Limit of expression involving modified Bessel functions of the first kind

Question

I am a scientist who is trying to come up with some analytic solutions for a system that I only have approximate answers to and I have run into the problem of finding the following limit.

$$ \lim_{\lambda \to \infty} \left( 2\sqrt{\lambda} \left[ \frac{I_0(2\sqrt{\lambda})}{I_1(2\sqrt{\lambda})} - \frac{I_1(2\sqrt{\lambda})}{I_0(2\sqrt{\lambda})} \right] \right).$$

Where $I_n(x)$ is modified Bessel function of the first kind.

I know this limit should be equal to 1 and Mathematica agrees as well but I do not know how to prove it. Any insights or direction would be helpful!

Answer 1

We calculate the equivalent limit

$$L=\lim_{x\to \infty}x\Big[\frac{I_0(x)}{I_1(x)}-\frac{I_1(x)}{I_0(x)}\Big]$$

The asymptotic expansions of the two modified Bessel functions are:

$$I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}(1+\frac{1}{8x}+\mathcal{O}(x^{-2}))+i\frac{e^{-x}}{\sqrt{2\pi x}}(1-\frac{1}{8x}+\mathcal{O}(x^{-2}))\\I_1(x)\sim\frac{e^x}{\sqrt{2\pi x}}(1-\frac{3}{8x}+\mathcal{O}(x^{-2}))-i\frac{e^{-x}}{\sqrt{2\pi x}}(1+\frac{3}{8x}+\mathcal{O}(x^{-2}))$$

from which we can easily deduce that

$$\frac{I_0(x)}{I_1(x)}\sim\Big[\frac{1+\frac{1}{8x}}{1-\frac{3}{8x}}+\mathcal{O(x^{-2})}\Big]\Big(1+i\mathcal{O}(\frac{e^{-2x}}{x})\Big)$$

and thus we can ignore imaginary parts in the calculation below, which reads:

$$L=\lim_{x\to\infty}\Big[x\Big(\frac{1+\frac{1}{8x}}{1-\frac{3}{8x}}-\frac{1-\frac{3}{8x}}{1+\frac{1}{8x}}\Big)+\mathcal{O}(1/x)\Big]=\lim_{8x\equiv y\to\infty}\frac{y(y-1)}{(y+1)(y-3)}=1$$

Answer 2

Too long for a comment.

Since you use Mathematica, you could notice that function $$f(x)=x^\color{red}{2}\Big[\frac{I_0(x)}{I_1(x)}-\frac{I_1(x)}{I_0(x)}\Big]$$ is almost a straight line with a slope of $1$.

Using the same approach as DinosaurEgg in his/her answer, using a few more terms $$x\Big[\frac{I_0(x)}{I_1(x)}-\frac{I_1(x)}{I_0(x)}\Big]=1+\frac{1}{2 x}+\frac{1}{2 x^2}+\frac{11}{16 x^3}+\frac{5}{4 x^4}+\cdots$$ which is a very good approximation even for small values of $x$.

Trying for $x=10$, the "exact" value is $1.0558548$ while the above truncated expression leads to $\frac{16893}{16000}=1.0558125$.