The value of `lim_(x to oo) (sqrt(x^(2) + x + 1) - sqrt(x^(2) - x + 1))` equals

Options

A. 0
B. 1
C. `-1`
D. `oo`

Answer

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

### Step 1: Multiply by the Conjugate
We start by multiplying and dividing the expression by the conjugate of the numerator:

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

### Step 2: Simplify the Numerator
Using the difference of squares, the numerator simplifies as follows:

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

This simplifies to:

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

### Step 3: Rewrite the Limit
Now we rewrite our limit:

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

### Step 4: Factor out \(x\) from the Square Roots
Next, we factor \(x^2\) out of the square roots in the denominator:

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

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

### Step 5: Substitute Back into the Limit
Substituting these back into our limit gives:

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

### Step 6: Evaluate the Limit
As \(x\) approaches infinity, the terms \(\frac{1}{x}\) and \(\frac{1}{x^2}\) approach 0:

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

### Final Answer
Thus, the value of the limit is:

\[ \boxed{1} \]