What are the possible limits of the iteration $x_{n+1}=\sqrt{x_n+3}$, $x_0=0$? Let $f(x)=\sqrt{x+3}$ for $x\ge -3$.
Consider the iteration $$x_{n+1}=f(x_n),x_0=0;n\ge 0$$ 
The possible limits of the iteration are 


*

*-1 

*3  

*0  

*$\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}}$
I think only option 4. is correct as it the only one satisfying $x^2-x-3=0$
 A: You're right, but you should be looking at the equation $x=\sqrt{x+3}$.
The equation $x=\sqrt{x+3}$ has only one solution, and it happens to be in $(-3,+\infty)$.
The equation $x^2-x-3=0$ has two solutions, both of which are in $(-3,+\infty)$.
The options given make the distinction above irrelevant but it might not be so in other questions.
A: You are right: The prospective limit would have to be a solution of
$$x^2=x+3,\quad x>0\ .\tag{1}$$
There is exactly one solution of $(1)$, namely
$$\xi:={1\over2}(1+\sqrt{13})\doteq2.3\ .$$
But the above argument does not prove that we actually have convergence. (For an example consider the sequence defined by $x_0:=2$, $x_{n+1}:=x_n^2$ $(n\geq0)$. Here $\xi=1$, but the sequence actually diverges.)  For a proof  that in the example of the question one indeed has  $\lim_{\to\infty}x_n=\xi$ put 
$$y_n:=\xi-x_n\quad(n\geq0)\ ,$$
with $y_0=\xi$. Then
$$y_{n+1}=\xi-x_{n+1}=\xi-\sqrt{x_n+3}=\xi-\sqrt{\xi^2-y_n}={y_n\over \xi+\sqrt{\xi^2-y_n}}\qquad(n\geq0)\ .$$
It is easy to see that $0<y_n\leq\xi<\xi^2$ implies
$$0<y_{n+1}<{y_n\over\xi}\leq1<\xi\ ,$$
so that we obtain $0<y_n\leq \xi^{1-n}$ $(n\geq0)$. This proves $\lim_{n\to\infty}y_n=0$ and hence our conjecture that $\lim_{\to\infty}x_n=\xi$. 
The obtained limit differs from all options given. Note that option 4. is just a folklore rewriting of the problem and cannot be considered as its solution.
A: IMO, the answer $4$ is rather poor, as it is just a rewriting of the recurrence relation and doesn't tell you much more.
A much better one would be the positive root of the characteristic equation,
$$x=\frac{\sqrt{13}+1}2.$$
