1. **State the problem:** Find the limit as $x$ approaches 3 of the expression $$\frac{x^2 - 9}{x - 3}$$. 2. **Recall the formula and rules:** When direct substitution leads to 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 $$x^2 - 9 = (x - 3)(x + 3)$$. 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:** $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$$.