If `f` be a function such that `f(9)=9` and `f'(9)=3`, then `lim_(xto9)(sqrt(f(x))-3)/(sqrt(x)-3)` is equal to

Options

A. 9
B. 3
C. 1
D. `1/3`

Answer

To solve the limit \( \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} \), we will follow these steps:

### Step 1: Substitute the limit
First, we substitute \( x = 9 \) into the limit expression:

\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = \frac{\sqrt{f(9)} - 3}{\sqrt{9} - 3} \]

Given that \( f(9) = 9 \), we have:

\[ \sqrt{f(9)} = \sqrt{9} = 3 \]

Thus, the limit becomes:

\[ \frac{3 - 3}{3 - 3} = \frac{0}{0} \]

This is an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This involves taking the derivative of the numerator and the derivative of the denominator.

The limit can be rewritten as:

\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} \]

Taking the derivatives:

1. **Numerator**: The derivative of \( \sqrt{f(x)} \) using the chain rule is:

\[ \frac{d}{dx}(\sqrt{f(x)}) = \frac{1}{2\sqrt{f(x)}} \cdot f'(x) \]

2. **Denominator**: The derivative of \( \sqrt{x} \) is:

\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 9} \frac{\frac{1}{2\sqrt{f(x)}} f'(x)}{\frac{1}{2\sqrt{x}}} \]

### Step 3: Simplify the expression
This simplifies to:

\[ \lim_{x \to 9} \frac{f'(x)}{\sqrt{f(x)} \cdot \sqrt{x}} \]

### Step 4: Substitute \( x = 9 \)
Now we can substitute \( x = 9 \):

\[ \frac{f'(9)}{\sqrt{f(9)} \cdot \sqrt{9}} = \frac{f'(9)}{\sqrt{9} \cdot 3} = \frac{f'(9)}{3 \cdot 3} = \frac{f'(9)}{9} \]

Given that \( f'(9) = 3 \):

\[ \frac{3}{9} = \frac{1}{3} \]

### Final Result
Thus, the limit is:

\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = \frac{1}{3} \]