Enter an exact answer. \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{x+5}= \) Number

To find the limit \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{x+5} \), we first notice that both the numerator and denominator become zero when \( x = -5 \), leading to the form \(\frac{0}{0}\). We can factor the numerator: \[ x^{2} - 25 = (x + 5)(x - 5) \] Now, we rewrite the limit: \[ \lim _{x \rightarrow-5} \frac{(x + 5)(x - 5)}{x + 5} \] We can cancel the \( (x + 5) \) term (as long as \( x \neq -5 \)): \[ \lim _{x \rightarrow-5} (x - 5) \] Now we can directly substitute \( x = -5 \): \[ -5 - 5 = -10 \] Thus, the exact answer is: \[ \lim _{x \rightarrow-5} \frac{x^{2}-25}{x+5} = -10 \]