Đáp án:$\lim (\sqrt[3]{n^3+3n+1}-\sqrt{n^2+4n})=-2$
Giải thích các bước giải:
$\lim (\sqrt[3]{n^3+3n+1}-\sqrt{n^2+4n})$
$=\lim (\sqrt[3]{n^3+3n+1}-n)-(\sqrt{n^2+4n}-n)$
$=\lim \dfrac{n^3+3n+1-n^3}{\sqrt[3]{n^3+3n+1}^2+\sqrt[3]{n^3+3n+1}.n+n^2}-\dfrac{n^2+4n-n^2}{\sqrt{n^2+4n}+n}$
$=\lim \dfrac{3n+1}{\sqrt[3]{n^3+3n+1}^2+\sqrt[3]{n^3+3n+1}.n+n^2}-\dfrac{4n}{\sqrt{n^2+4n}+n}$
$=\lim \dfrac{\dfrac{3}{n}+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n^2}+\dfrac{1}{n^3}}^2+\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}-\dfrac{4}{\sqrt{1+\dfrac{4}{n}}+1}$
$=\dfrac{0+0}{\sqrt[3]{1+0+0}^2+\sqrt[3]{1+0+0}+1}-\dfrac{4}{\sqrt{1+0}+1}$
$=-2$
