Asymptotic large order approximation for Bessel function expression How does one find the asymptotic large order approximation for $\sup_{0\le x\le\infty} \left(\sqrt{x} J_n(x)\right)$, where $J_n$ is the Bessel function of the first kind and order $n$.  This is NOT a problem of large argument expansion.
I have tried to using all of the standard recursion relations, integral and series representations, etc. with little success.  For example, the usual large order asymptotic expressions provide little insight here inasmuch as we are searching for the asymptotic, large-order behavior of the supremum of $\sqrt{x}J_n(x)$ over all real, non-negative $x$. 
Note:
I suspect that the answer might be $O(n^{\frac16})$ through, dare I admit it, empirical experiment.  But this "result" is not substantiated.  Any useful hints are greatly appreciated.

Let $\nu>0$ and denote $c_\nu=\sup_{x\ge 0}\sqrt{x}|J_\nu(x)|$. It is shown in this paper that
$$
c_\nu \leq \frac{{\sqrt {j_{\nu ,1} } }}{{\nu ^{1/3} }}2^{1/3} \sup_{x \in \mathbb R} \left| \operatorname{Ai}(x) \right| = \frac{{\sqrt {j_{\nu ,1} } }}{{\nu ^{1/3} }} \times 0.67488509 \ldots ,
$$
where $j_{\nu,1}$ is the first positive zero of $J_\nu(x)$ and $\operatorname{Ai}(x)$ is the Airy function. It is also proved that
$$
c_\nu \sim 2^{1/3} \sup_{x \in \mathbb R} \left| \operatorname{Ai}(x) \right|\times \nu^{1/6} = 0.67488509 \ldots \times \nu^{1/6}
$$
as $\nu \to +\infty$. Since $j_{\nu,1}\sim \nu$, the above upper bound is asymptotically sharp.
