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{Evaluating the limit gives}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right) = \sqrt {{\infty ^2} + \infty + 1} - \sqrt {{\infty ^2} - \infty } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \infty - \infty \cr
& {\text{Multiply the limit by the conjugate of }}\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right)\left( {\frac{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x} }}} \right) \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
& {\text{simplify}} \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{evaluating the limit}} \cr
& = \frac{{2\left( \infty \right) + 1}}{{\sqrt {{\infty ^2} + \infty + 1} - \sqrt {{\infty ^2} - \infty } }} = \frac{\infty }{\infty } \cr
& {\text{Multiply the numerator and denominator by }}\frac{1}{x} \cr
& = \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
& = \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}} }} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{2 + \frac{1}{\infty }}}{{\sqrt {1 + \frac{1}{\infty } + \frac{1}{{{\infty ^2}}}} + \sqrt {1 - \frac{1}{\infty }} }} \cr
& = \frac{{2 + 0}}{{\sqrt {1 + 0 + 0} + \sqrt {1 - 0} }} \cr
& = \frac{2}{2} \cr
& = 1 \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x} } \right) = 1 \cr} $$
