Tìm \(\mathop {\lim }\limits_{x \to a} {{{x^2} - \left( {a + 1} \right)x + a} \over {{x^3} - {a^3}}}\) ta được:
Giải chi tiết:
\(\eqalign{ & \mathop {\lim }\limits_{x \to a} {{{x^2} - \left( {a + 1} \right)x + a} \over {{x^3} - {a^3}}} = \mathop {\lim }\limits_{x \to a} {{{x^2} - ax - x + a} \over {{x^3} - {a^3}}} = \mathop {\lim }\limits_{x \to a} {{x\left( {x - a} \right) - \left( {x - a} \right)} \over {\left( {x - a} \right)\left( {{x^2} + ax + {a^2}} \right)}} = \mathop {\lim }\limits_{x \to a} {{\left( {x - a} \right)\left( {x - 1} \right)} \over {\left( {x - a} \right)\left( {{x^2} + ax + {a^2}} \right)}} \cr & = \mathop {\lim }\limits_{x \to a} {{x - 1} \over {{x^2} + ax + {a^2}}} = {{a - 1} \over {{a^2} + {a^2} + {a^2}}} = {{a - 1} \over {3{a^2}}} \cr} \)
Chọn A.