\(\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^3} - 1}}{{{x^2} - x}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^3} - 1}}{{{x^2} - x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left[ {{{\left( {1 + x} \right)}^2} + \left( {1 + x} \right) + 1} \right]}}{{x\left( {x - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^2} + \left( {1 + x} \right) + 1}}{{x - 1}} = \dfrac{{1 + 1 + 1}}{{ - 1}} = - 3\).
Chọn A.