Tính các giới hạn sau:

\(\mathop {\lim }\limits_{x \to - 1} \frac{{ - 4x + 1}}{{{x^2} - x + 3}}\);

\(\mathop {\lim }\limits_{x \to - 1} \frac{{ - 4x + 1}}{{{x^2} - x + 3}}\)\( = \frac{{\mathop {\lim }\limits_{x \to - 1} \left( { - 4x + 1} \right)}}{{\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - x + 3} \right)}}\)\( = \frac{{\mathop {\lim }\limits_{x \to - 1} \left( { - 4x} \right) + \mathop {\lim }\limits_{x \to - 1} 1}}{{\mathop {\lim }\limits_{x \to - 1} {x^2} - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 3}} = \frac{{4 + 1}}{{1 - \left( { - 1} \right) + 3}} = \frac{5}{5} = 1\).