Tìm giới hạn \(\mathop {\lim }\limits_{x \to  + \i...

C

- Hướng dẫn giải

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