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Question Number 162081 by cortano last updated on 26/Dec/21

$$\underset{x\to 0}{\lim }(\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}-\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}=?$$

Answered by mr W last updated on 26/Dec/21

$$\frac{1}{L}=\underset{x\to 0}{\lim }\frac{1}{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}-\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}+\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}}{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}-\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}+\sqrt{\frac{1}{x}-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}}{2\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x}}}}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{1+\sqrt{x+x\sqrt{x}}}+\sqrt{1-\sqrt{x+x\sqrt{x}}}}{2\sqrt{1+\sqrt{x}}}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{1+0}+\sqrt{1-0}}{2\sqrt{1+0}}=1$$$$$$$$\Rightarrow L=1$$

Commented by Tawa11 last updated on 26/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by blackmamba last updated on 26/Dec/21

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x-\sqrt{x+\sqrt{x}}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{2\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x-\sqrt{x+\sqrt{x}}}}$$$$=2\times \underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x}}}\right)}{\sqrt{x}(\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^2}}}}+\sqrt{1-\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^2}}}})}$$$$=2\times \frac{1}{1+1}=1$$

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