Tìm giới hạn \(A\; = \;\mathop {\lim \to + \infty } \left( {\sqrt {{x^2} + x + 1} - + x - 1}}}

* Hướng dẫn giải

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

vì \(\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty ,\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}}  - \sqrt[3]{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}} \right) = 1 - \sqrt[3]{2} < 0\)