The value of `underset(x to 2)lim (sqrt(1+sqrt(2+x))-sqrt3)/(x-2)` is

Options

A. `(1)/(8sqrt3)`
B. `1/(4sqrt3)`
C. 0
D. None of these

Answer

To solve the limit problem given by:

\[ \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 2 \) into the expression:

\[ \sqrt{1 + \sqrt{2 + 2}} - \sqrt{3} = \sqrt{1 + \sqrt{4}} - \sqrt{3} = \sqrt{1 + 2} - \sqrt{3} = \sqrt{3} - \sqrt{3} = 0 \]

The denominator also becomes \( 2 - 2 = 0 \). Thus, we have the indeterminate form \( \frac{0}{0} \).

**Hint for Step 1:** Always check the limit by direct substitution first to identify indeterminate forms.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

The limit now becomes:

\[ \lim_{x \to 2} \frac{\frac{d}{dx}(\sqrt{1 + \sqrt{2 + x}} - \sqrt{3})}{\frac{d}{dx}(x - 2)} \]

### Step 3: Differentiate the numerator
We differentiate the numerator:

1. The derivative of \( \sqrt{1 + \sqrt{2 + x}} \):
- Using the chain rule:
\[ \frac{d}{dx}(\sqrt{1 + \sqrt{2 + x}}) = \frac{1}{2\sqrt{1 + \sqrt{2 + x}}} \cdot \frac{d}{dx}(\sqrt{2 + x}) = \frac{1}{2\sqrt{1 + \sqrt{2 + x}}} \cdot \frac{1}{2\sqrt{2 + x}} \cdot 1 \]

2. The derivative of \( -\sqrt{3} \) is \( 0 \).

Thus, the derivative of the numerator is:

\[ \frac{1}{4\sqrt{1 + \sqrt{2 + x}} \cdot \sqrt{2 + x}} \]

### Step 4: Differentiate the denominator
The derivative of \( x - 2 \) is simply \( 1 \).

### Step 5: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 2} \frac{\frac{1}{4\sqrt{1 + \sqrt{2 + x}} \cdot \sqrt{2 + x}}}{1} = \lim_{x \to 2} \frac{1}{4\sqrt{1 + \sqrt{2 + x}} \cdot \sqrt{2 + x}} \]

### Step 6: Substitute \( x = 2 \) again
Now substitute \( x = 2 \):

\[ = \frac{1}{4\sqrt{1 + \sqrt{2 + 2}} \cdot \sqrt{2 + 2}} = \frac{1}{4\sqrt{1 + \sqrt{4}} \cdot \sqrt{4}} = \frac{1}{4\sqrt{1 + 2} \cdot 2} = \frac{1}{4 \cdot \sqrt{3} \cdot 2} = \frac{1}{8\sqrt{3}} \]

### Final Answer
Thus, the value of the limit is:

\[ \frac{1}{8\sqrt{3}} \]