\( \lim \rightarrow - 1 \quad \frac { x ^ { 2 } - 9 x - 10 } { x ^ { 2 } - 6 x - 7 } = \)

To evaluate the limit \( \lim_{x \to -1} \frac{x^2 - 9x - 10}{x^2 - 6x - 7} \), we first substitute \( x = -1 \) into the numerator and denominator: - The numerator: \[ (-1)^2 - 9(-1) - 10 = 1 + 9 - 10 = 0 \] - The denominator: \[ (-1)^2 - 6(-1) - 7 = 1 + 6 - 7 = 0 \] Since both the numerator and denominator equal zero, we have an indeterminate form \( \frac{0}{0} \). To resolve this, we factor both the numerator and the denominator. The numerator factors as follows: \[ x^2 - 9x - 10 = (x - 10)(x + 1) \] The denominator factors as follows: \[ x^2 - 6x - 7 = (x - 7)(x + 1) \] Now, we can rewrite the limit: \[ \frac{(x - 10)(x + 1)}{(x - 7)(x + 1)} \] Cancelling the common factor \( (x + 1) \) (valid since we aren't evaluating \( x = -1 \) directly now): \[ \lim_{x \to -1} \frac{x - 10}{x - 7} \] Now we can substitute \( x = -1 \): \[ \frac{-1 - 10}{-1 - 7} = \frac{-11}{-8} = \frac{11}{8} \] Thus, the limit is: \[ \frac{11}{8} \]