Đáp án:
\[\mathop {\lim }\limits_{x \to - 1} \frac{{x - 1 + \sqrt[3]{{2{x^3} + 10}}}}{{{x^2} + 5x + 4}} = \frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - 1} \frac{{x - 1 + \sqrt[3]{{2{x^3} + 10}}}}{{{x^2} + 5x + 4}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1 + \left( {\sqrt[3]{{2{x^3} + 10}} - 2} \right)}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1 + \frac{{2{x^3} + 10 - {2^3}}}{{{{\sqrt[3]{{2{x^3} + 10}}}^2} + 2\sqrt[3]{{2{x^3} + 10}} + 4}}}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1 + \frac{{2\left( {{x^3} + 1} \right)}}{{{{\sqrt[3]{{2{x^3} + 10}}}^2} + 2\sqrt[3]{{2{x^3} + 10}} + 4}}}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1 + \frac{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{{{\sqrt[3]{{2{x^3} + 10}}}^2} + 2\sqrt[3]{{2{x^3} + 10}} + 4}}}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{1 + \frac{{2\left( {{x^2} - x + 1} \right)}}{{{{\sqrt[3]{{2{x^3} + 10}}}^2} + 2\sqrt[3]{{2{x^3} + 10}} + 4}}}}{{x + 4}}\\
= \frac{{1 + \frac{{2.\left( {{{\left( { - 1} \right)}^2} - \left( { - 1} \right) + 1} \right)}}{{{{\sqrt[3]{{2{{\left( { - 1} \right)}^3} + 10}}}^2} + 2\sqrt[3]{{2{{\left( { - 1} \right)}^3} + 10}} + 4}}}}{{ - 1 + 4}}\\
= \frac{{1 + \frac{{2.3}}{{4.3}}}}{3}\\
= \frac{{1 + \frac{1}{2}}}{3}\\
= \frac{1}{2}
\end{array}\)
