Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + 2x - 3}}{{\sqrt[3]{{x + 7}} - 3x + {x^2}}} = - \frac{{60}}{{11}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + 2x - 3}}{{\sqrt[3]{{x + 7}} - 3x + {x^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - {x^2} + {x^2} - x + 3x - 3}}{{\left( {\sqrt[3]{{x + 7}} - 2} \right) + {x^2} - 3x + 2}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 3} \right)}}{{\frac{{x + 7 - {2^3}}}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}} + \left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 3} \right)}}{{\frac{{x - 1}}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}} + \left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x + 3}}{{\frac{1}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}} + x - 2}}\\
= \frac{{{1^2} + 1 + 3}}{{\frac{1}{{{{\sqrt[3]{{1 + 7}}}^2} + 2\sqrt[3]{{1 + 7}} + 4}} + 1 - 2}}\\
= \frac{5}{{\frac{1}{{12}} - 1}}\\
= - \frac{{60}}{{11}}
\end{array}\)
