In order to calculate the frequency-dependent friction in pipe flows, we need to calculate the complex ratio of Bessel functions $J_0/J_2$, particularly $N(x)=\frac{J_0(x j\sqrt{j})}{J_2(x j\sqrt{j} )}$.
Calculating this naively fails as both $J_0$ and $J_2$ quickly become very large, with standard algorithms (e.g. Matlab) reporting infinity for x>1000. However, the ratio remains reasonable and slowly, asymptotically approaches a magnitude of 1.
How can one calculate N for x>1000?
I understand you are asking for the large argument behaviour of the ratio of two Kelvin functions: $$\operatorname{ber}_{\nu}x+i\operatorname{bei}_{\nu}x=J_{\nu}\left(xe^{3\pi i/4% }\right) $$ Their modulus and phase have rather nice asymptotic expressions for large argument. With the notation $$\operatorname{ber}_{\nu}x+i\operatorname{bei}_{\nu}x=M_\nu(x)e^{i\theta_\nu(x)} $$ and $\mu=4\nu^2$ one has \begin{align} M_{\nu}\left(x\right)&=\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\left(1-% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)^{2}}{256}\frac{1}{x^{2}}-% \frac{(\mu-1)(\mu^{2}+14\mu-399)}{6144\sqrt{2}}\frac{1}{x^{3}}+O\left(\frac{1}% {x^{4}}\right)\right)\\ \theta_{\nu}\left(x\right)&=\frac{x}{\sqrt{2}}+\left(\frac{1}{2}\nu-\frac{1}{8}% \right)\pi+\frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{\mu-1}{16}\frac{1}{x^{2}}-% \frac{(\mu-1)(\mu-25)}{384\sqrt{2}}\frac{1}{x^{3}}+O\left(\frac{1}{x^{5}}% \right) \end{align} In the ratio $N(x)$, the diverging exponential term is eliminated and I suppose you can use these asymptotic formula for $x>1000$. The ratio asymptotically approaches -1.