<s>[INST] 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? 
 [/INST] Jonas produced the solution first, but I wanted to write up a solution for my own edification. 

For $x>1$ we can write the following inequalities,
$$x < x+a\sqrt{x} < (1+a)x\qquad (a>0),$$
This allows us to come up with an upper and lower bound for the numerator, 
$$\sqrt{x+\sqrt{x+\sqrt{x}}} < \sqrt{x+\sqrt{2x}}< \sqrt{(1+\sqrt{2})x} = \sqrt{x}\ \sqrt{(1+\sqrt{2})}$$
$$\sqrt{x+\sqrt{x+\sqrt{x}}} > \sqrt{x+\sqrt{x}} > \sqrt{x}$$
So our function, which we are taking the limit of, has the following bounds on it, 
$$\frac{\sqrt{x}}{x} <\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}<\frac{\sqrt{x}}{x}\sqrt{1+\sqrt{2}}, $$ 
as $x\rightarrow \infty$ the function $\sqrt{x}/x$ goes to $0$. Our function is bounded above by something which goes to zero and below by something which goes to zero it must also go to zero; this is called the squeeze theorem. 
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = 0.$$
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