Phương pháp giải:

Sử dụng các phương pháp tính giới hạn hàm số để tính các giới hạn và chọn đáp án đúng.

Lời giải chi tiết:

Ta có:

\(\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( {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( {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}\)

Chọn  B.