\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3 + \dfrac{2}{{{x^2}}} + \dfrac{5}{{{x^4}}}}}{{\dfrac{1}{{{x^4}}} - \dfrac{2}{x} + \dfrac{1}{{{x^3}}}}}\).
Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( {3 + \dfrac{2}{{{x^2}}} + \dfrac{5}{{{x^4}}}} \right) = 3 > 0\\\mathop {\lim }\limits_{x \to - \infty } \left( {\dfrac{1}{{{x^4}}} - \dfrac{2}{x} + \dfrac{1}{{{x^3}}}} \right) = 0\end{array} \right.\).
Vậy \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = + \infty \).
Chọn C.