Zadacha Kuznecov Predely 10-20

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Условие задачи

Вычислить предел функции:

<math>\lim_{x\to \frac{1}{3}} \frac{\sqrt[3]{\frac{x}{9}}-\frac{1}{3}} {\sqrt{\frac{1}{3}+x}- \sqrt{2x}}</math>

Решение

<math>\lim_{x\to \frac{1}{3}} \frac{\sqrt[3]{\frac{x}{9}}-\frac{1}{3}} {\sqrt{\frac{1}{3}+x}-\sqrt{2x}} = \lim_{x\to \frac{1}{3}} \frac{\left(\sqrt[3]{\frac{x}{9}}-\frac{1}{3}\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 \right)} {\left(\sqrt{\frac{1}{3}+x}-\sqrt{2x}\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 \right)} = </math>
<math>= \lim_{x\to \frac{1}{3}} \frac{\frac{x}{9}-\frac{1}{27}} {\left(\sqrt{\frac{1}{3}+x}-\sqrt{2x}\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 \right)} = </math>
<math>= \lim_{x\to \frac{1}{3}} \frac{-\frac{1}{9}\left(\frac{1}{3}-x\right)\left(\sqrt{\frac{1}{3}+x}+\sqrt{2x}\right)} {\left(\sqrt{\frac{1}{3}+x}-\sqrt{2x}\right)\left(\sqrt{\frac{1}{3}+x}+\sqrt{2x}\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 \right)} = </math>
<math>= \lim_{x\to \frac{1}{3}} \frac{-\frac{1}{9}\left(\frac{1}{3}-x\right)\left(\sqrt{\frac{1}{3}+x}+\sqrt{2x}\right)} {\left(\frac{1}{3}-x\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 \right)} = </math>
<math>= \lim_{x\to \frac{1}{3}} \frac{-\frac{1}{9}\left(\sqrt{\frac{1}{3}+x}+\sqrt{2x}\right)} {\sqrt[3]{\left(\frac{x}{9}\right)^2}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2} = \frac{-\frac{1}{9}\left(\sqrt{\frac{1}{3}+\frac{1}{3}}+\sqrt{2\cdot \frac{1}{3}}\right)} {\sqrt[3]{\left(\frac{1}{3} \cdot \frac{1}{9}\right)^2}+\sqrt[3]{\frac{1}{3} \cdot \frac{1}{9}} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2} = </math>
<math> = \frac{-\frac{1}{9}\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}\right)} {\sqrt[3]{\left(\frac{1}{27}\right)^2}+\sqrt[3]{\frac{1}{27}} \cdot \frac{1}{3} + \frac{1}{9}} = \frac{-\frac{1}{9} \cdot 2 \cdot \sqrt{\frac{2}{3}}} {\left(\frac{1}{3}\right)^2+\frac{1}{3} \cdot \frac{1}{3} + \frac{1}{9}} = \frac{-\frac{2}{9}\cdot \sqrt{\frac{2}{3}}} {\frac{3}{9}} = -\frac{2}{3} \sqrt{\frac{2}{3}}</math>