\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}}\)
A.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = 4 \).
B.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = - \infty \).
C.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = + \infty \).
D.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = \dfrac{1}{3} \).
A.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = 4 \).
B.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = - \infty \).
C.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = + \infty \).
D.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} + 2{x^2} + 5}}{{1 - 2{x^3} + x}} = \dfrac{1}{3} \).
Loga
Hóa Học lớp 12
