Evaluate the following limits : `lim_(x to infty)(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: Multiply by the Conjugate
To simplify the expression, we multiply and divide by the conjugate of the expression:
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 + 1} \right) \cdot \frac{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}} \]

### Step 2: Simplify the Expression
This gives us:
\[ \lim_{x \to \infty} \frac{(x^2 + x + 1) - (x^2 + 1)}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}} \]
Simplifying the numerator:
\[ (x^2 + x + 1) - (x^2 + 1) = x \]
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:
\[ \sqrt{x^2 + x + 1} = \sqrt{x^2(1 + \frac{1}{x} + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \]
\[ \sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x^2}} \]
Thus, the limit becomes:
\[ \lim_{x \to \infty} \frac{x}{x\left(\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}}\right)} \]
Cancelling \(x\) 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 4: Evaluate the Limit
As \(x \to \infty\), both \(\frac{1}{x}\) and \(\frac{1}{x^2}\) approach 0:
\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + 0 + 0} + \sqrt{1 + 0}} = \frac{1}{\sqrt{1} + \sqrt{1}} = \frac{1}{1 + 1} = \frac{1}{2} \]

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