I was wondering which areas this math problem involves in. I'm currently reading a course in calculus and I do not recognize the problem in my course literature .
Would be grateful if you could list the parts that you need to solve this problem.
Problem
Let $ (a_{n})_{n=1}^\infty $ designate the strictly growing sequence of all roots to $\tan x = x$ for $x> 0$.
Solve $\lim _{n\to \infty }\left((a_{n+1}) - (a_{n})\right)$
Perhaps this complete solution will help you to understand the methods.
(1). Preliminary work, some of which may be obvious:
(1-a). $\tan x$ is undefined when $(x-\pi /2)/\pi\in \Bbb Z.$ And if $n\in \Bbb N$ and $x-(n-1)\pi \in (\pi /2,\pi)$ then $\tan x < 0<x.$
(1-b).For $n\in \Bbb N$ let $A_n=[\pi n, \pi n+\pi /2).$ On $A_n$ the function $\tan x$ takes every value in $[0,\infty)$ while the function $f(x)=x$ is bounded and positive, so the function $g(x)=\tan x -f(x)$ takes both positive and negative values, and $g(x)$ is continuous on $A_n.$ So $g(x_n)=0$ for some $x_n\in A_n.$ And $g'(x)=-1+\sec^2 x>0$ for $x\in A_n$ except at the single point $x=\pi n ,$ so there cannot be more than one $x\in A_n$ for which $g(x)=0.$
(2). Let $x_n\in A_n$ with $g(x_n)=0.$ From (1-b) we know that $x_n$ is uniquely defined. Let $x_n=y_n +\pi n . $ We have $$x_{n+1}-x_n =(\pi (n+1)+y_{n+1})-(\pi n+y_n)=\pi +(y_{n+1}-y_n).$$ Now $y_n\in [0,\pi /2)$ and $\tan y_n\to \infty$ as $n\to \infty$ because $$\tan y_n=\tan x_n=x_n\in [\pi n, \pi n+\pi /2).$$ One of the properties of the $\tan $ function is that if $(y_n)_{n\in \Bbb N}$ is a sequence in $[0,\pi /2)$ with $\tan y_n\to \infty$ then $y_n\to \pi /2,$ which implies $y_{n+1}-y_n\to 0.$
Therefore $$x_{n+1}-x_n=((n+1)\pi +y_{n+1})-(n\pi +y_n)=$$ $$=\pi +(y_{n+1}-y_n)\to \pi \text { as } n\to \infty.$$