Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 2{\rm{x}}} - 2\sqrt {{x^2} + x} + x} \right)\).

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } x\left( {\sqrt {{x^2} + 2{\rm{x}}}  - 2\sqrt {{x^2} + x}  + x} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } x\left( {\dfrac{x}{{\sqrt {{x^2} + 2{\rm{x}}}  + \sqrt {{x^2}}  + x}} - \dfrac{x}{{x + \sqrt {{x^2} + x} }}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } {x^2}\left( {\dfrac{{x - \sqrt {{x^2} + 2{\rm{x}}} }}{{\left( {\sqrt {{x^2} + 2{\rm{x}}}  + \sqrt {{x^2} + x} } \right)\left( {x + \sqrt {{x^2} + x} } \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } {x^2}.\dfrac{{ - 2{\rm{x}}}}{{\left( {\sqrt {{x^2} + 2x}  + \sqrt {{x^2} + x} } \right)\left( {x + \sqrt {{x^2} + x} } \right)\left( {x + \sqrt {{x^2} + 2{\rm{x}}} } \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{ - 2{{\rm{x}}^3}}}{{{x^3}\left( {\sqrt {1 + \dfrac{2}{x}}  + \sqrt {1 + \dfrac{1}{x}} } \right)\left( {1 + \sqrt {1 + \dfrac{1}{x}} } \right)\left( {1 + \sqrt {1 + \dfrac{2}{x}} } \right)}}\\ = \dfrac{{ - 2}}{{2.2.2}} =  - \dfrac{1}{4}\end{array}\)