Tính \(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 6x + 5}}{{{x^3} + 2{x^2} - 1}}\) bằng?

* Hướng dẫn giải

\(\eqalign{
& \mathop {\lim }\limits_{x \to - 1} {{{x^2} + 6x + 5} \over {{x^3} + 2{x^2} - 1}} \cr
& = \mathop {\lim }\limits_{x \to - 1} {{\left( {x + 1} \right)\left( {x + 5} \right)} \over {\left( {x + 1} \right)\left( {{x^2} + x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to - 1} {{x + 5} \over {{x^2} + x - 1}} = {4 \over { - 1}} = - 4 \cr} \)