The problem asks to evaluate the limit $\lim_{x \to \infty} (\sqrt{x^2+x} - \sqrt{x^2-1})$.

1. Problem Description

The problem asks to evaluate the limit

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

2. Solution Steps

We can rewrite the expression by multiplying by the conjugate and dividing by the same conjugate:

$\sqrt{x^2+x} - \sqrt{x^2-1} = (\sqrt{x^2+x} - \sqrt{x^2-1}) \cdot \frac{\sqrt{x^2+x} + \sqrt{x^2-1}}{\sqrt{x^2+x} + \sqrt{x^2-1}}$

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

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

Now, we divide both the numerator and the denominator by $x$. Since $x \to \infty$, we can assume $x>0$, so $\sqrt{x^2} = x$.

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

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

Now we take the limit as $x \to \infty$.

$\lim_{x \to \infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}} + \sqrt{1-\frac{1}{x^2}}} = \frac{1+0}{\sqrt{1+0} + \sqrt{1-0}} = \frac{1}{\sqrt{1} + \sqrt{1}} = \frac{1}{1+1} = \frac{1}{2}$

3. Final Answer

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