Giới hạn \(I = \mathop {\lim }\limits_{x \to + \infty } \left( { - 2{x^3} + 4x + 5} \right)\) bằng

* Hướng dẫn giải

Ta có: \(I = \mathop {\lim }\limits_{x \to  + \infty } \left( { - 2{x^3} + 4x + 5} \right) = \mathop {\lim }\limits_{x \to  + \infty } {x^3}\left( { - 2 + \dfrac{4}{{{x^2}}} + \dfrac{5}{{{x^3}}}} \right)\).

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } {x^3} =  + \infty \\\mathop {\lim }\limits_{x \to  + \infty } \left( { - 2 + \dfrac{4}{{{x^2}}} + \dfrac{5}{{{x^3}}}} \right) =  - 2 < 0\end{array}\)

Vậy \(I = \mathop {\lim }\limits_{x \to  + \infty } \left( { - 2{x^3} + 4x + 5} \right) =  - \infty \).

Chọn A.