Tính $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}}$ bằng?

$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{(1 - \sqrt[3]{{x + 1}})\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - (x + 1)}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{3x\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{x + 1}} + {{\left( {\sqrt[3]{{x + 1}}} \right)}^2}} \right)}} \\= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3\left( {1 + \sqrt[3]{{0 + 1}} + {{\left( {\sqrt[3]{{0 + 1}}} \right)}^2}} \right)}} = \dfrac{{ - 1}}{9}\end{array}$