If `G(x)=-sqrt(25-x^(2))`, then `lim_(xrarr1) (G(x)-G(1))/(x-1)=?`

Options

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

Answer

To solve the limit problem given by \( G(x) = -\sqrt{25 - x^2} \), we need to find

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

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

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

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

### Step 2: Substitute \( G(x) \) and \( G(1) \) into the limit expression

Now we substitute \( G(x) \) and \( G(1) \) into the 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: Simplify the expression

This can be rewritten as:

\[ \lim_{x \to 1} \frac{\sqrt{24} - \sqrt{25 - x^2}}{x - 1}. \]

### Step 4: Check for indeterminate form

Now, if we substitute \( x = 1 \) directly into the expression, we get:

\[ \frac{\sqrt{24} - \sqrt{24}}{1 - 1} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

1. **Differentiate the numerator**: The derivative of \( \sqrt{24} - \sqrt{25 - x^2} \) is:

\[ 0 - \frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{25 - x^2}}. \]

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

\[ 1. \]

Now we can rewrite the limit as:

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

### Step 6: Evaluate the limit

Substituting \( x = 1 \):

\[ \frac{1}{\sqrt{25 - 1^2}} = \frac{1}{\sqrt{24}} = \frac{1}{2\sqrt{6}}. \]

### Final Answer

Thus, the limit is:

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