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 an alternative answer, consider the infinitely nested radical expression $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}$.
Evidently $y > \sqrt{x+\sqrt{x+\sqrt{x}}}$ for $x > 1$ and therefore $\frac{y}{x}$ is an upper bound for $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.  
Also $y > \sqrt{x}$ and therefore $\lim_{x\to\infty} y = \infty$
But we can express $x$ in terms of $y$ by using the substitution $y = \sqrt{x+y}$ in the infinitely nested expression, leading to $x = y(y-1)$.
Then $\lim_{x\to\infty}\frac{y}{x} = \lim_{y\to\infty} \frac{y}{y(y-1)} = \lim_{y\to\infty}\frac{1}{y-1} = 0$.
