What are the possible limits of the iteration $x_{n+1}=\sqrt{x_n+3}$, $x_0=0$?

Question

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. -1

2. 3

3. 0

4. $\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}}$

I think only option 4. is correct as it the only one satisfying $x^2-x-3=0$

Answer 1

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.

Answer 2

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.

Answer 3

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,

$$x=\frac{\sqrt{13}+1}2.$$