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Question Number 74766 by chess1 last updated on 30/Nov/19

Commented by mathmax by abdo last updated on 30/Nov/19

$$changement\frac{1}{x}=tleadyo{lim}_{t\to +\mathrm{\infty }}\sqrt{t+\sqrt{t+\sqrt{t}}}-\sqrt{t-\sqrt{t+\sqrt{t}}}$$$$={lim}_{t\to +\mathrm{\infty }}g\left(t\right)wehave\sqrt{t+\sqrt{t}}=\sqrt{t}\times \sqrt{1+\frac{1}{\sqrt{t}}}\sim \sqrt{t}(1+\frac{1}{2\sqrt{t}})$$$$=\sqrt{t}+\frac{1}{2}\Rightarrow \sqrt{t+\sqrt{t+\sqrt{t}}}\sim \sqrt{t+\sqrt{t}+\frac{1}{2}}also\sqrt{t-\sqrt{t+\sqrt{t}}}\sim \sqrt{t-\sqrt{t}-\frac{1}{2}}$$$$and\sqrt{t+\sqrt{t}+\frac{1}{2}}=\sqrt{t}\left(\sqrt{1+\frac{1}{\sqrt{t}}+\frac{1}{2t}}\right)\sim \sqrt{t}\{1+\frac{1}{2}(\frac{1}{\sqrt{t}}+\frac{1}{2t})\}$$$$=\sqrt{t}+\frac{1}{2}+\frac{1}{4\sqrt{t}}also\sqrt{t-\sqrt{t}-\frac{1}{2}}=\sqrt{t}\left(\sqrt{1-\frac{1}{\sqrt{t}}-\frac{1}{2t}}\right)$$$$\sim \sqrt{t}\{1-\frac{1}{2}(\frac{1}{\sqrt{t}}+\frac{1}{2t})\}=\sqrt{t}-\frac{1}{2}-\frac{1}{4\sqrt{t}}\Rightarrow$$$$f\left(t\right)\sim \sqrt{t}+\frac{1}{2}+\frac{1}{4\sqrt{t}}-\sqrt{t}+\frac{1}{2}+\frac{1}{4\sqrt{t}}=1+\frac{1}{2\sqrt{t}}\to 1(t\to +\mathrm{\infty })\Rightarrow$$$${lim}_{x\to 0^{+}}(\sqrt{....}-\sqrt{....})=1$$

Commented by chess1 last updated on 30/Nov/19

$$\mathrm{thanks}$$

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