1. **Problem:** Find the limit $$\lim_{x \to 3} \frac{x^3 - 27}{x - 3}$$. 2. **Formula and rules:** This is a limit of the form $$\frac{f(x) - f(a)}{x - a}$$ where $$f(x) = x^3$$ and $$a = 3$$. This resembles the definition of the derivative: $$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$$. 3. **Intermediate work:** Factor the numerator using the difference of cubes: $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$. 4. Substitute into the limit expression: $$\lim_{x \to 3} \frac{(x - 3)(x^2 + 3x + 9)}{x - 3}$$. 5. Cancel the common factor $$x - 3$$: $$\lim_{x \to 3} \frac{\cancel{(x - 3)}(x^2 + 3x + 9)}{\cancel{x - 3}} = \lim_{x \to 3} (x^2 + 3x + 9)$$. 6. Evaluate the limit by direct substitution: $$3^2 + 3 \times 3 + 9 = 9 + 9 + 9 = 27$$. **Final answer:** $$27$$.