$1=\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$ $$\lim_{n \to  0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$ 
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?
 A: As $\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}\ge0$ and $\sqrt{1+4n^2}\ge1$
$\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\dfrac{1+\sqrt{1+4n^2}}2$ 
A: 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.
A: You have discovered that the limit of a pointwise convergent sequence of functions is not necessarily continuous.
Define, over $[0,\infty)$, the functions
$$
f_0(x)=0,
\qquad
f_{n+1}(x)=\sqrt{x+f_n(x)}
$$
So $f_1(x)=\sqrt{x}$, $f_2(x)=\sqrt{x+\sqrt{x}}$, and so on.
Clearly, $f_n(0)=0$, for every $n$, and $f_n(x)$ is continuous. On the other hand, for $x>0$, we have
$$
\lim_{n\to\infty}f_n(x)=\frac{1+\sqrt{1+4x}}{2}
$$
(see $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$)
So the sequence of functions $f_n(x)$ converges pointwise to the function
$$
f(x)=\begin{cases}
0 & \text{if $x=0$}\\[6px]
\dfrac{1+\sqrt{1+4x}}{2} & \text{if $x>0$}
\end{cases}
$$
and this function is not continuous:
$$
\lim_{x\to0}\lim_{n\to\infty}f_n(x)=
\lim_{x\to0}\frac{1+\sqrt{1+4x}}{2}=1
$$
You can't exchange the limits:
$$
\lim_{n\to\infty}\lim_{x\to0}f_n(x)=
\lim_{n\to\infty}0=0
$$
