Задача Кузнецов Пределы 10-19

Условие задачи

Вычислить предел функции:

$\lim_{x\to \frac{1}{2}} \frac{\sqrt[3]{\frac{x}{4}}-\frac{1}{2}} {\sqrt{\frac{1}{2}+x}-\sqrt{2x}}$

Решение

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

$= \lim_{x\to \frac{1}{2}} \frac{\frac{x}{4}-\frac{1}{8}} {\left(\sqrt{\frac{1}{2}+x}-\sqrt{2x}\right)\left(\sqrt[3]{\left(\frac{x}{4}\right)^2}+\sqrt[3]{\frac{x}{4}} \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 \right)} =$

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

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

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

$= \frac{-\frac{1}{4}\left(1+1\right)} {\sqrt[3]{\left(\frac{1}{8}\right)^2}+\sqrt[3]{\frac{1}{8}} \cdot \frac{1}{2} + \frac{1}{4}} = \frac{-\frac{2}{4}} {\left(\frac{1}{2}\right)^2+\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4}} = \frac{-\frac{1}{2}} {\frac{3}{4}} = -\frac{2}{3}$