Compute the limit Consider an equation
$$
   \tan (x) = \frac{a x}{x^2+b}
$$
where $a,b \neq 0$. Plotting $\tan(x)$ and function on the RHS we can see that this equation has infinitely many positive solutions $x_{n}$ and $x_{n} \sim \pi n$ as $n \to \infty$, i.e. 
$$
   \lim\limits_{n \to \infty} \frac{x_{n}}{n} = \pi.
$$
But is it possible to show the latter equality analitically?
 A: Let us assume $a,b>0$. The proof can be easily adapted if $a,b$ are not positive. 
Consider $f(x) = \dfrac{ax}{x^2+b} - \tan(x)$. Note that $f(x)$ is nice except at $x = m \pi + \pi/2$ i.e. $$f((m \pi + \pi/2)^-) = - \infty \,\,\,\,\,\, f((m \pi + \pi/2)^+) = \infty$$ 
$$f'(x) = \dfrac{a(b-x^2)}{(b+x^2)^2} - \sec^2(x)$$ For large enough $x$ i.e. $x > \sqrt{ab}$, we have that $f'(x) < 0$.
Hence, for large enough $n$ within $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right)$, the function is decreasing and changes sign. Hence, there is exactly one root in this interval.
Further, we have that $f(n \pi) = \dfrac{an \pi}{n^2 \pi^2+b} > 0$.
\begin{align}
f \left(n \pi + \dfrac{a}{n \pi} \right) & = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(n \pi + \dfrac{a}{n \pi} \right)\\
& = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2} - \tan\left(\dfrac{a}{n \pi} \right)\\
& = \dfrac{a}{\left(n \pi + \dfrac{a}{n \pi} \right)} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq \dfrac{a}{n \pi} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq 0.
\end{align}
Hence, the root in the interval $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right) $ in fact lies within $\left( n \pi, n \pi + \dfrac{a}{n \pi} \right)$.
Hence, we have that $$\left \vert x_{n+k} - n\pi  \right \vert \leq \dfrac{a}{n \pi}$$ where $k$ is a fixed natural number and takes into account some initial ugliness in the function where there could be possibly more or less roots. Hence,
$$\left \vert \dfrac{x_{n+k}}{n} - \pi \right \vert \leq \dfrac{a}{n^2 \pi}$$
This gives your desired result.
A: Without loss of generality we may assume that $a>0$. Assume that $x$ is "large enough" depending on $a,b$. Then $\frac{a x}{x^2+b}$ is strictly decreasing on $(x_0,\infty)$ so it meets $\tan(x)$ exactly one on the intervals $(n\pi,n\pi+\pi/2)$. Denote it by $x_n$. Then $\lim_{n\to\infty} x_n=\infty$. Furthermore $\lim_{n\to\infty} \frac{a x_n}{x_n^2+b}=0$. It implies $\lim_{n\to\infty}\tan(x_n)=0$. Hence $x_n-n\pi=o(1)$. From this $\frac{x_n}{n}=\pi+o\left(\frac{1}{n}\right)$.
