Can we refine this asymptotic for Laguerre polynomials? I just found an interesting and useful limit for Laguerre polynomials:
$$\lim_{n \to \infty} L_n \left( \frac{2r}{n+1/2} \right)=J_0(2 \sqrt{2r})$$
I'm using specifically this form of the argument because it's the one I'm working with in the application. Of course, we can set any fixed number instead of $1/2$ in the denominator.
I found this limit in a paper, which references G. Szego. Orthogonal Polynomials. Amer. Math. Soc. Colloq. Publ. 23, Amer. Math. Soc. Providence,
RI, 1975. Fourth Edition., , Theorem 8.1.3.
While the limit is useful for very large orders and smallish $r$, I would really like to know if there's a refinement that could be applied to derive an asymptotic expansion, which would depend on $n$.
Here's an illustration which shows that the limit is not that good for larger $r$ (though it does approximate the roots better than the magnitude):

Not sure how we could refine this asymptotic or how the original limit was derived (as I don't have the linked book).
One way is considering the differential equations for both functions.
There's also an interesting result from Gradshteyn-Ryzhik:
$$L_n(z)= \frac{2}{n!} e^z \int_0^\infty e^{-t^2} t^{2n+1} J_0(2t \sqrt{z}) dt$$
Which may or may not be related to the limit above.
 A: From the given representation, we can express
\begin{equation}
 L_n(\frac{2r}{n+1/2})= \frac{2}{n!} e^{\frac{2r}{n+1/2}} \int_0^\infty e^{-t^2} t^{2n+1} J_0\left(2t \sqrt{\frac{2r}{n+1/2}}\right) \,dt
\end{equation} 
changing $ t=\sqrt{u\left( n+1/2 \right)}$,
\begin{equation}
 L_n(\frac{2r}{n+1/2})= \frac{(n+1/2)^{n+1}}{n!} e^{\frac{2r}{n+1/2}}\int_0^\infty e^{-u\left( n+1/2 \right)}u^{n+1/2}J_0\left( 2\sqrt{2ru} \right)\frac{du}{\sqrt{u}}
\end{equation} 
or
\begin{equation}
  L_n(\frac{2r}{n+1/2})= \frac{(n+1/2)^{n+1}}{n!} e^{\frac{2r}{n+1/2}}\int_0^\infty e^{-\left( n+1/2 \right)\left( u-\ln u \right)}J_0\left( 2\sqrt{2ru} \right)\frac{du}{\sqrt{u}}
\end{equation} 
The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
$u-\ln(u)\sim 1+(u-1)^2/2$ and $u^{-1/2}J_0\left( 2\sqrt{2ru}\right)\sim J_0\left( 2\sqrt{2r}\right)$, then
\begin{equation}
  L_n(\frac{2r}{n+1/2})\sim \sqrt{2\pi}\frac{(n+1/2)^{n+1/2}}{n!} e^{\frac{2r}{n+1/2}}
  e^{-(n+1/2)}J_0\left( 2\sqrt{2r}\right)
\end{equation} 
Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(\frac{2r}{n+1/2})\sim J_0\left( 2\sqrt{2r}\right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.
