1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$$. 2. **Recall the formula and rules:** When direct substitution results in an indeterminate form like $$\frac{0}{0}$$, we try to simplify the expression. 3. **Simplify the numerator:** Notice that $$x^2 - 9$$ is a difference of squares, which factors as $$x^2 - 9 = (x - 3)(x + 3)$$. 4. **Rewrite the limit:** $$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$$ 5. **Cancel common factors:** Since $$x \neq 3$$ in the limit process, we can cancel $$x - 3$$: $$\lim_{x \to 3} \frac{\cancel{(x - 3)}(x + 3)}{\cancel{x - 3}} = \lim_{x \to 3} (x + 3)$$ 6. **Evaluate the simplified limit:** Substitute $$x = 3$$: $$3 + 3 = 6$$ 7. **Final answer:** $$\boxed{6}$$