Approximate solution to a transcendental equation in the limit of a variable I have the following transcendental equation:
$$2\cot{x}=\frac{kx}{hL}-\frac{hL}{kx}\tag1$$
I would like to inquire whether an approximate solution to $(1)$ can be developed in the limit $h\rightarrow0$.
Some typical values are: $k=16, L=0.25,h=0.001$. Using numerical methods following solutions are obtained for this set of parameters:
x = 0.00559016, 3.1416, 6.28319, 9.42478, 12.5664, 15.708, ....
 A: When $h \to 0$, the second term on the right-hand side becomes negligible, and the first-term tends to infinity. Hence, the equation approximates the equation $2 \cot x \to \infty$, which occurs when $x$ is an integer multiple of $\pi$. Indeed, this is what you see in the numerical solutions: all of them are (very!) close to integer multiples of $\pi$.
Another observation: it depends on which side of zero you are approaching it from! It seems like you are approaching zero from the right, so your roots will always be slightly greater than the integer multiples of $\pi$. This is because $\lim_{x \to n \pi} \cot x$ is actually undefined because the limit tends to negative infinity from the left and positive infinity from the right. But when $h$ tends to zero from the right, the right hand side of your expression tends to positive infinity. So the solutions will always lie slightly above the integer multiples of $\pi$.
A: Changing notations, you want to find the zero's of function
$$f(x)=\cot(x)-a x+\frac 1{ax}$$ where $a$ is potentially large.
Instead, in order to remove the discontinuities, I prefer to consider the function
$$g(x)=x \cos (x)-a \left(x^2-1\right) \sin (x)$$ Developing as series around $x=k \pi$ and using later series reversion, we have as an approximation
$$\color{blue}{x_k=t+\frac{t}{a \left(t^2-1\right)}-\frac{t \left(t^2+1\right)}{a^2 \left(t^2-1\right)^3}-\frac{t \left(t^6-8 t^4-14 t^2-3\right)}{3 a^3 \left(t^2-1\right)^5}+O\left(\frac{1}{a^4}\right)}$$ where $\color{red}{t=k \pi}$.
Using your numbers $(k=16, L=0.25,h=0.001)$ gives $a=32000$. Let us use only the first and second term since $a$ is effectively very large. The first roots will then be
$$\{0,3.1416037,6.2831904,9.4247813,12.566373,15.707965,18.849558\}$$ which seems to be OK.
Notice that I missed the very small first root because of the transformation form $f(x)$ to $g(x)$.
Edit
For the small root, we need a special treatment
$$\cot(x)-a x+\frac 1{ax}=0 \implies a=\frac{\left(\frac{\sqrt{5 x^2-3 x^2 \cos (2 x)}}{\sqrt{2}}+x \cos (x)\right) \csc (x)}{2 x^2}$$ Developing as series around $x=0$ gives
$$a=\frac{1}{x^2}+\frac{2}{3}-\frac{31 x^2}{45}+\frac{1069 x^4}{945}+O\left(x^6\right)$$ Now, series reversion
$$x=\frac 1{\sqrt a} \left( 1+\frac{1}{3 a}-\frac{8}{45 a^2}+\frac{53}{630
   a^3}+O\left(\frac{1}{a^4}\right)\right)$$ which, for $a=32000$ gives $x=0.005590228174$ while the exact solution is exactly the same.
