If `G(x) = - sqrt(25 -x^(2))`, then `lim_(x to 1) (G(x)-G(1))/(x-1)` has the value of:

Options

A. `1//24`
B. `1//5`
C. `-sqrt(24)`
D. none of these

Answer

To solve the limit problem, we need to find the value of

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

where \( G(x) = -\sqrt{25 - x^2} \).

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

First, we need to evaluate \( G(1) \):

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

### Step 2: Set up the limit expression

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

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

### Step 3: Check for indeterminate form

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 can 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 \( G(x) - G(1) \):
- \( G'(x) = -\frac{d}{dx}(\sqrt{25 - x^2}) = -\frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{25 - x^2}} \)

2. Differentiate the denominator \( x - 1 \):
- The derivative is simply \( 1 \).

Now we can rewrite the limit:

\[ \lim_{x \to 1} \frac{G'(x)}{1} = G'(1) \]

### Step 5: Calculate \( G'(1) \)

Now we evaluate \( G'(1) \):

\[ G'(1) = \frac{1}{\sqrt{25 - 1^2}} = \frac{1}{\sqrt{24}} = \frac{1}{2\sqrt{6}} \]

### Final Answer

Thus, the value of the limit is:

\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} = \frac{1}{2\sqrt{6}} \]