1. **Problem:** Find the value of the limit $$\lim_{x \to 3} \frac{x^3 - 27}{x - 3}$$. 2. **Recall the formula:** 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 of $$f(x)$$ at $$x = 3$$. 3. **Factor the numerator:** Use the difference of cubes formula: $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$. 4. **Simplify the expression:** $$\frac{x^3 - 27}{x - 3} = \frac{(x - 3)(x^2 + 3x + 9)}{x - 3}$$ Cancel the common factor: $$\frac{\cancel{(x - 3)}(x^2 + 3x + 9)}{\cancel{(x - 3)}} = x^2 + 3x + 9$$. 5. **Evaluate the limit:** Substitute $$x = 3$$: $$3^2 + 3 \times 3 + 9 = 9 + 9 + 9 = 27$$. 6. **Conclusion:** The limit is $$27$$. The limit theorem used is the factorization and cancellation of common factors to resolve the indeterminate form.