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