Evaluate : `lim_(x to oo) sqrt(x^(2)+x +1) - sqrt(x^(2)+1)`

Answer

To evaluate 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
To simplify the expression, we can multiply and divide by the conjugate:

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

### Step 2: Simplify the numerator
Using the difference of squares, the numerator becomes:

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

This simplifies to:

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

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

\[ \sqrt{x^2 + x + 1} = \sqrt{x^2(1 + \frac{1}{x} + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}, \]
and
\[ \sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x^2}}. \]

### Step 4: Substitute back into the limit
Substituting these back, we have:

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

### Step 5: Cancel \(x\)
We can cancel \(x\) from 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 6: Evaluate the limit as \(x\) approaches infinity
As \(x\) approaches infinity, \(\frac{1}{x}\) and \(\frac{1}{x^2}\) 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
Thus, the limit is

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