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?

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 $x=0^+$.

