Determine \( \lim _{x \rightarrow 9} \frac{9-x}{3-\sqrt{x}} \)

To evaluate the limit \[ \lim _{x \rightarrow 9} \frac{9-x}{3-\sqrt{x}}, \] we first substitute \( x = 9 \): \[ \frac{9-9}{3-\sqrt{9}} = \frac{0}{3-3} = \frac{0}{0}. \] This results in an indeterminate form, so we need to simplify the expression. We can start by rationalizing the denominator. The expression can be rewritten as: \[ \frac{9-x}{3-\sqrt{x}} \cdot \frac{3+\sqrt{x}}{3+\sqrt{x}}. \] Now, we multiply and simplify: \[ \lim_{x \rightarrow 9} \frac{(9-x)(3+\sqrt{x})}{(3-\sqrt{x})(3+\sqrt{x})}. \] The denominator simplifies as follows: \[ (3-\sqrt{x})(3+\sqrt{x}) = 3^2 - (\sqrt{x})^2 = 9 - x. \] Thus, our expression simplifies to: \[ \lim_{x \rightarrow 9} \frac{(9-x)(3+\sqrt{x})}{9-x}. \] Since \( 9-x \) is in both the numerator and the denominator, we can cancel it, where \( x \neq 9 \): \[ \lim_{x \rightarrow 9} (3 + \sqrt{x}). \] Now, substituting \( x = 9 \) into the simplified expression gives: \[ 3 + \sqrt{9} = 3 + 3 = 6. \] Thus, the limit is \[ \boxed{6}. \]