If `f(9)=9" and f'"(9)=4` then `Lt_(x to 9)(sqrt(f(x))-3)/(sqrt(x)-3)` is equal to

Options

A. 2
B. 3
C. 4
D. `-4`

Answer

To solve the limit problem \(\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3}\), given that \(f(9) = 9\) and \(f'(9) = 4\), we can follow these steps:

### Step-by-Step Solution:

1. **Substitution**: Start by substituting \(x = 9\) into the limit expression.
\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = \frac{\sqrt{f(9)} - 3}{\sqrt{9} - 3} = \frac{\sqrt{9} - 3}{3 - 3} = \frac{3 - 3}{0} = \frac{0}{0} \]
This is an indeterminate form.

**Hint**: If you encounter a \(0/0\) form, consider using L'Hôpital's rule.

2. **Applying L'Hôpital's Rule**: Since we have an indeterminate form, we can apply L'Hôpital's rule, which states that we can take the derivative of the numerator and the denominator separately.
\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{\frac{d}{dx}(\sqrt{f(x)})}{\frac{d}{dx}(\sqrt{x})} \]

3. **Differentiate the Numerator and Denominator**:
- The derivative of the numerator \(\sqrt{f(x)}\) is:
\[ \frac{d}{dx}(\sqrt{f(x)}) = \frac{f'(x)}{2\sqrt{f(x)}} \]
- The derivative of the denominator \(\sqrt{x}\) is:
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

4. **Substituting Back into the Limit**:
\[ \lim_{x \to 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to 9} \frac{f'(x) \cdot 2\sqrt{x}}{2\sqrt{f(x)}} \]
This simplifies to:
\[ \lim_{x \to 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}} \]

5. **Evaluate the Limit**: Now substitute \(x = 9\):
\[ = \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{3} = 4 \]

### Final Answer:
Thus, the limit is:
\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = 4 \]