Trong các mệnh đề sau, mệnh đề nào sai?

Đáp án đúng là: B

+) \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - x + 1} + x - 2} \right)\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {{x^2} - x + 1} \right) - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2} - x + 1} - \left( {x - 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{3x - 3}}{{\sqrt {{x^2} - x + 1} - \left( {x - 2} \right)}}\)

\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{3 - \frac{3}{x}}}{{ - \sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} - \left( {1 - \frac{2}{x}} \right)}} = - \frac{3}{2}\)

(vì \[\mathop {\lim }\limits_{x \to - \infty } \frac{3}{x} = 0;\mathop {\lim }\limits_{x \to - \infty } \frac{2}{x} = 0;\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x} = 0;\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}} = 0\]). Đáp án A đúng.

+) \(\mathop {\lim }\limits_{x \to - {1^ - }} \left( {3x + 2} \right) = - 1\); \(\mathop {\lim }\limits_{x \to - {1^ - }} \left( {x + 1} \right) = 0\) mà \(x \to - {1^ - }\) nên \(x + 1 < 0\).

Do đó \(\mathop {\lim }\limits_{x \to - {1^ - }} \frac{{3x + 2}}{{x + 1}} = + \infty \). Suy ra đáp án B sai.

+) \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} + x - 2} \right)\)\( = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} + 1 - \frac{2}{x}} \right) = + \infty \)

(vì \(\mathop {\lim }\limits_{x \to + \infty } x = + \infty \) và \( = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} + 1 - \frac{2}{x}} \right) = 2 > 0\)). Vậy đáp án C đúng.

+) \(\mathop {\lim }\limits_{x \to - {1^ + }} \left( {3x + 2} \right) = - 1\); \(\mathop {\lim }\limits_{x \to - {1^ + }} \left( {x + 1} \right) = 0\) mà \(x \to - {1^ + }\) nên \(x + 1 > 0\).

Do đó \(\mathop {\lim }\limits_{x \to - {1^ + }} \frac{{3x + 2}}{{x + 1}} = - \infty \). Suy ra đáp án D đúng.