1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$$. 2. **Recall the formula and rules:** The expression is a rational function. Direct substitution of $x=3$ gives $$\frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0}$$ which is indeterminate. We use algebraic simplification to resolve this. 3. **Factor the numerator:** $$x^2 - 9 = (x - 3)(x + 3)$$. 4. **Simplify the expression:** $$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3}$$. Since $x \neq 3$ in the limit process, we can cancel $x - 3$: $$= x + 3$$. 5. **Evaluate the limit:** Substitute $x = 3$ into the simplified expression: $$3 + 3 = 6$$. **Final answer:** $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$$.