Convergence of the sequence $f_{1}\left(x\right)=\sqrt{x} $ and  $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ 
Let $\left\{ f_{n}\right\} $ denote the set of functions on
  $[0,\infty) $ given by $f_{1}\left(x\right)=\sqrt{x} $ and
  $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ for $n\ge1 $.
  Prove that this sequence is convergent and find the limit function.

We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n} $. Does this fact still apply on $C[0,\infty) $, the set of continuous functions on $[0,\infty) $. If yes, what bound would we use?

You know that the sequence is increasing. Now notice that if $f(x)$ is the positive solution of $y^2=x+y$, we have $f_n(x)<f(x) \ \forall n \in \mathbb{N}$. In fact, notice that this hold for $n=1$  and assuming for $n-1$ we have 
$$f^{2}_{n+1}(x) = x +f_n(x) < x+f(x) = f^{2}(x).$$ Then there exist $\lim_n f_n(x)$ and taking the limit in $f^{2}_{n+1}(x) = x + f_n(x)$, we find that $\lim_n f_n(x)= f(x)$.
