Tính \(\mathop {\lim }\limits_{x \to + \infty } \frac{{2018}}{{2{x^3} - 5{x^5}}}\) có giá trị bằng:
Giải chi tiết:
Ta có: \(\mathop {\lim }\limits_{x \to + \infty } \left( {2{x^3} - 5{x^5}} \right) = \mathop {\lim }\limits_{x \to + \infty } {x^5}\left( {\frac{2}{{{x^2}}} - 5} \right) = - \infty .\)
Vì \(\mathop {\lim }\limits_{x \to + \infty } \left( {2{x^3} - 5{x^5}} \right) = - \infty \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{2018}}{{2{x^3} - 5{x^5}}} = 0.\)
Chọn C.