Cho hàm số \(f\left( x \right) = \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^4} + {x^2} + 1}}} \).

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  + \infty } \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^4} + {x^2} + 1}}}  = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{\left( {x - 1} \right){{\left( {x + 2} \right)}^2}}}{{{x^4} + {x^2} + 1}}}  = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} - \frac{2}{{{x^4}}}}}{{1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}}  = 0\)