Đáp án:
\[L = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3x + 2}}\left( {{x^3} - 3x + 1} \right) - 6}}{{{x^2} - 4}} = \frac{{75}}{{16}}\]
Giải thích các bước giải:
\(\begin{array}{l}
L = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3x + 2}}\left( {{x^3} - 3x + 1} \right) - 6}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3x + 2}}\left( {{x^3} - 3x - 2} \right) + 3\sqrt[3]{{3x + 2}} - 6}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3x + 2}}\left( {{x^3} - 2{x^2} + 2{x^2} - 4x + x - 2} \right) + 3.\left( {\sqrt[3]{{3x + 2}} - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3x + 2}}\left( {x - 2} \right){{\left( {x + 1} \right)}^2} + 3.\frac{{3x + 2 - {2^3}}}{{{{\sqrt[3]{{3x + 2}}}^2} + 2\sqrt[3]{{3x + 2}} + 4}}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\sqrt[3]{{3x + 2}}{{\left( {x + 1} \right)}^2}}}{{x + 2}} + \frac{9}{{\left( {x + 2} \right)\left( {{{\sqrt[3]{{3x + 2}}}^2} + 2\sqrt[3]{{3x + 2}} + 4} \right)}}} \right]\\
= \frac{{\sqrt[3]{{3.2 + 2}}.{{\left( {2 + 1} \right)}^2}}}{{2 + 2}} + \frac{9}{{\left( {2 + 2} \right).\left( {{{\sqrt[3]{{3.2 + 2}}}^2} + 2.\sqrt[3]{{3.2 + 2}} + 4} \right)}}\\
= \frac{{2.9}}{4} + \frac{9}{{4.12}}\\
= \frac{{75}}{{16}}
\end{array}\)
