How to rigorously prove that the equation $\tan x=x$ has an infinite number of solutions Is it true that  $\tan x=x$ has an infinitely many solutions.
It seems intuitively true,
But how to rigorously prove it?
I can prove that $\lim_{x \rightarrow (\frac{\pi}{2} + \pi k)^-} \tan x = \infty$ and
$\lim_{x \rightarrow (\frac{\pi}{2} + \pi k)^+} \tan x = -\infty$ so I can use this without a proof but what next?
Thanks!
 A: Hint: Apply IVP (Intermediate Value Property) for the function $\tan x -x$ to each of the intervals $(n\pi-\frac {\pi} 2, n\pi+\frac {\pi} 2)$.
A: For all interval of $x$ defined as $\left] n\pi-\frac{\pi}{2},n\pi+\frac{\pi}{2}  \right[; \forall n \in \mathbb{Z}$ we have:

*

*The function $y_1=x$ is continuous because $\lim_{x \to \alpha} y_1=\alpha$.

*The function $y_2=tan(x)$ is continuous because $\lim_{x^+ \to \alpha} y_2=tan(\alpha)$ and $\lim_{x^- \to \alpha} y_2=tan(\alpha)$.

*So interval of $y_1$ is $\left] n\pi-\frac{\pi}{2},n\pi+\frac{\pi}{2}  \right[$.

*The interval of $y_2$ is $\left] -\infty,\infty\right[$, becaus $\lim_{x^- \to n\pi-\frac{\pi}{2}}y_2=-\infty$ and $\lim_{x^+ \to n\pi+\frac{\pi}{2}}y_2=\infty$.

*The values of $x=tan(x) \in  \left] n\pi-\frac{\pi}{2},n\pi+\frac{\pi}{2}  \right[ \cap \left] -\infty,+\infty\right[$ in this interval exists at least one  value because the intersection is not empty and the both functions are continuous (you can prove there is only one value because de both functions are monotone in the interval).

Because there are infinity intervals with at least one value, there are infinity many solutions.
