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