Evaluate $\lim_{n\to\infty}n(n\pi+\pi/2-x_n)$

Question

Where $x_n$ is the solution of $\tan(x)=x$ in the interval $(n\pi ,n\pi +\frac\pi2)$ , $n\ge0$.

Any hints on how to approach this problem with elementary methods in the first place? The answer should be $\frac1\pi$. $$\lim_{n\to\infty}n\biggl(n\pi+\frac\pi2-x_n\biggr)$$

Answer 1

$x_n$ is the solution of the fixed-point equation $$x_n=n\pi+\arctan(x_n).$$ For large $n$, this will be close to the pole of the tangent at $n\pi+\frac\pi2$, so this formula can be modified to $$ x_n=n\pi+\frac\pi2-\arctan\left(\frac1{x_n}\right)=n\pi+\frac\pi2-\frac1{n\pi+\frac\pi2}+O(n^{-2}). $$ as $\arctan(u)=u+O(u^3)$ for $u\approx 0$ and $\frac1{x_n}=\frac1{n\pi+\frac\pi2}+O(n^{-2})$.

This approximation $x_n^0=(n+\frac12)\pi-\left((n+\frac12)\pi\right)^{-1}$ of $x_n$ is even good for low values of $n$, as the numerical experiment shows \begin{array}{r|ll} n&x_n^0&x_n\\\hline 1 & 4.500182389595496 & 4.493409457909064 \\ 2 & 7.726657679500967 & 7.725251836937707 \\ 3 & 10.904628605797479 & 10.904121659428899 \\ 4 & 14.066431410891004 & 14.066193912831473 \\ 5 & 17.220885069983172 & 17.22075527193077 \\ 6 & 20.371381496613076 & 20.37130295928756 \\ 7 & 23.51950358376561 & 23.519452498689006 \\ 8 & 26.666089333609268 & 26.66605425881267 \\ 9 & 29.811623905294212 & 29.81159879089296 \\ \end{array}