Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw  some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help? 

For large $x$ you have $\sqrt x<x$ and so
$$
\sqrt{x+\sqrt{x+\sqrt{x}}}<\sqrt{x+\sqrt{2x}}<\sqrt{3x}.
$$
Since $\sqrt{x+\sqrt{x+\sqrt{x}}}/x<\sqrt{3x}/x\to0$ when $x\to\infty$, your limit is zero.
