Ta có
$\underset{x \to -1}{\lim} \dfrac{\sqrt[3]{x} + 1}{x + 1} = \underset{x \to -1}{\lim} \dfrac{\sqrt[3]{x} + 1}{(\sqrt[3]{x} + 1)(\sqrt[3]{x^2} - \sqrt[3]{x} + 1)} = \underset{x \to -1}{\lim} \dfrac{1}{\sqrt[3]{x^2} - \sqrt[3]{x} + 1} = \dfrac{1}{1 -(-1) + 1} = \dfrac{1}{3}$