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Đáp án:

Ta có:

$\begin{array}{l}B=\lim \left(\sqrt[3]{n^3+9n^2}-n\right)\\ =\lim \frac{\left(\sqrt[3]{n^3+9n^2}-n\right)\left(\sqrt[3]{{\left(n^3+9n^2\right)}^2}+n\sqrt[3]{n^3+9n^2}+n^2\right)}{\sqrt[3]{{\left(n^3+9n^2\right)}^2}+n\sqrt[3]{n^3+9n^2}+n^2}\\ =\lim \frac{n^3+9n^2-n^3}{\sqrt[3]{{\left(n^3+9n^2\right)}^2}+n\sqrt[3]{n^3+9n^2}+n^2}\\ =\lim \frac{9n^2}{\sqrt[3]{{\left(n^3+9n^2\right)}^2}+n\sqrt[3]{n^3+9n^2}+n^2}\\ =\lim \frac{9n^2}{n^2\sqrt[3]{{\left(1+\frac{9}{n}\right)}^2}+n^2.\sqrt[3]{1+\text{\hspace{0.17em}}\frac{9}{n}}+n^2}\\ =\lim \frac{9}{\sqrt[3]{{\left(1+\frac{9}{n}\right)}^2}+\sqrt[3]{1+\frac{9}{n}}+1}=\frac{9}{1+1+1}=3\end{array}$

Đáp án cần chọn là: D