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Question Number 117037 by bemath last updated on 09/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}-\sqrt{\mathrm{x}}=?$$

Answered by Lordose last updated on 09/Oct/20

$$\frac{1}{2}$$

Answered by bobhans last updated on 09/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}-\sqrt{\mathrm{x}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{x}(1+\frac{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}{\mathrm{x}})}-\sqrt{\mathrm{x}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{x}}[\sqrt{1+\sqrt{\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{{\mathrm{x}}^3}}}}-1]=$$$$\mathrm{let}\sqrt{\mathrm{x}}=\frac{1}{\mathrm{t}}\mathrm{or}\mathrm{x}=\frac{1}{{\mathrm{t}}^2}\mathrm{where}\mathrm{t}\to 0$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{1+\sqrt{{\mathrm{t}}^2+\sqrt{{\mathrm{t}}^6}}}-1}{\mathrm{t}}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{1}{\sqrt{1+\sqrt{{\mathrm{t}}^2+\sqrt{{\mathrm{t}}^6}}}+1}\times \underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{{\mathrm{t}}^2+\sqrt{{\mathrm{t}}^6}}}{\mathrm{t}}=$$$$\frac{1}{2}\times \underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{t}\sqrt{1+\sqrt{{\mathrm{t}}^2}}}{\mathrm{t}}=\frac{1}{2}\times 1=\frac{1}{2}$$

Answered by Dwaipayan Shikari last updated on 09/Oct/20

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

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