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