\(\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^3} + 1}}{{{x^2} + x}}\) bằng 

* Hướng dẫn giải

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