`lim_(x->oo)(sqrt(x+sqrt(x))-sqrt(x))`equals

Options

A. `1/2`
B. 2
C. 3
D. 4

Answer

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

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

### Step 2: Identify the indeterminate form
As \( x \) approaches infinity, both \( \sqrt{x + \sqrt{x}} \) and \( \sqrt{x} \) approach infinity, leading to the indeterminate form \( \infty - \infty \).

### Step 3: Rationalize the expression
To resolve the indeterminate form, we can rationalize the expression by multiplying and dividing 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 simplifies to:
\[ \lim_{x \to \infty} \frac{(x + \sqrt{x}) - x}{\sqrt{x + \sqrt{x}} + \sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \]

### Step 4: Simplify the numerator and denominator
Now, we have:
\[ \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \]
We can factor out \( \sqrt{x} \) from the numerator and denominator:
\[ = \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1 + \frac{1}{\sqrt{x}}} + 1\right)} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{\sqrt{x}}} + 1} \]

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

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x} \right) = \frac{1}{2} \]

---