Tìm giới hạn \(\mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{1}{{{x^3} - 1}} - \dfrac{1}{{x - 1}}} \right)\)
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{1}{{{x^3} - 1}} - \dfrac{1}{{x - 1}}} \right)\\ = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{1}{{x - 1}}} \right)\\ = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{ - {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\end{array}\)
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = + \infty \\\mathop {\lim }\limits_{x \to {1^ + }} \left( { - {x^2} - x} \right) = - 2\end{array}\)
Suy ra: \(\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{ - {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = - \infty \)