If `f(9)=9" and "f'(9)=4," then "underset(xto9)("lim")(sqrt(f(x))-3)/(sqrt(x)-3)=`

Options

A. `-4`
B. 4
C. 2
D. 1

Answer

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

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

\[ \frac{\sqrt{f(9)} - 3}{\sqrt{9} - 3} = \frac{\sqrt{9} - 3}{3 - 3} = \frac{3 - 3}{0} = \frac{0}{0} \]

This gives us an indeterminate form \(0/0\).

**Hint**: When you encounter \(0/0\), consider using L'Hôpital's Rule.

### Step 2: Apply 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 derivative of the denominator:

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

### Step 3: Differentiate the numerator and denominator
Now we differentiate the numerator and the denominator:

- The derivative of \(\sqrt{f(x)}\) using the chain rule is:
\[ \frac{1}{2\sqrt{f(x)}} \cdot f'(x) \]

- The derivative of \(\sqrt{x}\) is:
\[ \frac{1}{2\sqrt{x}} \]

Thus, we have:

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

### Step 4: Simplify the expression
The \(2\) in the numerator and denominator cancels out:

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

### Step 5: Substitute \(x = 9\) again
Now we substitute \(x = 9\) into the simplified limit:

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

Given that \(f(9) = 9\) and \(f'(9) = 4\):

\[ \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4 \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} = 4 \]

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