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Question Number 108131 by mathdave last updated on 14/Aug/20

Commented by Her_Majesty last updated on 14/Aug/20

$$justatry$$$$forx\to +\mathrm{\infty }wehave\frac{x+\sqrt{x}}{x}=1+\frac{1}{\sqrt{x}}\to 1$$$$\Rightarrow \sqrt{x+\sqrt{x+\sqrt{x}}}\sim \sqrt{x+1}$$$$\Rightarrow thelimit=1$$$$or:\sqrt{x+\sqrt{x}}\sim \frac{1}{2}+\sqrt{x}whenx\to +\mathrm{\infty }$$$$\Rightarrow \sqrt{x+\sqrt{x+\sqrt{x}}}=\sqrt{x+\frac{1}{2}+\sqrt{x}}\sim \sqrt{x+\sqrt{x}}\sim \frac{1}{2}+\sqrt{x}$$$$and\sqrt{x+1}\sim \sqrt{x}$$$$\Rightarrow thelimit=1$$

Answered by abdomathmax last updated on 14/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}{\sqrt{\mathrm{x}+1}}\Rightarrow \mathrm{for}\mathrm{x}>0\mathrm{we}\mathrm{get}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}}.\sqrt{1+\sqrt{\frac{\mathrm{x}+\sqrt{\mathrm{x}}}{{\mathrm{x}}^2}}}}{\sqrt{\mathrm{x}}\sqrt{1+\frac{1}{\mathrm{x}}}}=\frac{\sqrt{1+\sqrt{\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}\sqrt{\mathrm{x}}}}}}{\sqrt{1+\frac{1}{\mathrm{x}}}}$$$$\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=1$$$$$$

Commented by Her_Majesty last updated on 14/Aug/20

$$great!$$

Commented by abdomathmax last updated on 15/Aug/20

$$\mathrm{thanks}$$

Answered by bemath last updated on 15/Aug/20

$$\sqrt{\underset{x\to \mathrm{\infty }}{\lim }\frac{x+\sqrt{x+\sqrt{x}}}{x+1}}=\sqrt{\underset{x\to \mathrm{\infty }}{\lim }\frac{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^2}}}}{1+\frac{1}{x}}}$$$$=\sqrt{\frac{1+0}{1+0}}=1$$

Answered by 1549442205PVT last updated on 15/Aug/20

$$\mathrm{Put}\frac{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}=\mathrm{y}\Rightarrow {\mathrm{y}}^2=\frac{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}{\mathrm{x}+\sqrt{\mathrm{x}}}$$$$=\frac{1+\sqrt{\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{{\mathrm{x}}^3}}}}{1+\sqrt{\frac{1}{\mathrm{x}}}}\to 1\mathrm{when}\mathrm{x}\to \mathrm{\infty }$$$$\Rightarrow \underset{\mathbf{x}\to \mathrm{\infty }}{\mathbf{lim}}\left({\mathbf{y}}^2\right)=1\Rightarrow \underset{\mathbf{x}\to \mathrm{\infty }}{\mathbf{lim}\mathit{y}}=1(\mathbf{as}\mathbf{y}>0)$$

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