Задача №1975
Условие
Найти предел \(\lim_{x\to{0+0}}\left(\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}-\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}\right)\).
Решение
\[
\lim_{x\to{0+0}}\left(\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}-\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}\right)=|\infty-\infty|=\\
=\lim_{x\to{0+0}}\frac{\left(\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}-\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}\right)\cdot\left(\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}+\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}\right)}{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}+\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}}=\\
=2\cdot\lim_{x\to{0+0}}\frac{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}+\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}}=\\
=2\cdot\lim_{x\to{0+0}}\frac{\sqrt{1+\sqrt{x}}}{\sqrt{1+\sqrt{x+x\sqrt{x}}}+\sqrt{1-\sqrt{x+x\sqrt{x}}}}
=2\cdot\frac{1}{1+1}=1.
\]
Ответ:
1