Let \( f(x)=\frac{x-8}{x^{2}-8 x} \). Find the indicated quantities, if they exist. \( \begin{array}{ll}\text { (A) } \lim _{x \rightarrow 0} f(x) & \text { (B) } \lim _{x \rightarrow 8} f(x) \\ \text { (C) } \lim _{x \rightarrow 64} f(x)\end{array} \) A. \( \lim _{x \rightarrow 0} \frac{x-8}{x^{2}-8 x}=\square \) (Type an integer or a simplified fraction.) B. The limit does not exist. (B) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \lim _{x \rightarrow 8} \frac{x-8}{x^{2}-8 x}=\square \) (Type an integer or a simplified fraction.) B. The limit does not exist. (C) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. lim \( \frac{x-8}{x \rightarrow 64} x^{2}-8 x \) \( \begin{array}{l}\text { B. The limit does not exist. }\end{array} \)

To find the requested limits of the function \( f(x) = \frac{x-8}{x^2 - 8x} \), we'll analyze each limit step by step. A. For \( \lim_{x \rightarrow 0} f(x) \): \[ f(0) = \frac{0-8}{0^2 - 8 \cdot 0} = \frac{-8}{0} \] As the denominator approaches zero and the numerator is non-zero, the limit does not exist. Hence, we can state this limit as either positive or negative infinity depending on the direction. B. For \( \lim_{x \rightarrow 8} f(x) \): \[ f(8) = \frac{8-8}{8^2 - 8 \cdot 8} = \frac{0}{0} \] This indicates an indeterminate form, so we must simplify \( f(x) \): \[ f(x) = \frac{x-8}{x(x-8)} = \frac{1}{x} \quad (x \neq 8) \] Now, we can compute the limit: \[ \lim_{x \rightarrow 8} f(x) = \lim_{x \rightarrow 8} \frac{1}{x} = \frac{1}{8} \] C. For \( \lim_{x \rightarrow 64} f(x) \): Using the simplified function: \[ \lim_{x \rightarrow 64} f(x) = \lim_{x \rightarrow 64} \frac{1}{x} = \frac{1}{64} \] In summary: A. \( \lim_{x \rightarrow 0} f(x) = \text{does not exist} \) (since it approaches infinity) B. \( \lim_{x \rightarrow 8} f(x) = \frac{1}{8} \) C. \( \lim_{x \rightarrow 64} f(x) = \frac{1}{64} \)