\(\mathop {\lim }\limits_{x \to - \infty } \left( {3{x^3} - 5{x^2} + 7} \right)\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - \infty } \left( {3{x^3} - 5{x^2} + 7} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( {3 - \dfrac{5}{x} + \dfrac{7}{{{x^2}}}} \right)\).
Ta có: \(\mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty ;\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {3 - \dfrac{5}{x} + \dfrac{7}{{{x^2}}}} \right) = 3 > 0\).
Vậy \(\mathop {\lim }\limits_{x \to - \infty } \left( {3{x^3} - 5{x^2} + 7} \right) = - \infty \).
Chọn B.