Limit of $n(n\pi + \frac{\pi}{2} - x_n)$, where $x_n$ is the solution of $x = \tan(x)$ in the interval $(n \pi, n\pi + \frac{\pi}{2})$

Question

Let $x_n$ denote the solution to $x = \tan(x)$ in the interval $(n \pi,n \pi +\frac{\pi}{2}).$ Find

$$ \lim_{n \to \infty}n \left(n\pi + \frac{\pi}{2} - x_n\right)$$

Answer 1

Let $f(y)=\tan y.$ Then $f(0)=0$ and $f'(0)=1.$ So for $y\in (0,\pi/2)$ let $y=yg(y)+\tan y$ where $\lim_{y\to 0}g(y)=0.$

Let $y_n=n\pi+\pi/2-x_n.$ Then $$0<y_n<\tan y_n=\frac {1}{\tan x_n}=\frac {1}{x_n}<\frac {1}{n\pi}.$$ So $0<\frac {y_n}{1/n\pi}<1.$ So $g(y_n)\to 0$ as $n\to \infty.$

Now $$ny_n=n \tan y_n+ny_ng(y_n)=$$ $$=\frac {n}{\tan x_n}+ \frac {y_n}{1/n\pi} \frac {g(y_n)}{\pi}=$$ $$=\frac {n}{x_n}+\frac {y_n}{1/n\pi} \frac {g(y_n)}{\pi}$$ and we have $$\frac {1}{\pi (1+1/2n)}= \frac {n}{n\pi+\pi/2}<\frac {n}{x_n}<\frac {n}{n\pi}=\frac {1}{\pi}.$$

Answer 2

$x_n\in(n\pi,n\pi+\frac{\pi}{2})\implies n\pi+\frac{\pi}{2}-x_n\in(0,\frac{\pi}{2})$ and $\frac{1}{n\pi+\frac{\pi}{2}}<\frac{1}{x_n}<\frac{1}{n\pi}$. \begin{align*} \tan\left(n\pi+\frac{\pi}{2}-x_n\right) &=\tan\left(\frac{\pi}{2} - x_n\right)\\ &=\frac{1}{\tan x_n}\\ &=\frac{1}{x_n}. \end{align*} So $$n\pi+\frac{\pi}{2}-x_n=\arctan\frac{1}{x_n}\sim\frac{1}{x_n}\quad (n\to\infty).$$ $$\lim_{n\to\infty}n\left(n\pi+\frac{\pi}{2}-x_n\right) =\lim_{n\to\infty}n\arctan\frac{1}{x_n} =\lim_{n\to\infty}\frac{n}{x_n}=\frac{1}{\pi}.$$

Answer 3

http://blitiri.blogspot.com/2012/11/solving-tan-x-x.html

We say that $x_n = \frac{(2n+1)\pi}{2}-\epsilon_n$ with $\epsilon_n$ being an error term.

Thus our limit is actually $\lim_{n\to\infty}n\epsilon_n$.

You can find that $\frac{2}{(2n+1)\pi-2\epsilon_n}=\frac{1}{x_n} = \cot(x_n) = \tan(\epsilon_n) = \epsilon_n + O(\epsilon_n^3)$ by using trig identities.

Hence in our limit:

$\lim_{n\to \infty}n \epsilon_n = \lim \frac{2n}{(2n+1)\pi-2\epsilon_n} - nO(\epsilon_n^3)) = \lim \frac{2}{2\pi+\frac{\pi}{2n}-2 \epsilon_n/n} -nO(\epsilon_n)^3 = \frac{1}{\pi}$

My argument for $\epsilon_n/n \to 0$ is that $\epsilon_n \to 0$ in any case. Furthermore you could do it rigiously, but since $\epsilon_n \sim \frac{2}{(2n+1)}\pi$, the third order limit is zero.