Evaluate \( \lim _{x \rightarrow-2} \frac{x^{2}-5 x-14}{x^{2}+10 x+16} \). If the corresponding limit is infinity, then insert INF in the blank. If the corresponding limit is negative infinity, then insert -INF in the blank. If the corresponding limit does not exist, then insert. DNE in the blank. Otherwise, write the number as an integer or a reduced fraction.

To evaluate the limit \( \lim _{x \rightarrow -2} \frac{x^{2}-5x-14}{x^{2}+10x+16} \), we will first substitute \( x = -2 \) directly into the function. Calculating the numerator: \[ x^2 - 5x - 14 = (-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0 \] Calculating the denominator: \[ x^2 + 10x + 16 = (-2)^2 + 10(-2) + 16 = 4 - 20 + 16 = 0 \] Since both the numerator and the denominator evaluate to \( 0 \), we have a \( \frac{0}{0} \) indeterminate form. We can use algebraic manipulation to simplify the expression. Factoring the numerator: \[ x^2 - 5x - 14 = (x - 7)(x + 2) \] Factoring the denominator: \[ x^2 + 10x + 16 = (x + 2)(x + 8) \] Now we can rewrite the limit: \[ \frac{(x - 7)(x + 2)}{(x + 2)(x + 8)} \] We can cancel the \( (x + 2) \) term in the numerator and denominator (for \( x \neq -2 \)): \[ \lim _{x \rightarrow -2} \frac{x - 7}{x + 8} \] Now we substitute \( x = -2 \) again: \[ \frac{-2 - 7}{-2 + 8} = \frac{-9}{6} = -\frac{3}{2} \] Thus, the limit is: \[ \lim _{x \rightarrow -2} \frac{x^{2}-5x-14}{x^{2}+10x+16} = -\frac{3}{2} \] Therefore, the answer is: \(-\frac{3}{2}\)