Trong các mệnh đề sau, mệnh đề nào sai

B

- Hướng dẫn giải

\(\begin{array}{l}
 + )\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - x + 1}  + x - 2} \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {\sqrt {{x^2} - x + 1}  + x - 2} \right)\left( {\sqrt {{x^2} - x + 1}  - x + 2} \right)}}{{\sqrt {{x^2} - x + 1}  - x + 2}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^2} - x + 1 - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2} - x + 1}  - x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{3x - 3}}{{\sqrt {{x^2} - x + 1}  - x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{3 - \frac{3}{x}}}{{ - \sqrt {1 - \frac{2}{x} + \frac{1}{{{x^2}}}}  - 1 + \frac{2}{x}}} =  - \frac{3}{2}\\
 + )\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{3x - 2}}{{x + 1}} =  + \infty \,\,\,\left( {{\rm{do}}\,\,\,\,\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( {3x - 2} \right) =  - 5\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( {x + 1} \right) = 0;x - 1 < 0
\end{array} \right.} \right)\\
 + )\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} - x + 1}  + x - 2} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {\sqrt {{x^2} - x + 1}  + x - 2} \right)\left( {\sqrt {{x^2} - x + 1}  - x + 2} \right)}}{{\sqrt {{x^2} - x + 1}  - x + 2}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} - x + 1 - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2} - x + 1}  - x + 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{3x - 3}}{{\sqrt {{x^2} - x + 1}  - x + 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{3 - \frac{3}{x}}}{{\sqrt {1 - \frac{2}{x} + \frac{1}{{{x^2}}}}  - 1 + \frac{2}{x}}} =  + \infty \\
 + )\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{3x - 2}}{{x + 1}} =  - \infty \,\,\,\left( {{\rm{do}}\,\,\,\,\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {3x - 2} \right) =  - 5\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {x + 1} \right) = 0;x - 1 > 0
\end{array} \right.} \right)
\end{array}\)