8. \( \lim _{x \rightarrow 1} \frac{x^{2}-4 x-x-4}{x^{2}+2 x-3} \)

To evaluate the limit \( \lim _{x \rightarrow 1} \frac{x^{2}-4 x-x-4}{x^{2}+2 x-3} \), first simplify the expression. Start with the numerator: \[ x^2 - 4x - x - 4 = x^2 - 5x - 4. \] Next, simplify the denominator: \[ x^2 + 2x - 3 = (x + 3)(x - 1). \] Now, rewrite the limit: \[ \lim_{x \rightarrow 1} \frac{x^2 - 5x - 4}{(x + 3)(x - 1)}. \] Substituting \( x = 1 \) into the numerator: \[ 1^2 - 5(1) - 4 = 1 - 5 - 4 = -8. \] In the denominator: \[ (1 + 3)(1 - 1) = 4 \cdot 0 = 0. \] Since we have a \( \frac{-8}{0} \) form, it indicates we need to look into the values of \( x \) approaching 1 from both sides. The denominator approaches 0 while the numerator approaches -8. Therefore, the limit behaves as follows: For \( x \) approaching 1 from the left (\( x \to 1^- \)): \[ \frac{-8}{0^-} \to +\infty. \] For \( x \) approaching 1 from the right (\( x \to 1^+ \)): \[ \frac{-8}{0^+} \to -\infty. \] Since the left-hand limit and the right-hand limit are not equal, the limit does not exist. Thus, the final answer is: \[ \text{The limit does not exist.} \]