Let \( f(x)=\frac{x^{2}-2 x-48}{x-8} \). Find a) \( \lim _{x \rightarrow 8} f(x) \), b) \( \operatorname{limim}_{x \rightarrow 0} f(x) \), and \( \left.c\right) \lim _{x \rightarrow-6} f(x) \). a) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \lim _{x \rightarrow 8} f(x)=\square \) (Simplify your answer.) B. The limit does not exist.

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 0}\left(\frac{x^{2}-2x-48}{x-8}\right)\) - step1: Rewrite the expression: \(\frac{\lim _{x\rightarrow 0}\left(x^{2}-2x-48\right)}{\lim _{x\rightarrow 0}\left(x-8\right)}\) - step2: Calculate: \(\frac{-48}{\lim _{x\rightarrow 0}\left(x-8\right)}\) - step3: Calculate: \(\frac{-48}{-8}\) - step4: Calculate: \(6\) Calculate the limit \( \lim_{x\rightarrow 8} \frac{x^{2}-2 x-48}{x-8} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow 8}\left(\frac{x^{2}-2x-48}{x-8}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow 8}\left(x+6\right)\) - step2: Rewrite the expression: \(\lim _{x\rightarrow 8}\left(x\right)+\lim _{x\rightarrow 8}\left(6\right)\) - step3: Calculate: \(8+6\) - step4: Calculate: \(14\) Calculate the limit \( \lim_{x\rightarrow -6} \frac{x^{2}-2 x-48}{x-8} \). Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow -6}\left(\frac{x^{2}-2x-48}{x-8}\right)\) - step1: Rewrite the expression: \(\frac{\lim _{x\rightarrow -6}\left(x^{2}-2x-48\right)}{\lim _{x\rightarrow -6}\left(x-8\right)}\) - step2: Calculate: \(\frac{0}{\lim _{x\rightarrow -6}\left(x-8\right)}\) - step3: Calculate: \(\frac{0}{-14}\) - step4: Calculate: \(0\) Let's solve the limits step by step. ### a) \( \lim _{x \rightarrow 8} f(x) \) To find this limit, we first substitute \( x = 8 \) into the function: \[ f(8) = \frac{8^{2} - 2 \cdot 8 - 48}{8 - 8} = \frac{64 - 16 - 48}{0} = \frac{0}{0} \] This is an indeterminate form, so we need to simplify the function. We can factor the numerator: \[ x^{2} - 2x - 48 = (x - 8)(x + 6) \] Thus, we can rewrite \( f(x) \): \[ f(x) = \frac{(x - 8)(x + 6)}{x - 8} \] For \( x \neq 8 \), this simplifies to: \[ f(x) = x + 6 \] Now we can find the limit as \( x \) approaches 8: \[ \lim_{x \rightarrow 8} f(x) = 8 + 6 = 14 \] So, the answer is: A. \( \lim _{x \rightarrow 8} f(x) = 14 \) ### b) \( \lim _{x \rightarrow 0} f(x) \) Now we substitute \( x = 0 \): \[ f(0) = \frac{0^{2} - 2 \cdot 0 - 48}{0 - 8} = \frac{-48}{-8} = 6 \] So, the limit is: \[ \lim_{x \rightarrow 0} f(x) = 6 \] ### c) \( \lim _{x \rightarrow -6} f(x) \) Now we substitute \( x = -6 \): \[ f(-6) = \frac{(-6)^{2} - 2 \cdot (-6) - 48}{-6 - 8} = \frac{36 + 12 - 48}{-14} = \frac{0}{-14} = 0 \] So, the limit is: \[ \lim_{x \rightarrow -6} f(x) = 0 \] ### Summary of Results: - a) \( \lim _{x \rightarrow 8} f(x) = 14 \) - b) \( \lim _{x \rightarrow 0} f(x) = 6 \) - c) \( \lim _{x \rightarrow -6} f(x) = 0 \)