Convergence of the sequence $x_n=\tan x_{n-1}$ I'm stuck with this problem:

Consider the sequence $x_n=\tan x_{n-1}$ with arbitrary $x_1$. Does this sequence converge?

I tried to apply Cauchy criterion. We have $|\tan x-\tan y|\ge |x-y|$ for all $x,y$ in the domain of the function $\tan x$, but the domain of $\tan x$ is not "continuous" because it does not contain numbers of the form $n\cdot \frac{\pi}{2}$. So all I can say about this problem is I guess that the sequence diverges, without any logical explanation.
Thank you very much.  
 A: There will be infinitely many initial values $x_1$ such that $x_n$ is eventually constant.  But that's the only way it can converge, because $|\tan(x) - \tan(y)| > |x - y|$ for $x \ne y$ in the same interval $((n-1/2) \pi, (n+1/2) \pi)$ (not for all $x$, $y$ in the domain: for example $\tan(n\pi) = 0$ for all integers $n$).
A: To complement Robert's idea, assume that $(x_{n})$ converges.
Denoting $L = \lim_{n\to\infty} x_{n}$, then $L$ is a solution of the equation $L = \tan L$. It is not hard to show that $L$ cannot be a half-integer multiple of $\pi$: $L \notin \pi \Bbb{Z} + \tfrac{1}{2}\pi$.
Now let $\epsilon > 0$ be small enough so that
$$ (L-\epsilon, L+\epsilon) \subset (n\pi-\tfrac{1}{2}\pi, n\pi+\tfrac{1}{2}\pi) $$
for some $n \in \Bbb{Z}$, and let $N$ be such that $|x_{n} - L| < \epsilon$ whenever $n \geq N$. Then by the Mean Value Theorem (this is applicable since both $x_{n}$ and $L$ lie in the open interval on which $\tan$ is defined and differentiable),
$$ |x_{n+1} - L| = \left|\tan x_{n} - \tan L\right| \geq \left|x_{n} - L\right|. $$
Since this is true for any $n \geq N$, we must have
$$ |x_{n} - L| \geq |x_{N} - L|. $$
Therefore, in order to have convergence, we must have $x_{N} = x_{N+1} = \cdots = L$. That is, $(x_{n})$ must become eventually constant.
