Kết quả của giới hạn \(\mathop {\lim \to + \infty } \left( {\sqrt {{x^2} + x} - - {x^2}}}} \right)\) là:

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - x + x - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - x} \right)\\ + \mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} + x - {x^2}}}{{\sqrt {{x^2} + x}  + x}}\\ + \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} - {x^3} + {x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {{\sqrt[3]{{{x^3} - {x^2}}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{x}{{\sqrt {{x^2} + x}  + x}}\\ + \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {{\sqrt[3]{{{x^3} - {x^2}}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt {1 + \frac{1}{x}}  + 1}}\\ + \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{1 + \sqrt[3]{{1 - \frac{1}{x}}} + {{\sqrt[3]{{1 - \frac{1}{x}}}}^2}}}\\ = \frac{1}{{1 + 1}} + \frac{1}{{1 + 1 + 1}} = \frac{5}{6}.\end{array}\)