Câu hỏi 17

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

\(\begin{array}{l} = \mathop {\lim }\limits_{x \to  - \,\infty } \frac{{x\left( {2 - \frac{1}{x}} \right)\left( {\left| x \right|\sqrt {1 + \frac{1}{{{x^2}}}}  + 1} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to  - \,\infty } \frac{{\left( {2 - \frac{1}{x}} \right)\left( { - \,x\sqrt {1 + \frac{1}{{{x^2}}}}  + 1} \right)}}{x}\\ = \mathop {\lim }\limits_{x \to  - \,\infty } \left( {2 - \frac{1}{x}} \right)\left( { - \,\sqrt {1 + \frac{1}{{{x^2}}}}  + \frac{1}{x}} \right) =  - \,2.\end{array}\)

Chọn B