Answer
$6$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right) \cr
& \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} = \frac{{9 - 9}}{{3 - \sqrt 9 }} = \frac{0}{0}{\text{Indeterminate}} \cr
& \cr
& {\text{Rationalizing }} \cr
& \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} \times \frac{{3 + \sqrt x }}{{3 + \sqrt x }} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{\left( {9 - x} \right)\left( {3 + \sqrt x } \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}} \cr
& {\text{Recall that }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{\left( {9 - x} \right)\left( {3 + \sqrt x } \right)}}{{{{\left( 3 \right)}^2} - {{\left( {\sqrt x } \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& = \mathop {\lim }\limits_{x \to 9} \frac{{\left( {9 - x} \right)\left( {3 + \sqrt x } \right)}}{{9 - x}} \cr
& = \mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right) \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right) = 3 + \sqrt 9 = 6 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} = 6 \cr} $$
