Cho hàm số \(f\left( x \right) = \sqrt {{x^2} - x + 2} \).

* Hướng dẫn giải

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

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