 3)^{1/3}}{(x - 3)} = \frac{3}{(x + 1)} (x - 3)^{-2/3}$$. 7. **Evaluate the limit:** As $$x \to 3$$, $$x + 1 \to 4$$ and $$ (x - 3)^{-2/3} \to \infty$$ because the exponent is negative and the base approaches zero. Therefore, $$\lim_{x \to 3} \frac{3}{4} (x - 3)^{-2/3} = \infty$$. 8. **Conclusion:** The limit diverges to infinity. **Final answer:** $$\infty$$.