1. **State the problem:** Find the limit $$\lim_{x \to 7} \frac{x^2 - 6x - 7}{2x^2 - 5x - 63}$$. 2. **Recall the limit rule:** If direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, factor numerator and denominator to simplify. 3. **Substitute $x=7$ directly:** $$\text{Numerator} = 7^2 - 6 \times 7 - 7 = 49 - 42 - 7 = 0$$ $$\text{Denominator} = 2 \times 7^2 - 5 \times 7 - 63 = 2 \times 49 - 35 - 63 = 98 - 35 - 63 = 0$$ Direct substitution gives $$\frac{0}{0}$$, so factorization is needed. 4. **Factor numerator:** $$x^2 - 6x - 7 = (x - 7)(x + 1)$$ 5. **Factor denominator:** $$2x^2 - 5x - 63$$ Find factors of $$2 \times (-63) = -126$$ that sum to $$-5$$: $$9$$ and $$-14$$. Rewrite: $$2x^2 + 9x - 14x - 63 = (2x^2 + 9x) - (14x + 63) = x(2x + 9) - 7(2x + 9) = (x - 7)(2x + 9)$$ 6. **Simplify the expression:** $$\frac{(x - 7)(x + 1)}{(x - 7)(2x + 9)} = \frac{x + 1}{2x + 9}, \quad x \neq 7$$ 7. **Evaluate the limit by substitution:** $$\lim_{x \to 7} \frac{x + 1}{2x + 9} = \frac{7 + 1}{2 \times 7 + 9} = \frac{8}{14 + 9} = \frac{8}{23}$$ **Final answer:** $$\boxed{\frac{8}{23}}$$