23. \( \lim _{x \rightarrow 4} \frac{x^{2}-9 x+20}{x^{2}-3 x-4} \)

To find the limit \( \lim _{x \rightarrow 4} \frac{x^{2}-9 x+20}{x^{2}-3 x-4} \), we first substitute \( x = 4 \) into the expression. Calculating the numerator: \[ x^2 - 9x + 20 = 4^2 - 9 \cdot 4 + 20 = 16 - 36 + 20 = 0. \] Now calculating the denominator: \[ x^2 - 3x - 4 = 4^2 - 3 \cdot 4 - 4 = 16 - 12 - 4 = 0. \] Since both the numerator and the denominator approach \( 0 \) as \( x \) approaches \( 4 \), we use L'Hôpital's Rule or factorization to evaluate the limit. First, we factor both the numerator and the denominator. The numerator \( x^2 - 9x + 20 \) can be factored as: \[ (x-4)(x-5), \] and the denominator \( x^2 - 3x - 4 \) can be factored as: \[ (x-4)(x+1). \] So, we rewrite the limit as: \[ \lim_{x \to 4} \frac{(x-4)(x-5)}{(x-4)(x+1)}. \] For \( x \neq 4 \), we can cancel the \( (x-4) \) term: \[ \lim_{x \to 4} \frac{x-5}{x+1}. \] Now substituting \( x = 4 \) gives: \[ \frac{4-5}{4+1} = \frac{-1}{5}. \] Thus, the limit is: \[ \boxed{-\frac{1}{5}}. \]