Cho dãy số

- Hướng dẫn giải

Ta có:

$$\begin{gathered} \begin{array}{l}{u}_n=\sqrt[3]{n+2}-\sqrt[3]{n}=\frac{n+2-n}{{\left(\sqrt[3]{n+2}\right)}^2+\sqrt[3]{n+2}.\sqrt[3]{n}+{\left(\sqrt[3]{n}\right)}^2}\\ =\frac{2}{{\left(\sqrt[3]{n\left(1+\frac{2}{n}\right)}\right)}^2+\sqrt[3]{n\left(1+\frac{2}{n}\right)}.\sqrt[3]{n}+{\left(\sqrt[3]{n}\right)}^2}\end{array} \\ =\frac{2}{{\left(\sqrt[3]{n}\right)}^2\left[{\left(\sqrt[3]{1+\frac{2}{n}}\right)}^2+\sqrt[3]{1+\frac{2}{n}}+1\right]} \end{gathered}$$

Vì $\lim \frac{2}{3{\left(\sqrt[3]{n}\right)}^2}=0$ và $\lim \frac{1}{{\left(\sqrt[3]{1+\frac{2}{n}}\right)}^2+\sqrt[3]{1+\frac{2}{n}}+1}=\frac{1}{3}$

Do đó $\lim u_n=0$

Chọn đáp án B.