Giải thích các bước giải:
\(\begin{array}{l}
f,\\
\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt[3]{{{x^3} + 9}} - \sqrt {{x^2} + 3} }}{{{x^4} + {x^2} - 2}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {\sqrt[3]{{{x^3} + 9}} - 2} \right) + \left( {2 - \sqrt {{x^2} + 3} } \right)}}{{\left( {{x^4} + {x^3}} \right) - \left( {{x^3} + {x^2}} \right) + \left( {2{x^2} + 2x} \right) - \left( {2x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\frac{{\left( {{x^3} + 9} \right) - {2^3}}}{{{{\sqrt[3]{{{x^3} + 9}}}^2} + 2.\sqrt[3]{{{x^3} + 9}} + {2^2}}} + \frac{{{2^2} - \left( {{x^2} + 3} \right)}}{{2 + \sqrt {{x^2} + 3} }}}}{{\left( {x + 1} \right)\left( {{x^3} - {x^2} + 2x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\frac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{{{\sqrt[3]{{{x^3} + 9}}}^2} + 2.\sqrt[3]{{{x^3} + 9}} + 4}} - \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{2 + \sqrt {{x^2} + 3} }}}}{{\left( {x + 1} \right)\left( {{x^3} - {x^2} + 2x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\frac{{{x^2} - x + 1}}{{{{\sqrt[3]{{{x^3} + 9}}}^2} + 2.\sqrt[3]{{{x^3} + 9}} + 4}} - \frac{{x - 1}}{{2 + \sqrt {{x^2} + 3} }}}}{{\left( {{x^3} - {x^2} + 2x - 2} \right)}}\\
= \dfrac{{\frac{{{{\left( { - 1} \right)}^2} - \left( { - 1} \right) + 1}}{{{{\sqrt[3]{{{{\left( { - 1} \right)}^3} + 9}}}^2} + 2.\sqrt[3]{{{{\left( { - 1} \right)}^3} + 9}} + 4}} - \frac{{\left( { - 1} \right) - 1}}{{2 + \sqrt {{{\left( { - 1} \right)}^2} + 3} }}}}{{{{\left( { - 1} \right)}^3} - {{\left( { - 1} \right)}^2} + 2.\left( { - 1} \right) - 2}}\\
= \dfrac{{\frac{3}{{12}} + \frac{2}{4}}}{{ - 6}}\\
= - \frac{1}{8}
\end{array}\)
