\(\mathop {\lim }\limits_{x \to  - \infty } \frac{...

B

- Hướng dẫn giải

Ta có \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - x}  - \sqrt {4{x^2} + 1} }}{{2x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\left( {\sqrt {1 - \frac{1}{x}}  - \sqrt {4 + \frac{1}{{{x^2}}}} } \right)}}{{x\left( {2 + \frac{3}{x}} \right)}}\) \( = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \left( {\sqrt {1 - \frac{1}{x}}  - \sqrt {4 + \frac{1}{{{x^2}}}} } \right)}}{{\left( {2 + \frac{3}{x}} \right)}}\) \( = \frac{1}{2}\)