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

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 64}\left(\frac{x-8}{x^{2}-8x}\right)\) - step1: Rewrite the expression: \(\frac{\lim _{x\rightarrow 64}\left(x-8\right)}{\lim _{x\rightarrow 64}\left(x^{2}-8x\right)}\) - step2: Calculate: \(\frac{56}{\lim _{x\rightarrow 64}\left(x^{2}-8x\right)}\) - step3: Calculate: \(\frac{56}{3584}\) - step4: Calculate: \(\frac{1}{64}\) Calculate the limit \( \lim_{x\rightarrow 8} \frac{x-8}{x^{2}-8 x} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow 8}\left(\frac{x-8}{x^{2}-8x}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow 8}\left(\frac{1}{x}\right)\) - step2: Rewrite the expression: \(\frac{\lim _{x\rightarrow 8}\left(1\right)}{\lim _{x\rightarrow 8}\left(x\right)}\) - step3: Evaluate: \(\frac{1}{\lim _{x\rightarrow 8}\left(x\right)}\) - step4: Evaluate: \(\frac{1}{8}\) Calculate the limit \( \lim_{x\rightarrow 0} \frac{x-8}{x^{2}-8 x} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow 0}\left(\frac{x-8}{x^{2}-8x}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow 0}\left(\frac{1}{x}\right)\) - step2: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow 0^{-}}\left(\frac{1}{x}\right)\\&\lim _{x\rightarrow 0^{+}}\left(\frac{1}{x}\right)\end{align}\) - step3: Evaluate the left-hand limit: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step4: Limit does not exist: \(\textrm{The limit does not exist}\) Let's analyze the limits step by step: ### (A) \( \lim_{x \rightarrow 0} f(x) \) For this limit, we have: \[ f(x) = \frac{x-8}{x^{2}-8x} \] Substituting \( x = 0 \): \[ f(0) = \frac{0-8}{0^{2}-8 \cdot 0} = \frac{-8}{0} \] Since the denominator approaches zero while the numerator is non-zero, the limit does not exist. **Answer:** A. \( \lim_{x \rightarrow 0} \frac{x-8}{x^{2}-8 x} = \text{The limit does not exist} \) ### (B) \( \lim_{x \rightarrow 8} f(x) \) Now, substituting \( x = 8 \): \[ f(8) = \frac{8-8}{8^{2}-8 \cdot 8} = \frac{0}{0} \] This is an indeterminate form, so we need to simplify \( f(x) \): \[ f(x) = \frac{x-8}{x(x-8)} = \frac{1}{x} \quad \text{for } x \neq 8 \] Now, we can find the limit as \( x \) approaches 8: \[ \lim_{x \rightarrow 8} f(x) = \lim_{x \rightarrow 8} \frac{1}{x} = \frac{1}{8} \] **Answer:** B. \( \lim_{x \rightarrow 8} \frac{x-8}{x^{2}-8 x} = \frac{1}{8} \) ### (C) \( \lim_{x \rightarrow 64} f(x) \) Using the simplified form: \[ \lim_{x \rightarrow 64} f(x) = \lim_{x \rightarrow 64} \frac{1}{x} = \frac{1}{64} \] **Answer:** C. \( \lim_{x \rightarrow 64} \frac{x-8}{x^{2}-8 x} = \frac{1}{64} \) ### Summary of Results: - (A) The limit does not exist. - (B) \( \frac{1}{8} \) - (C) \( \frac{1}{64} \)