The original question was to find the value this function approaches as $x$ goes to infinity given that the limit exists. This is easy to figure and turns out to be $(1+5^{1/2})/2$.
I was though, thinking, if it were possible to come up with such a continuous function over non-negative numbers first and then find the limit. I did try a bit by assuming the function is continuous and taking the derivative and somehow managing a solvable differential equation but didn't reach anywhere. It'd be amazing to see if such a continuous function exists or not!!
$f : [0,\infty) \mapsto [0,\infty)$
$f(x+1)=(1+f(x))^{1/2}$
given $f(0)=0$
Assume $f$ is such a function. Then $f(x)\ge -1$ for all $x$ and $f(1)=1$.
Conversely, let $f_0\colon [0,1]\to[-1,\infty)$ be any continuous function with $f_0(0)=0$, $f_0(1)=1$. Then we can define recursively functions $f_n\colon [n,n+1]\to [-1,\infty)$ by letting $f_n(x)=\sqrt{f_{n-1}(x-1)+1}$. Then $f_n$ is continuous and we have $f_n(n+1)=\sqrt{f_n(n)+1}$. Therefore the function $f\colon [0,\infty)\to[-1,\infty)$ given by $f(x)=f_{\lfloor x\rfloor}(x)$ is continuous, has $f(0)=0$ and $f(x+1)=\sqrt{f(x)+1}$ for all $x$.
As $f_0$ is continuous, it is bounded, say $m_0\le f_0(x)\le M_0$ for all $x\in[0,1]$. And if $m_n\le f(x)\le M_n$ for $x\in[n,n+1]$, then $m_{n+1}:=\sqrt{m_n+1}\le f(x)\le \sqrt{M_n+1}=:M_{n+1}$ for $x\in[n+1,n+2]$. With $\phi:=\frac{\sqrt 5+1}2$, we may assume wlog that $m_0<\phi<M_0$. For $0\le x<y$, we have $$0<\sqrt{y+1}-\sqrt{x+1}=\frac{y-x}{\sqrt{y+1}+\sqrt{x+1}}<\frac{y-x}2.$$ Therefore (noting that $m_1\ge 0$), we have $m_n<\phi<M_n$ for all $n$ and $M_n-m_n\to 0$. We conclude that $\lim_{x\to\infty}f(x)=\phi$.