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