`lim_(xtooo) [sqrt(x+sqrt(x+sqrt(x)))-sqrt(x)]` is equal to

Options

A. 1
B. 0
C. 2
D. none of these

Explanation

To solve the limit \( \lim_{x \to \infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ L = \lim_{x \to \infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \]

### Step 2: Rationalize the expression
To simplify the expression, we can rationalize it by multiplying and dividing by the conjugate:
\[ L = \lim_{x \to \infty} \frac{\left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \left( \sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x} \right)}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \]
This simplifies to:
\[ L = \lim_{x \to \infty} \frac{x + \sqrt{x + \sqrt{x}} - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \]

### Step 3: Simplify the numerator
Next, we simplify the numerator:
\[ \sqrt{x + \sqrt{x}} = \sqrt{x(1 + \frac{1}{\sqrt{x}})} = \sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}} \]

### Step 4: Simplify the denominator
Now, we simplify the denominator:
\[ \sqrt{x + \sqrt{x + \sqrt{x}}} = \sqrt{x(1 + \frac{\sqrt{x + \sqrt{x}}}{x})} = \sqrt{x} \sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} \]
And
\[ \sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x} = \sqrt{x} \left( \sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} + 1 \right) \]

### Step 5: Substitute back into the limit
We substitute these back into our limit:
\[ L = \lim_{x \to \infty} \frac{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{x} \left( \sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} + 1 \right)} \]
This simplifies to:
\[ L = \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} + 1} \]

### Step 6: Evaluate the limit
As \( x \to \infty \):
- \( \frac{1}{\sqrt{x}} \to 0 \)
- \( \frac{\sqrt{x + \sqrt{x}}}{x} \to 0 \)

Thus, we have:
\[ L = \frac{\sqrt{1 + 0}}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \boxed{1} \]