# Limit of the ratio of two modified Bessel functions

Could you help me evaluate the following limit: $$\lim_{x\to \infty} \frac{K_1(x)}{K_0(x)}$$

where $K_\nu$ is the modified Bessel function of the second kind of order $\nu$. Both $K_1$ and $K_0$ approaches $0$ as $x\to\infty$, so we have an indeterminate form. L'Hospital doesn't seem to help because the derivative of $K_\nu$ is $K_{\nu +1}$, and all the orders of the modified Bessel function of the second kind approach $0$ as $x\to\infty$.

Another related limit that I am also struggling is: $$\lim_{x\to 0} \frac{K_1(x)}{K_0(x)}$$

• That ratio is $1$. Just use the large argument asymptotic expansions. – Mark Viola May 23 '18 at 23:31
• @MarkViola I tried that approach, but didn't get the result that I expected for a related calculation. However, if you are sure that the ratio is $1$ for large $x$, then I will double check my other results. – A Slow Learner May 23 '18 at 23:33

For large values of $x$ $$K_1(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}+\frac 3 8\frac 1 {x^{ 3/2}}-\frac{15}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$ $$K_0(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}-\frac 1 8\frac 1 {x^{ 3/2}}+\frac{9}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$ making $$\frac{K_1(x)}{K_0(x)}=1+\frac{1}{2 x}-\frac{1}{8 x^2}+O\left(\frac{1}{x^3}\right)$$
For small values of $x$ $$K_1(x)=\frac{1}{x}+\frac{1}{4} x (2 \log (x)+2 \gamma -1-2 \log (2))+O\left(x^3\right)$$ $$K_0(x)=-\log (x)-\gamma +\log (2)+\frac{1}{4} x^2 (-\log (x)-\gamma +1+\log (2))+O\left(x^3\right)$$ making $$\frac{K_1(x)}{K_0(x)}=\frac{-1}{x (\log (x)+\gamma -\log (2))}+O\left(x\right)$$
For large values of $x$ $$K_n(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}+\frac {4n^2-1} 8\frac 1 {x^{ 3/2}}+\frac{16 n^4-40 n^2+9}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$ making $$\frac{K_m(x)}{K_n(x)}=1+\frac{(m-n) (m+n)}{2 x}+\frac{(m-n) (m+n) \left(m^2-n^2-2\right)}{8 x^2}+O\left(\frac{1}{x^3}\right)$$