Giới hạn \(\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} - 3x + 1} \right)\) bằng

* Hướng dẫn giải

Ta có: \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^2} - 3x + 1} \right)\)\( = \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^2}\left( {1 - \frac{3}{x} + \frac{1}{{{x^2}}}} \right)} \right]\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } {x^2} =  + \infty \) và \(\mathop {\lim }\limits_{x \to  - \infty } \left( {1 - \frac{3}{x} + \frac{1}{{{x^2}}}} \right) = 1\) nên \(\mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^2}\left( {1 - \frac{3}{x} + \frac{1}{{{x^2}}}} \right)} \right] =  + \infty \).

Vậy \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^2} - 3x + 1} \right) =  + \infty \).