Connection between Legendre polynomial and Bessel function In Abramovitz and Stegun (Eq. 9.1.71) I found this curious relation
$$\lim_{\nu\to\infty} \left[ \nu^\mu P_\nu^{-\mu}\left(\cos \frac{x}{\nu} \right) \right]= J_\mu(x) \qquad(1)$$
valid for $x>0$.
In fact it can be used to obtain a rather good approximation
$$ P_\nu^{-\mu}(\cos\theta) \approx  \frac{1}{\nu^\mu} J_\mu(\nu \theta)$$
of the Legendre polynomial in terms of a Bessel function for small $\theta$ (but $\nu\theta$ potentially large). This relation is a way to understand the eikonal approximation of wave scattering (which is the reason I noted it in the first place).
As I am looking into the eikonal approximation, I would appreciate if somebody could help me   proving equation (1)?
 A: Consider the differential equation for the associated Legendre polynomials, 
$$(1-z^2)w''(z) - 2z w(z) + \left(\nu(\nu+1) - \frac{\mu^2}{1-z^2}\right)w(z) = 0.$$ 
Change variables. 
Let $z = \cos \frac{x}{\nu}$. 
(Notice, for example, that $\frac{d}{dz} = -\frac{\nu}{\sin\frac{x}{\nu}} \frac{d}{dx}$.) 
In the limit $\nu\to\infty$ the DE takes the form 
$$x^2 w''(x) + x w'(x) + (x^2-\mu^2)w(x) = 0$$
which is, of course, Bessel's equation. 
Therefore,
$$\lim_{\nu\to\infty} P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \propto J_\mu(x).$$
Since it is getting late, I leave it as an exercise to find the constant. 
Addendum: 
The argument above tells us that in the limit $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is some combination of solutions to Bessel's equation.
The singular solution $Y_\mu(x)$ is ruled out since $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is not singular at $x=0$.
Using the integral representation for $-1<z<1$ and $\mathrm{Re}\,\mu > 0$, 
$$P_\nu^{-\mu}(z) = \frac{(1-z^2)^{-\mu/2}}{\Gamma(\mu)} 
\int_z^1 d t\, P_\nu(t)(t-z)^{\mu-1},$$
we find for $x\ll 1 \ll \nu$ that 
$$P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \sim 
\frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2\nu}\right)^\mu.$$
(Here we exploit the fact that for $x\ll 1$, $P_\nu\left(\cos \frac{x}{\nu}\right) = 1+O(x^2)$.)
But for small $x$ we have 
$$J_\mu(x) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2}\right)^\mu$$
and so 
$$\lim_{\nu\to\infty} \left[\nu^\mu P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)\right] = J_\mu(x).$$
