Задача №1961

\[ \lim_{x\to{0}}\frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{x+2\sqrt[3]{x^4}}=\left|\frac{0}{0}\right|=\\ =\lim_{x\to{0}}\frac{\left(\sqrt[3]{27+x}-\sqrt[3]{27-x}\right)\cdot\left(\sqrt[3]{(27+x)^2}+\sqrt[3]{27+x}\cdot\sqrt[3]{27-x}+\sqrt[3]{(27-x)^2}\right)}{\left(x+2\sqrt[3]{x^4}\right)\cdot\left(\sqrt[3]{(27+x)^2}+\sqrt[3]{27+x}\cdot\sqrt[3]{27-x}+\sqrt[3]{(27-x)^2}\right)}=\\ =\lim_{x\to{0}}\frac{2x}{x\cdot\left(1+2\sqrt[3]{x}\right)\cdot\left(\sqrt[3]{(27+x)^2}+\sqrt[3]{27+x}\cdot\sqrt[3]{27-x}+\sqrt[3]{(27-x)^2}\right)}=\\ =\lim_{x\to{0}}\frac{2}{\left(1+2\sqrt[3]{x}\right)\cdot\left(\sqrt[3]{(27+x)^2}+\sqrt[3]{27+x}\cdot\sqrt[3]{27-x}+\sqrt[3]{(27-x)^2}\right)}=\frac{2}{27}. \]