Đáp án đúng: B
Giải chi tiết:\(\begin{array}{l} \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {2{x^2} + 1} \right)\left( {2{x^3} + x} \right)}}{{\left( {2{x^4} + x} \right)\left( {x + 1} \right)}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{{2{x^2} + 1}}{{{x^2}}}.\dfrac{{2{x^3} + x}}{{{x^3}}}}}{{\dfrac{{2{x^4} + x}}{{{x^4}}}.\dfrac{{x + 1}}{x}}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {2 + \dfrac{1}{{{x^2}}}} \right)\left( {2 + \dfrac{1}{{{x^2}}}} \right)}}{{\left( {2 + \dfrac{1}{{{x^3}}}} \right)\left( {1 + \dfrac{1}{x}} \right)}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2.2}}{{2.1}} = 2 \end{array}\)
Chọn B.
