Question:

<p><strong>Step 1:</strong> Recognize the indeterminate form 0/0. Multiply numerator and denominator by the conjugate of the numerator:</p><p>$$\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} \cdot \frac{\sqrt{f(x)} + 3}{\sqrt{f(x)} + 3} = \lim_{x \to 9} \frac{f(x) - 9}{(\sqrt{x} - 3)(\sqrt{f(x)} + 3)}$$</p><p><strong>Step 2:</strong> Multiply numerator and denominator by the conjugate of the denominator's first factor:</p><p>$$= \lim_{x \to 9} \frac{(f(x) - 9)(\sqrt{x} + 3)}{(x - 9)(\sqrt{f(x)} + 3)}$$</p><p><strong>Step 3:</strong> Rewrite as a product of two limits using the derivative definition:</p><p>$$= \lim_{x \to 9} \frac{f(x) - 9}{x - 9} \cdot \lim_{x \to 9} \frac{\sqrt{x} + 3}{\sqrt{f(x)} + 3}$$</p><p><strong>Step 4:</strong> Apply known values: $\lim_{x \to 9} \frac{f(x) - 9}{x - 9} = f'(9) = 4$ and $\lim_{x \to 9} \frac{\sqrt{x} + 3}{\sqrt{f(x)} + 3} = \frac{6}{6} = 1$</p><p>∴ Answer: $4 \times 1 = \boxed{4}$</p>