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Đáp án:

Ta có: 

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

Đáp án cần chọn là: C