Задача №1964

\[ \begin{aligned} & \sqrt[3]{1+\frac{x}{3}}=\left(1+\frac{x}{3}\right)^\frac{1}{3}=1+\frac{x}{3\cdot{3}}+o(x)=1+\frac{x}{9}+o(x);\\ & \sqrt[4]{1+\frac{x}{4}}=\left(1+\frac{x}{4}\right)^\frac{1}{4}=1+\frac{x}{4\cdot{4}}+o(x)=1+\frac{x}{16}+o(x);\\ & \sqrt{1-\frac{x}{2}}=\left(1+\left(-\frac{x}{2}\right)\right)^\frac{1}{2}=1-\frac{x}{2\cdot{2}}+o(x)=1-\frac{x}{4}+o(x). \end{aligned} \]

\[ \lim_{x\to{0}}\frac{\sqrt[3]{1+\frac{x}{3}}-\sqrt[4]{1+\frac{x}{4}}}{1-\sqrt{1-\frac{x}{2}}}=\left|\frac{0}{0}\right| =\lim_{x\to{0}}\frac{1+\frac{x}{9}+o(x)-\left(1+\frac{x}{16}+o(x)\right)}{1-\left(1-\frac{x}{4}+o(x)\right)}=\\ =\lim_{x\to{0}}\frac{7x+o(x)}{36x+o(x)} =\left|\begin{aligned}&7x+o(x)\sim{7x};\\&36x+o(x)\sim{36x}.\end{aligned}\right|=\lim_{x\to{0}}\frac{7x}{36x}=\frac{7}{36}. \]