1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{6 + x - x^2}{12 - x - x^2}$$. 2. **Recall the limit rule:** If direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, try to simplify the expression. 3. **Substitute $x=3$ directly:** $$\frac{6 + 3 - 3^2}{12 - 3 - 3^2} = \frac{6 + 3 - 9}{12 - 3 - 9} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** - Numerator: $$6 + x - x^2 = -x^2 + x + 6 = -(x^2 - x - 6)$$ - Factor inside parentheses: $$x^2 - x - 6 = (x - 3)(x + 2)$$ - So numerator: $$-(x - 3)(x + 2)$$ - Denominator: $$12 - x - x^2 = -x^2 - x + 12 = -(x^2 + x - 12)$$ - Factor inside parentheses: $$x^2 + x - 12 = (x - 3)(x + 4)$$ - So denominator: $$-(x - 3)(x + 4)$$ 5. **Simplify the fraction:** $$\frac{-(x - 3)(x + 2)}{-(x - 3)(x + 4)} = \frac{(x - 3)(x + 2)}{(x - 3)(x + 4)}$$ Cancel out the common factor $(x - 3)$ (valid since $x \to 3$, $x \neq 3$): $$= \frac{x + 2}{x + 4}$$ 6. **Evaluate the simplified limit:** $$\lim_{x \to 3} \frac{x + 2}{x + 4} = \frac{3 + 2}{3 + 4} = \frac{5}{7}$$ **Final answer:** $$\boxed{\frac{5}{7}}$$