Approximate an equation with Bessel functions when one variable tends to 0 Consider:
$$- \frac{\mu_1 \epsilon_1 J_{n - 1}(p) J_{n + 1}(p)}{p^2 J_n^2 (p)} + \frac{(\mu_1 \epsilon_2 + \mu_2 \epsilon_1) J'_n (p) K'_n (q)}{pq J_n(p) K_n(q)} + \frac{\mu_2 \epsilon_2 K_{n - 1}(q) K_{n + 1}(q)}{q^2 K_n^2 (q)} = n^2 \frac{\mu_1 \epsilon_1 + \mu_2 \epsilon_2}{p^2 q^2}$$
where $n > 1$ is an integer; $\mu_{1,2}$, $\epsilon_{1,2}$ are constant, $J$ is the Bessel function of the first kind and $K$ is the Modified Bessel function of the second kind. $p, q \in \mathbb{R}$ are the variables.
In the limit for $q \to 0$, this equation should become:
$$(\mu_1 \epsilon_2 + \mu_2 \epsilon_1) \frac{p J_{n - 1}(p)}{J_n(p)} = n(\epsilon_1 - \epsilon_2)(\mu_2 - \mu_1) + \frac{\epsilon_2 \mu_2}{n - 1}p^2$$
But how is it possible?

My attempt: for $q \to 0$,
$$\frac{K_{n-1}(q)}{q K_n(q)} \to \frac{1}{2(n - 1)}$$
which looks like the $(n - 1)$ term in the above solution; but also
$$\frac{K_{n+1}(q)}{q K_n(q)} \to \infty$$
so, the whole third addend should diverge, as well as the RHS; as regards the second addend:
$$\frac{K'_n(q)}{q K_n(q)} \to -\infty$$
and with three potentially diverging terms I am not able to proceed.

If anyone is curious, this is from a Physics problem, shown in S. A. Schelkunoff, Electromnagnetic Waves (D. Van Nostrand Company, Inc., Princeton, New Jersey, 1943), p. 425 (pdf, p. 440). It seems that there are no physical considerations between the full equation and its approximation, so this is the reason why I'm posting this here.
 A: *

*Rewrite the equation as:


$$- \mu_1 \epsilon_1 \frac{J_{n - 1}(p)}{p J_n (p)} \frac{J_{n + 1}(p)}{p J_n (p)} + (\mu_1 \epsilon_2 + \mu_2 \epsilon_1) \frac{ J'_n (p)}{p J_n(p)} \frac{K'_n (q)}{q K_n(q)} + \mu_2 \epsilon_2 \frac{K_{n - 1}(q)}{q K_n (q)} \frac{K_{n + 1}(q)}{q K_n (q)} = n^2 \frac{\mu_1 \epsilon_1 + \mu_2 \epsilon_2}{p^2 q^2}$$


*Use:


$$\frac{J_{n + 1}(p)}{p J_n(p)} = - \frac{J_{n - 1}(p)}{p J_n(p)} + \frac{2n}{p^2}$$
$$\frac{K_{n + 1}(q)}{q K_n(q)} = \frac{K_{n - 1}(q)}{q K_n(q)} + \frac{2n}{q^2}$$
$$\frac{J'_n(p)}{p J_n(p)} = \frac{J_{n - 1}(p)}{p J_n(p)} - \frac{n}{p^2}$$
$$\frac{K'_n(q)}{q K_n(q)} = - \frac{K_{n - 1}(q)}{q K_n(q)} - \frac{n}{q^2}$$
(they can be obtained from the Bessel functions recurrence and derivatives relations)
Replace the above identities in the equation. Create a single term from all the $n^2 / (p^2 q^2)$ addends.


*Consider $q \to 0$ and use the limiting form mentioned in the question:


$$\frac{K_{n-1}(q)}{q K_n(q)} \to \frac{1}{2(n - 1)}$$


*Multiply both sides by $q^2$. If $q \to 0$, all the terms with $q^2$ at the numerator go to $0$. The equation reduces to:


$$n (\mu_1 \epsilon_2 + \mu_2 \epsilon_1) \frac{J_{n - 1}(p)}{p J_n(p)} = (\epsilon_1 - \epsilon_2)(\mu_2 - \mu_1) \frac{n^2}{p^2} + \mu_2 \epsilon_2 \frac{n}{n - 1}$$


*Multiply both sides by $p^2 / n$.

