Đáp án: $\dfrac43$
Giải thích các bước giải:
Ta có:
$A=\lim_{n\to +\infty}\dfrac{\sqrt[3]{2n^2-n^3}+n}{\sqrt{n^2+n}-n}$
$\to A=\lim_{n\to +\infty}\dfrac{\dfrac{2n^2-n^3+n^3}{(\sqrt[3]{2n^2-n^3})^2-n\sqrt[3]{2n^2-n^3}+n^2}}{\dfrac{n^2+n-n^2}{\sqrt{n^2+n}+n}}$
$\to A=\lim_{n\to +\infty}\dfrac{\dfrac{2n^2}{(\sqrt[3]{2n^2-n^3})^2-n\sqrt[3]{2n^2-n^3}+n^2}}{\dfrac{n}{\sqrt{n^2+n}+n}}$
$\to A=\lim_{n\to +\infty}\dfrac{2n(\sqrt{n^2+n}+n)}{(\sqrt[3]{2n^2-n^3})^2-n\sqrt[3]{2n^2-n^3}+n^2}$
$\to A=\lim_{n\to +\infty}\dfrac{2(\sqrt{1+\dfrac1n}+1)}{(\sqrt[3]{\dfrac2n-1})^2-\sqrt[3]{\dfrac2n-1}+1}$
$\to A=\dfrac{2(\sqrt{1+0}+1)}{(\sqrt[3]{0-1})^2-\sqrt[3]{0-1}+1}$
$\to A=\dfrac43$
