Let $I_{v}(\cdot)$ be the modified Bessel function of the first kind of order $v$. Does anybody knows an approximation for $\frac{I_{\frac{v}{2}}(\frac{v}{2}x)}{I_{\frac{v}{2}-1}(\frac{v}{2}x)}$ that holds uniformly with respect to $x$ as $v\rightarrow\infty$?
Thanks for any help.
This is not an answer but it is too long for a comment.
I do not know how this has been obtained but, if you look here, a similar question has been asked four says ago (unfortunately, it did not receive any comment or answer). This post asked to confirm that when $v\to \infty$ $$R_v=\frac{I_{\frac{v}{2}}(\frac{v}{2}x)}{I_{\frac{v}{2}-1}(\frac{v}{2}x)}=\frac{x}{1+\sqrt{1+x^{2}}}+O\left(\frac{1}{v}\right)$$ which seems to be quite good when compared numerically even for small values of $v$.
Computing the Taylor expansion of $R_{100}$ we get $$R_{100}=\frac{x}{2}-\frac{25 x^3}{204}+\frac{625 x^5}{10608}-\frac{62500 x^7}{1792089}+O\left(x^9\right)$$ while $$\frac{x}{1+\sqrt{1+x^{2}}}=\frac{x}{2}-\frac{x^3}{8}+\frac{x^5}{16}-\frac{5 x^7}{128}+O\left(x^{9}\right)$$ which show very similar coefficients.
Edit
If you look at this paper and, in particular at equation $(3.3a)$ $$\frac{I_{n}(x)}{I_{n-1}(x)}=\frac {x}{2n+x\frac{I_{n+1}(x)}{I_{n}(x)}} $$ you then end with $$\frac{I_{n}(nx)}{I_{n-1}(nx)}=\frac {x}{2+x\frac{I_{n+1}(nx)}{I_{n}(nx)}} $$ So, solving for $L$ equation $$L=\frac x{2+x L}\implies L=\frac{x}{1+\sqrt{1+x^{2}}}$$