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, so it factors as $$ (x - 3)(x + 3) $$. 4. **Rewrite the expression:** $$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3}$$ 5. **Cancel common factors:** $$\frac{\cancel{(x - 3)}(x + 3)}{\cancel{(x - 3)}} = x + 3$$ 6. **Evaluate the limit:** Now substitute $$x = 3$$: $$3 + 3 = 6$$ **Final answer:** $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$$