Find each limit for the function \( f(x)=\frac{x^{2}-2 x-15}{x-5} \). Use \( -\infty \) and \( \infty \) when appropriale. (A) \( \lim _{x \rightarrow 5^{-}} f(x) \) (B) \( \lim _{x \rightarrow 5^{+}} f(x) \) (C) \( \lim _{x \rightarrow 5} f(x) \) (A) Select the correct choice and, if necessary, till in the answer box in your choice. A. \( \lim _{x \rightarrow 5^{-}} f(x)=\square \) (Simplify your answer.) B. The limit does not exist and is neither - \( \infty \) nor \( \infty \).

To find the limits for the function \( f(x)=\frac{x^{2}-2x-15}{x-5} \), we can first simplify the function. We begin by factoring the numerator: \[ x^{2} - 2x - 15 = (x - 5)(x + 3) \] Now, substituting this back into \( f(x) \) gives us: \[ f(x) = \frac{(x-5)(x+3)}{x-5} \] As long as \( x \neq 5 \), we can simplify it to: \[ f(x) = x + 3 \] Now we can calculate the limits. (A) \( \lim _{x \rightarrow 5^{-}} f(x) \): As \( x \) approaches 5 from the left, we substitute \( x \) into the simplified function: \[ \lim _{x \rightarrow 5^{-}} f(x) = 5 + 3 = 8 \] So, A. \( \lim _{x \rightarrow 5^{-}} f(x)=8 \). (B) \( \lim _{x \rightarrow 5^{+}} f(x) \): As \( x \) approaches 5 from the right, we substitute \( x \) into the simplified function: \[ \lim _{x \rightarrow 5^{+}} f(x) = 5 + 3 = 8 \] (C) \( \lim _{x \rightarrow 5} f(x) \): This limit is also evaluated the same way for both sides: \[ \lim _{x \rightarrow 5} f(x) = 5 + 3 = 8 \] Thus, we conclude: A. \( \lim _{x \rightarrow 5^{-}} f(x) = 8 \) B. The limit does not exist and is neither - \( \infty \) nor \( \infty \). is incorrect; instead, both one-sided limits exist and equal 8. Overall, the final limits are: A. \( \lim _{x \rightarrow 5^{-}} f(x) = 8 \) B. \( \lim _{x \rightarrow 5^{+}} f(x) = 8 \) C. \( \lim _{x \rightarrow 5} f(x) = 8 \)