Asymptotics of Bessel function $J_n(z)$ for $|z| \gg n$ but $|z|\not\gg n^2$ Let $\{n_j\}_{j=1}^\infty$ be a sequence of integers such that $n_j \to +\infty$, let $z_j = x_j + iy_j$ be a sequence of complex numbers approaching the real line, i.e. $x_j \to + \infty$ and $y_j \to 0^+$. Let $J_n$ denote the Bessel function. I am trying to prove
$$\lim_{j\to +\infty} x_j \left[ J^2_{n_j}(z_j) - J^2_{n_j}(x_j) \right] = 0 \tag{1}$$
under the assumption
$$x_j \gg n_j$$
i.e. the argument of the Bessel function is much larger than its order. Under the stronger assumption
$$x_j \gg n_j^2$$
the equation $(1)$ follows from the large argument asymptotics of Bessel functions and error estimates. E.g. for the Hankel function DLMF (10.17.13) says
$$H_n^{(1)}(z) = \left( \frac{2}{\pi z} \right)^{\frac{1}{2}} e^{i \omega} (1+R_1(n,z)), \qquad \omega = z - \frac{\pi n}{2} - \frac{\pi}{4} \tag{2}$$
and for the remainder term we have the estimate (say $0\leq \arg z \leq \pi/2$)
$$|R_1(n,z)| \leq C \frac{n^2}{|z|} e^{\frac{n^2}{|z|}}$$
where $C>0$ is a constant independent of $n,z$. Analogous bound holds for $H_n^{(2)}(z)$. Thus, in the regime $|z| \gg n^2$, the above remainder term goes to zero, and expressing $J_n$ in terms of Hankel functions and substituting the above expansion in $(1)$ we prove the result.
So one way to finish proving $(1)$ is to ask what is the asymptotics of $J_n(z)$, or of $H_n^{(1)}(z)$, in the regime
$$|z|\gg n, \qquad \text{and} \qquad |z|\not\gg n^2 \tag{3}$$
I would think that $(2)$ should still hold, but couldn't find it anywhere. Is $(2)$ known to hold under $(3)$, i.e. are there error bounds in terms of the quantity $\frac{n}{|z|}$? If not what is the asymptotic behavior in the regime $(3)$? I'd appreciate any help and direction.
 A: In terms of the modulus function $M_\nu  (z)$ and phase function $\theta _\nu  (z)$, we can write
$$
H_\nu ^{(1)} (z) = M_\nu  (z)e^{i\theta _\nu  (z)}, \quad
J_\nu  (z) = M_\nu  (z)\cos \theta _\nu  (z).
$$
It is known that
\begin{align*}
M_{\nu}^{2}\left(z\right)\sim \;&\frac{2}{\pi z}\left(1  +\frac{1}{2}\frac{\mu-1}{(2z)^{2}}+\frac{1\cdot 3}{2\cdot 4}\frac{(\mu-1)(\mu-9)}{(2z)^{4}}\right.\\ & \qquad\qquad\qquad\quad\;\;\,\left.+\frac{1 
\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{(\mu-1)(\mu-9)(\mu-25)}{(2z)^{6}}+\ldots 
\right)
\end{align*}
and
\begin{align*}
\theta_{\nu}\left(z\right)\sim z-\left(\frac{1}{2}\nu+\frac{1}{4}\right)\pi & + 
\frac{\mu-1}{2(4z)}+\frac{(\mu-1)(\mu-25)}{6(4z)^{3}}+\frac{(\mu-1)(\mu^{2}-11 
4\mu+1073)}{5(4z)^{5}}\\& +\frac{(\mu-1)(5\mu^{3}-1535\mu^{2}+54703\mu-3 75733)}{ 
14(4z)^{7}}+\ldots,
\end{align*}
as $z\to \infty$ in the sector $|\arg z|\leq\pi-\delta<\pi$, with $\mu =4\nu^2$. It is seen that in order for the terms to decay for large $z$, we need only $\nu =o(z)$. See Section 5.11.4 in Y. L. Luke's book The Special Functions and their Approximations, Vol. 1 and Section 13.75 in G. N. Watson's book A Treatise on the Theory of Bessel Functions.
You may also try using the known uniform asymptotic expansions.
