\(\lim_{x\to\infty}\left(\frac{1}{\sqrt[3]{n^3+1}-n}\right)=\lim_{x\to\infty}\left(\frac{1}{\frac{n^3+1-n^3}{\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2}}\right)\)
\(=\lim_{x\to\infty}\left(\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2\right)=\lim_{x\to\infty}\left\lbrack n^2\left(\sqrt[3]{\left(1+\frac{1}{n^3}\right)^2}+\sqrt[3]{1+\frac{1}{n^3}}+1\right)\right\rbrack=+\infty\)
\(\lim_{x\to\infty}\left(\sqrt[3]{n^3-2n^2}-n\right)\)
\(=\lim_{x\to\infty}\frac{n^3-2n^2-n^3}{\sqrt[3]{\left(n^3-2n^2\right)^2}+n\cdot\sqrt[3]{n^3-2n^2}+n^2}\)
\(=\lim_{x\to\infty}\frac{-2n^2}{n^2\cdot\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}=\lim_{x\to\infty}\frac{-2}{\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}\)
=-∞
