Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right) \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right) = \infty - \infty \cr
& {\text{Rationalizing}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right) \times \frac{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\sqrt {{x^2} + x + 1} } \right)}^2} - {{\left( {\sqrt {{x^2} - x} } \right)}^2}}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + x + 1 - {x^2} + x}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2x + 1}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2x + 1}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} = \frac{\infty }{\infty } \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2x + 1}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2x}}{x} + \frac{1}{x}}}{{\sqrt {\frac{{{x^2}}}{{{x^2}}} + \frac{x}{{{x^2}}} + \frac{1}{{{x^2}}}} + \sqrt {\frac{{{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}}} }} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{1}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{1}{{{x^2}}}} }} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{1}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{1}{{{x^2}}}} }} = \frac{{2 + \frac{1}{\infty }}}{{\sqrt {1 + \frac{1}{\infty } + \frac{1}{{{\infty ^2}}}} + \sqrt {1 - \frac{1}{{{\infty ^2}}}} }} \cr
& {\text{ }} = \frac{2}{{\sqrt 1 + \sqrt 1 }} \cr
& {\text{ }} = 1 \cr} $$
