If `f(x )=sqrt(25-x^(2))`, then what is `underset(xto1)lim(f(x)-f(1))/(x-1)` equal to

Options

A. `1/24`
B. ` 1/5`
C. `-sqrt(24)`
D. `1/(sqrt24)`

Answer

To solve the limit problem given by the function \( f(x) = \sqrt{25 - x^2} \), we need to evaluate:

\[ \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} \]

### Step 1: Calculate \( f(1) \)

First, we need to find the value of \( f(1) \):

\[ f(1) = \sqrt{25 - 1^2} = \sqrt{25 - 1} = \sqrt{24} \]

### Step 2: Set up the limit expression

Now we can substitute \( f(1) \) back into our limit expression:

\[ \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1} \frac{\sqrt{25 - x^2} - \sqrt{24}}{x - 1} \]

### Step 3: Check the form of the limit

If we directly substitute \( x = 1 \) into the limit, we get:

\[ \frac{\sqrt{25 - 1^2} - \sqrt{24}}{1 - 1} = \frac{\sqrt{24} - \sqrt{24}}{0} = \frac{0}{0} \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we take the derivative of the numerator and the derivative of the denominator:

1. **Numerator**: \( \sqrt{25 - x^2} \)
- Derivative:
\[ \frac{d}{dx}(\sqrt{25 - x^2}) = \frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{25 - x^2}} \]

2. **Denominator**: \( x - 1 \)
- Derivative:
\[ \frac{d}{dx}(x - 1) = 1 \]

Now we can rewrite the limit using these derivatives:

\[ \lim_{x \to 1} \frac{\frac{-x}{\sqrt{25 - x^2}}}{1} = \lim_{x \to 1} \frac{-x}{\sqrt{25 - x^2}} \]

### Step 5: Substitute \( x = 1 \)

Now we substitute \( x = 1 \):

\[ \frac{-1}{\sqrt{25 - 1^2}} = \frac{-1}{\sqrt{25 - 1}} = \frac{-1}{\sqrt{24}} \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = \frac{-1}{\sqrt{24}} \]