\(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} - x + 1} }}{{x + 1}}\) bằng
A. 2 | B. -2 | C. 1 | D. -1 |
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} - x + 1} }}{{x + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{1 + \frac{1}{x}}} = - 2
\end{array}\)
Đáp án: B
-- Mod Toán 11