If $\tan(x_n \pi) = -x_n$ and $x_0=0$, show that $$\lim_{n\to \infty} (n-x_n)=\frac{1}{2}$$
I am not getting any idea from where to start this. Can you give me some hint?
EDIT: $x_n$ is the nth solution of $y=\tan(x_n\pi)$ and $y=-x_n$ starting from $n=0$ where $x_n=0$
On
Take $n>1$.
First I shall show that $x_n \in (n-1,n)$
Using the Intermediate value theorem, we have that $ x_m\in (n-\frac{1}{2},n)$ for some integers $m$. Since $(\tan(x\pi)+x)'=\pi\sec^2(x\pi)+x>0$ in that interval, we have $x_m$ is the only solution in that interval. Also, there exist no solution in $(n-1,n-\frac{1}{2})$ because $\tan(x\pi)>0$ in that interval.
Thus, $x_m$ is the only solution in $(n-1,n)$, and by induction we can show that $n=m$.
Let $k=n-x_n$, note that $0<k<1$. The given equation can be written as: $$\tan((x_n -n)\pi) = -x_n$$ $$\Leftrightarrow \tan(-k\pi)=-(n-k)$$ $$\Leftrightarrow k\pi =\arctan(n-k)$$ Thus, as $n \to\infty$, $k\pi=\arctan(n-k)\to \frac{\pi}{2}$ (Note that $k<1$).
As a result, we have $k=\frac{1}{2}$, which was to be proved
Note that for $n\geq 1$, by the intermediate value theorem, $x_n\in (n-1/2,n)$ and therefore $n-x_n\in (0,1/2)$. It turns out that $n-x_n$ is increasing (prove it!). Then $n-x_n$ is convergent to some real number $L\in (0,1/2]$. Now, if $L<1/2$ then, as $n\to+\infty$, $$+\infty\leftarrow x_n=-\tan(x_n \pi)=\tan(-x_n \pi)=\tan((n-x_n) \pi)\rightarrow \tan(L \pi)\in\mathbb{R}^+$$ which is a contradiction. Hence we may conclude that $L=1/2$.