`lim_(xrarroo)(sqrt(x+sqrtx)-sqrtx)` is

Options

A. `(1)/(2)`
B. 0
C. infinite
D. 1

Answer

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

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

### Step 2: Rationalize the expression
To simplify the expression, we can multiply and divide by the conjugate:
\[ \lim_{x \to \infty} \frac{\left( \sqrt{x + \sqrt{x}} - \sqrt{x} \right) \left( \sqrt{x + \sqrt{x}} + \sqrt{x} \right)}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \]
This gives us:
\[ \lim_{x \to \infty} \frac{(x + \sqrt{x}) - x}{\sqrt{x + \sqrt{x}} + \sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \]

### Step 3: Simplify the denominator
Now, we can simplify the denominator:
\[ \sqrt{x + \sqrt{x}} + \sqrt{x} = \sqrt{x(1 + \frac{1}{\sqrt{x}})} + \sqrt{x} = \sqrt{x} \left( \sqrt{1 + \frac{1}{\sqrt{x}}} + 1 \right) \]

### Step 4: Substitute back into the limit
Substituting this back into our limit, we have:
\[ \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \left( \sqrt{1 + \frac{1}{\sqrt{x}}} + 1 \right)} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{\sqrt{x}}} + 1} \]

### Step 5: Evaluate the limit
As \( x \to \infty \), \( \frac{1}{\sqrt{x}} \to 0 \). Therefore, we can evaluate the limit:
\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \boxed{\frac{1}{2}} \]