If `G(x)=sqrt(25-x^(2))` then what is `lim_(xto1)(G(x)-G(1))/(x-1)` equal to ?

Options

A. `-(1)/(2sqrt(6))`
B. `(1)/(5)`
C. `-(1)/(6)`
D. `(1)/(sqrt(6))`

Answer

To solve the limit problem \( \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} \) where \( G(x) = \sqrt{25 - x^2} \), we will follow these steps:

### Step 1: Evaluate \( G(1) \)
First, we need to find \( G(1) \):
\[ G(1) = \sqrt{25 - 1^2} = \sqrt{25 - 1} = \sqrt{24} \]

### Step 2: Rewrite the limit
Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} = \lim_{x \to 1} \frac{\sqrt{25 - x^2} - \sqrt{24}}{x - 1} \]

### Step 3: Substitute \( x = 1 \)
If we directly substitute \( x = 1 \) into the limit, we get:
\[ \frac{\sqrt{25 - 1^2} - \sqrt{24}}{1 - 1} = \frac{\sqrt{24} - \sqrt{24}}{0} = \frac{0}{0} \]
This is an indeterminate form, so we will apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
1. Differentiate the numerator:
\[ \frac{d}{dx}(\sqrt{25 - x^2}) = \frac{-x}{\sqrt{25 - x^2}} \]
2. Differentiate the denominator:
\[ \frac{d}{dx}(x - 1) = 1 \]

Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{\frac{-x}{\sqrt{25 - x^2}}}{1} = \lim_{x \to 1} \frac{-x}{\sqrt{25 - x^2}} \]

### Step 5: Evaluate the limit again
Now substitute \( x = 1 \):
\[ \frac{-1}{\sqrt{25 - 1^2}} = \frac{-1}{\sqrt{24}} = \frac{-1}{2\sqrt{6}} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} = \frac{-1}{2\sqrt{6}} \]