Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt[3]{{{x^2} + x + 8}} + x - 1}}{{\sqrt[3]{{x + 2}} - \sqrt[3]{{{x^2} - 3x - 3}}}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt[3]{{{x^2} + x + 8}} - 2 + x + 1}}{{\sqrt[3]{{x + 2}} - 1 + 1 - \sqrt[3]{{{x^2} - 3x - 3}}}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\dfrac{{{x^2} + x + 8 - 8}}{{{{\left( {\sqrt[3]{{{x^2} + x + 8}}} \right)}^2} + 2\sqrt[3]{{{x^2} + x + 8}} + 4}} + x + 1}}{{\dfrac{{x + 2 - 1}}{{{{\left( {\sqrt[3]{{x + 2}}} \right)}^2} + \sqrt[3]{{x + 2}} + 1}} + \dfrac{{1 - {x^2} + 3x + 3}}{{1 + \sqrt[3]{{{x^2} - 3x - 3}} + {{\left( {\sqrt[3]{{{x^2} - 3x - 3}}} \right)}^2}}}}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\dfrac{x}{{{{\left( {\sqrt[3]{{{x^2} + x + 8}}} \right)}^2} + 2\sqrt[3]{{{x^2} + x + 8}} + 4}} + 1}}{{\dfrac{1}{{{{\left( {\sqrt[3]{{x + 2}}} \right)}^2} + \sqrt[3]{{x + 2}} + 1}} + \dfrac{{4 - x}}{{1 + \sqrt[3]{{{x^2} - 3x - 3}} + {{\left( {\sqrt[3]{{{x^2} - 3x - 3}}} \right)}^2}}}}}\\
= \dfrac{{\dfrac{{ - 1}}{{{{\left( {\sqrt[3]{{{1^2} - 1 + 8}}} \right)}^2} + 2\sqrt[3]{{{1^2} - 1 + 8}} + 4}} + 1}}{{\dfrac{1}{{{{\left( {\sqrt[3]{{ - 1 + 2}}} \right)}^2} + \sqrt[3]{{ - 1 + 2}} + 1}} + \dfrac{{4 + 1}}{{1 + \sqrt[3]{{{1^2} + 3.1 - 3}} + {{\left( {\sqrt[3]{{{1^2} + 3.1 - 3}}} \right)}^2}}}}}\\
= \dfrac{{\dfrac{{ - 1}}{{12}} + 1}}{{\dfrac{1}{3} + 1}}\\
= \dfrac{{11}}{{12}}.\dfrac{3}{4} = \dfrac{{11}}{{16}}
\end{array}$
