$x_n$ is the $n$'th positive solution to $x=\tan(x)$. Find $\lim_{n\to\infty}\left(x_n-x_{n-1}\right)$

$x_n$ is the $n$'th positive solution to $x=\tan(x)$. Find $\lim_{n\to\infty}\left(x_n-x_{n-1}\right)$.

Solution 1:

HINT

Show that $n \pi < x_n < n \pi + \dfrac{\pi}2$, where $x_0 = 0$ by considering the function $f(x) = x - \tan(x)$, which is continuous in the interval $n \pi < x_n < n \pi + \dfrac{\pi}2$.

Now consider $y_n = n \pi + \dfrac{\pi}2 - x_n$. We have $y_n \in (0,\pi/2)$. Since $x_n = \tan(x_n)$, we have $$n \pi + \dfrac{\pi}2 - y_n = \cot(y_n) \in \left(n \pi, n \pi + \dfrac{\pi}2\right) \implies \underbrace{\tan(y_n) \in \left(\dfrac1{n \pi + \dfrac{\pi}2}, \dfrac1{n \pi}\right) \implies y_n < \dfrac1{n \pi}}_{\text{Because $0 < y_n < \tan(y_n)$}}$$ Hence, we have $$\left \vert x_n - \left(n \pi + \dfrac{\pi}2\right) \right \vert < \dfrac1{n \pi}$$ Conclude what you want from this.

Solution 2:

HINT: Sketch $x = \mathrm{tan}(x)$ and think about the asymptotes of tan$(x)$.