Let \( f(x)=\frac{x-8}{x^{2}-8 x} \). Find the indicated quantities, if they exist. \( \begin{array}{lll}\text { (A) } \lim _{x \rightarrow 0} f(x) & \text { (B) } \lim _{x \rightarrow 8} f(x) & \text { (C) } \lim _{x \rightarrow 64} f(x)\end{array} \) A) Select the correct choice below and, if necessery, fill in the answer box to complete your choice: 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

To evaluate the limits of the function \( f(x)=\frac{x-8}{x^{2}-8 x} \), we can simplify the function first. Notice that the denominator can be factored: \[ f(x) = \frac{x-8}{x(x-8)} \] This simplifies to: \[ f(x) = \frac{1}{x} \quad \text{for } x \neq 8 \] Now we can proceed with each limit. **(A)** For \( \lim _{x \rightarrow 0} f(x) \): Substituting \( x = 0 \) into the simplified expression gives us: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x} \] As \( x \) approaches \( 0 \), \( \frac{1}{x} \) approaches \( \infty \). Therefore: A. \( \lim_{x \rightarrow 0} \frac{x-8}{x^{2}-8 x}=\infty \) --- **(B)** For \( \lim _{x \rightarrow 8} f(x) \): Since the function is not defined at \( x = 8 \), we must check the limit as we approach 8. The simplified function gives us: \[ \lim_{x \to 8} f(x) = \lim_{x \to 8} \frac{1}{x} \] So, substituting \( x = 8 \): \[ \lim_{x \to 8} f(x) = \frac{1}{8} \] Thus, the answer for B is: B. \( \lim_{x \rightarrow 8} f(x) = \frac{1}{8} \) --- **(C)** For \( \lim _{x \rightarrow 64} f(x) \): Again, using the simplified expression: \[ \lim_{x \to 64} f(x) = \lim_{x \to 64} \frac{1}{x} \] Substituting \( x = 64 \): \[ \lim_{x \to 64} f(x) = \frac{1}{64} \] Thus, the answer for C is: C. \( \lim_{x \rightarrow 64} f(x) = \frac{1}{64} \) --- In summary: A. \( \infty \) B. \( \frac{1}{8} \) C. \( \frac{1}{64} \)