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?
 A: 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 $\lim_n f_n(x)= f(x)$.
A: In a sense, the answer is "yes". You can apply the Monotone Convergence Theorem pointwise.  That is, for a fixed value of $x$, show that the sequence of nonnegative numbers $\bigl(f_n(x)\bigr)$ is bounded above and increasing. Then of course for each $x$, the sequence $\bigl(f_n(x)\bigr)$ converges to some $f(x)$. (You can find $f(x)$ explicitly by taking the limits of both sides of your defining relation for $f_n$.)
Towards showing that $\bigl(f_n(x)\bigr)$ is indeed bounded above, try using the bound $1+2\sqrt x$. 
Perhaps better, as suggested by Did, would be to formally find the limit fuction first, and then show that this serves as an upper bound.
A: Hints:


*

*For every $x\gt0$, the function $u_x:t\mapsto \sqrt{x+t}$ is continuous hence every convergent sequence defined by $x_{n+1}=u_x(x_n)$ for every $n\geqslant0$ has limit $z_x$ such that $u_x(z_x)=z_x$. Here, $z_x=\frac12(1+\sqrt{1+4x})$.

*For every $x\gt0$, $z_x-u_x(t)=c_x(t)\cdot(z_x-t)$, with $c_x(t)=1/(z_x+t)$ hence $0\lt c_x(t)\leqslant 1/z_x$.

*For every $x\gt0$, applying the preceding remark to any sequence $(x_n)_n$ such that $x_0\leqslant z_x$ and $x_{n+1}=u_x(x_n)$ for every $n\geqslant0$ shows that $z_x-z_x^{-n}\cdot(z_x-x_0)\leqslant x_n\leqslant z_x$ for every $n\geqslant0$.

*For every $x\gt0$, $z_x\gt1$.


Conclusion: 


*

*For every $x\gt0$, $f_n(x)\to z_x$. On the other hand, $f_n(0)=0$ for every $n\geqslant0$ hence $f_n(0)\to0\ne z_0$ and the limit is discontinuous at $x=0^+$.

A: Clearly, for each $x \ge 0$ the sequence $\{f_n(x)\}_n$ is increasing.
Let 
$$
g:[0,\infty) \to \mathbb{R},\ g(x)=
\left\{
\begin{array}{cc}
0 & \text{ if } x=0\cr
\frac{1+\sqrt{1+4x}}{2} & \text{ if } x>0
\end{array}
\right..
$$
I claim that $f_n(x) \le g(x)$ for every $x \ge 0$. Indeed $f_n(0)=0=g(0)$ for every $n$, and $f_1(x) \le g(x)$ for every $x > 0$. If I suppose that $f_n(x) \le g(x)$ for $n \ge 1$ and $x > 0$, then
$$
f_{n+1}^2(x)-g^2(x)=x+f_n(x)-\frac{1+2x+\sqrt{1+4x}}{2}=f_n(x)-g(x)\le 0,
$$
i.e. $f_{n+1}(x)\le g(x)$. Hence $f_n(x)\le g(x)$ for every $n \ge 1$ and $x > 0$.
Thus for each $x \ge 0$, the sequence $\{f_n(x)\}_n$ is convergent (being increasing and bounded from above), and its (pointwise) limit $f: [0,\infty) \to \mathbb{R}, x \mapsto f(x)$ satisfies $f(x)=\sqrt{x+f(x)}$, i.e. $f(x)=g(x)$ for every $x \ge 0$.
