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:
$$
\lim_{x\to 0} \frac{K_1(x)}{K_0(x)}
$$
 A: 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 
$$\frac{K_1(x)}{K_0(x)}=\frac{-1}{x (\log (x)+\gamma -\log (2))}+O\left(x\right)$$
Edit
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
$$\frac{K_m(x)}{K_n(x)}=1+\frac{(m-n) (m+n)}{2 x}+\frac{(m-n) (m+n) \left(m^2-n^2-2\right)}{8
   x^2}+O\left(\frac{1}{x^3}\right)$$
