Tính lim x → 0 1 − căn bậc ba của (x + 1) /3 x bằng?

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

Đáp án cần chọn là: D