Задача №1976

\[ \lim_{x\to\infty}\left(\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3-x^2+1}\right)=\left|\infty-\infty\right|=\\ =\lim_{x\to\infty}\frac{\left(\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3-x^2+1}\right)\cdot\left(\sqrt[3]{\left(x^3+x^2+1\right)^2}+\sqrt[3]{x^3+x^2+1}\cdot\sqrt[3]{x^3-x^2+1}+\sqrt[3]{\left(x^3-x^2+1\right)^2}\right)}{\sqrt[3]{\left(x^3+x^2+1\right)^2}+\sqrt[3]{x^3+x^2+1}\cdot\sqrt[3]{x^3-x^2+1}+\sqrt[3]{\left(x^3-x^2+1\right)^2}}=\\ =\lim_{x\to\infty}\frac{2x^2}{\sqrt[3]{\left(x^3+x^2+1\right)^2}+\sqrt[3]{x^3+x^2+1}\cdot\sqrt[3]{x^3-x^2+1}+\sqrt[3]{\left(x^3-x^2+1\right)^2}}=\\ =\lim_{x\to\infty}\frac{2}{\sqrt[3]{\left(1+\frac{1}{x}+\frac{1}{x^3}\right)^2}+\sqrt[3]{\left(x^3+x^2+1\right)\cdot\left(1-\frac{1}{x}+\frac{1}{x^3}\right)}+\sqrt[3]{\left(1-\frac{1}{x}+\frac{1}{x^3}\right)^2}}=\frac{2}{3}. \]