$$ \operatorname { im } _ { \rightarrow 9 } ( \frac { x - 9 } { \sqrt { x } - 3 } ) $$

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The limit is 6. To find the limit $$ \lim_{x \to 9} \left( \frac{x - 9}{\sqrt{x} - 3} \right), $$ we first substitute $$ x = 9 $$: $$ \frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{3 - 3} = \frac{0}{0}, $$ which is an indeterminate form. To resolve this, we can simplify the expression. Notice that $$ \sqrt{x} - 3 $$ can be factored using the difference of squares. We can multiply the numerator and the denominator by $$ \sqrt{x} + 3 $$: $$ \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}. $$ The denominator simplifies as follows: $$ (\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9. $$ Thus, we have: $$ \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}. $$ For $$ x \neq 9 $$, we can cancel $$ x - 9 $$: $$ \sqrt{x} + 3. $$ Now we can take the limit as $$ x $$ approaches 9: $$ \lim_{x \to 9} (\sqrt{x} + 3) = \sqrt{9} + 3 = 3 + 3 = 6. $$ Therefore, the limit is $$ \boxed{6}. $$

\( \operatorname { im } _ { \rightarrow 9 } ( \frac { x - 9 } { \sqrt { x } - 3 } ) \)

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\( \operatorname { im } _ { \rightarrow 9 } ( \frac { x - 9 } { \sqrt { x } - 3 } ) \)

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