Question

Помогите с решением.

Answer 1

Ответ:

$-\frac{2\sqrt{6} }{9}$

Объяснение:

$\lim_{x \to \frac{1}{3} } \frac{\sqrt[3]{\frac{x}{9}} - \frac{1}{3} }{\sqrt[]{\frac{1}{3}+ x} -\sqrt{2x} } =[\frac{0}{0}] = \lim_{x \to \frac{1}{3} } \frac{(\sqrt[3]{\frac{x}{9}} - \frac{1}{3})( \sqrt[]{\frac{1}{3}+ x} +\sqrt{2x})\frac{1}{9} }{\frac{1}{3^3}- \frac{x}{9} } } =$

$=\lim_{x \to \frac{1}{3} } \frac{(\sqrt[3]{\frac{x}{9}} - \frac{1}{3})( \sqrt[]{\frac{1}{3}+ x} +\sqrt{2x})\frac{1}{9} }{(\frac{1}{3}- \sqrt[3]{\frac{x}{9} })(\frac{1}{9}+ \frac{\sqrt[3]{\frac{x}{9} }}{3} +\sqrt[3]{\frac{x^2}{9^2} }) } = -\lim_{x \to \frac{1}{3} } \frac{ \sqrt[]{\frac{1}{3}+ x} +\sqrt{2x} }{1+ 3\sqrt[3]{\frac{x}{9} } +9\sqrt[3]{\frac{x^2}{9^2} }}=\\=-\frac{2\sqrt{\frac{2}{3} } }{1+1+ 1} = -\frac{2}{3} \sqrt{\frac{2}{3} } = -\frac{2\sqrt{6} }{9}$