Tính lim từ x đến 0 1-căn bậc ba (x+1)/3x bằng? A.-1/3 B.0 C.1/3 D.-1/9

* Hướng dẫn giải

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