`lim _(xrarr oo) (sqrt(x^2+x+1)-sqrt(x^2+1))=`

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. `1`
D. `-1`

Explanation

To solve the limit \( \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 + 1} \right) \), we can follow these steps:

### Step 1: Rationalize the expression
We start by multiplying and dividing by the conjugate of the expression:

\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 + 1} \right) \cdot \frac{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}} \]

This gives us:

\[ \lim_{x \to \infty} \frac{(\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}} \]

### Step 2: Simplify the numerator
The numerator simplifies as follows:

\[ \sqrt{x^2 + x + 1}^2 - \sqrt{x^2 + 1}^2 = (x^2 + x + 1) - (x^2 + 1) = x \]

So we have:

\[ \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}} \]

### Step 3: Factor out \(x\) from the square roots
Next, we factor \(x\) out of the square roots in the denominator:

\[ \sqrt{x^2 + x + 1} = x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \quad \text{and} \quad \sqrt{x^2 + 1} = x\sqrt{1 + \frac{1}{x^2}} \]

Substituting these into the limit gives us:

\[ \lim_{x \to \infty} \frac{x}{x\left(\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}}\right)} \]

### Step 4: Cancel \(x\)
We can cancel \(x\) in the numerator and denominator:

\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}}} \]

### Step 5: Apply the limit
As \(x\) approaches infinity, \(\frac{1}{x}\) and \(\frac{1}{x^2}\) both approach 0. Thus, we have:

\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + 0 + 0} + \sqrt{1 + 0}} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
The limit is:

\[ \boxed{\frac{1}{2}} \]