If `G(x)=-sqrt(25-x)`. Then `lim_(xrarr1) (G(x)-G(1))/(x-1)` has the value

Options

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

Answer

To solve the limit problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Define the function and the limit
We have the function:
\[ G(x) = -\sqrt{25 - x} \]
We need to evaluate the limit:
\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} \]

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

### Step 3: Substitute into the limit
Now, substituting \( G(1) \) into the limit gives:
\[ \lim_{x \to 1} \frac{G(x) - (-\sqrt{24})}{x - 1} = \lim_{x \to 1} \frac{G(x) + \sqrt{24}}{x - 1} \]

### Step 4: Check for indeterminate form
If we substitute \( x = 1 \) directly into the limit:
\[ G(1) + \sqrt{24} = -\sqrt{24} + \sqrt{24} = 0 \]
And the denominator becomes \( 1 - 1 = 0 \). Thus, we have the indeterminate form \( \frac{0}{0} \).

### Step 5: Apply L'Hôpital's Rule
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} = \lim_{x \to 1} \frac{G'(x)}{1} \]
We need to find \( G'(x) \).

### Step 6: Differentiate \( G(x) \)
To differentiate \( G(x) \):
\[ G(x) = -\sqrt{25 - x} = -(25 - x)^{1/2} \]
Using the chain rule:
\[ G'(x) = -\frac{1}{2}(25 - x)^{-1/2} \cdot (-1) = \frac{1}{2\sqrt{25 - x}} \]

### Step 7: Evaluate \( G'(1) \)
Now we evaluate \( G'(1) \):
\[ G'(1) = \frac{1}{2\sqrt{25 - 1}} = \frac{1}{2\sqrt{24}} = \frac{1}{2 \cdot 2\sqrt{6}} = \frac{1}{4\sqrt{6}} \]

### Step 8: Conclusion
Thus, the value of the limit is:
\[ \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} = G'(1) = \frac{1}{4\sqrt{6}} \]

### Final Answer:
The value of the limit is:
\[ \frac{1}{4\sqrt{6}} \]

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