$1=\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$ $$\lim_{n \to  0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$ 
this look amazingly, square root of zero is $1$.
$$s=\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$$
$$s=\sqrt{n+s}$$
$$s=\frac{ 1\pm \sqrt{1+4n^2}}{2}$$
$$ \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\frac{1\pm \sqrt{1+4n^2}}{2}$$
$$\lim_{n \to  0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\lim_{n \to  0}\frac{1\pm \sqrt{1+4n^2}}{2}$$
But there are two answer $0$ and $1$, which is correct?

First, you must prove if the limit exist.
If $s=1$, then $s=\sqrt{n+s}$ means $1=\sqrt{n+1}$
If $s=0$, then $s=\sqrt{n+s}$ means $0=\sqrt{n+0}$
When n=0, both are correct, but the limit is unique, so the limit doesn't exist.
