1. **Problem:** Evaluate $$\lim_{x \to -1} \frac{x^2 - x - 2}{x^3 - 6x^2 - 7x}$$ 2. **Formula and rules:** When direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, factor numerator and denominator to simplify. 3. **Step-by-step solution:** - Substitute $$x = -1$$: $$(-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0$$ $$(-1)^3 - 6(-1)^2 - 7(-1) = -1 - 6 + 7 = 0$$ So, we have $$\frac{0}{0}$$ indeterminate form. - Factor numerator: $$x^2 - x - 2 = (x - 2)(x + 1)$$ - Factor denominator: $$x^3 - 6x^2 - 7x = x(x^2 - 6x - 7)$$ Factor quadratic: $$x^2 - 6x - 7 = (x - 7)(x + 1)$$ So denominator is: $$x(x - 7)(x + 1)$$ - Simplify the expression: $$\frac{(x - 2)(x + 1)}{x(x - 7)(x + 1)} = \frac{x - 2}{x(x - 7)}$$ (cancel $$x + 1$$) - Substitute $$x = -1$$ now: $$\frac{-1 - 2}{-1(-1 - 7)} = \frac{-3}{-1 \times -8} = \frac{-3}{8}$$ 4. **Answer:** $$\lim_{x \to -1} \frac{x^2 - x - 2}{x^3 - 6x^2 - 7x} = -\frac{3}{8}$$ This completes the solution for the first problem using the concept of indeterminate forms and algebraic simplification.