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

(begin{array}{l}mathop {lim }limits_{x to  + infty } xleft( {sqrt {{x^2} + 2{rm{x}}}  – 2sqrt {{x^2} + x}  + x} right)\ = mathop {lim }limits_{x to  + infty } xleft( {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})