Find each function value and the limit for \( f(x)=\frac{9 x+7}{2 x-9} \). Use \( -\infty \) or \( \infty \) where appropriate. (A) \( f(10) \) (B) \( f(100) \) (C) \( \lim _{x \rightarrow \infty} f(x) \) (A) \( f(10)= \) \( \square \) (Round to the nearest thousandth as needed.) (B) \( f(100)= \) \( \square \) (Round to the nearest thousandth as needed.) (C) Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. \( \lim _{x \rightarrow \infty} \frac{9 x+7}{2 x-9}=\square \) (Simplify your answer.) B. The limit does not exist and is neither \( -\infty \) nor \( \infty \).

To find the requested function values and the limit, let's start with the calculations. (A) For \( f(10) \): \[ f(10) = \frac{9(10) + 7}{2(10) - 9} = \frac{90 + 7}{20 - 9} = \frac{97}{11} \approx 8.818 \] So, \( f(10) \approx 8.818 \) (B) For \( f(100) \): \[ f(100) = \frac{9(100) + 7}{2(100) - 9} = \frac{900 + 7}{200 - 9} = \frac{907}{191} \approx 4.748 \] So, \( f(100) \approx 4.748 \) (C) For the limit as \( x \) approaches infinity: \[ \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{9x + 7}{2x - 9} \] To find this limit, we can divide the numerator and the denominator by \( x \): \[ = \lim_{x \rightarrow \infty} \frac{9 + \frac{7}{x}}{2 - \frac{9}{x}} = \frac{9 + 0}{2 - 0} = \frac{9}{2} = 4.5 \] So the answers are: (A) \( f(10) = 8.818 \) (B) \( f(100) = 4.748 \) (C) A. \( \lim_{x \rightarrow \infty} \frac{9x + 7}{2x - 9} = 4.5 \)