\( \mathop { \lim } \limits_{x \to - \infty } \dfrac{1}{{ \sqrt {{x^2} + x + 1} - x}} \)
A.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = + \infty\).
B.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = - \infty\).
C.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = - \dfrac{1}{4}\).
D.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = 0\).
A.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = + \infty\).
B.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = - \infty\).
C.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = - \dfrac{1}{4}\).
D.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\sqrt {{x^2} + x + 1} - x}} = 0\).
Loga
Hóa Học lớp 12
