1. **State the problem:** Find the limit $$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$$. 2. **Recognize the indeterminate form:** Substituting $x=9$ directly gives $$\frac{\sqrt{9} - 3}{9 - 9} = \frac{3 - 3}{0} = \frac{0}{0}$$ which is indeterminate. 3. **Use algebraic manipulation:** Multiply numerator and denominator by the conjugate of the numerator to simplify: $$\frac{\sqrt{x} - 3}{x - 9} \times \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$$ 4. **Simplify numerator using difference of squares:** $$ (\sqrt{x})^2 - 3^2 = x - 9 $$ So the expression becomes: $$ \frac{x - 9}{(x - 9)(\sqrt{x} + 3)} $$ 5. **Cancel common factor $x - 9$:** $$ \frac{\cancel{x - 9}}{\cancel{x - 9}(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3} $$ 6. **Evaluate the limit by direct substitution:** $$ \lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} $$ **Final answer:** $$\boxed{\frac{1}{6}}$$