Obtaining the Airy kernel from the Christoffel-Darboux formula with asymptotic Hermite polynomials Let the Kernel associated to a family of orthogonal polynomial $p_n(x)$ with weight $w(x)$ be defined as
$$K_N(x,y):=\frac{\sqrt{w(x)w(y)}}{\int w(x) p_{N-1}(x)p_{N-1}(x)dx} \frac{p_N(x)p_{N-1}(y)-p_{N-1}(x) p_N(y)}{x-y} $$
(this is known as the Christoffel-Darboux formula)
I'm trying to reproduce the result of Forrester (p.723 eq.3.6). Basically, using the polynomials
$$p_n(x):=2^{-n}H_n(x)\\w(x):=e^{-x^2}$$
and the asymptotic expansion
$$ e^{-x^2/2}H_n(x) \sim\pi^{1/4}2^{n/2+1/4} \Gamma(n-1)^{1/2} n^{-1/12} \left(\pi \text{Ai}(t)+\mathcal{O}(n^{-2/3})\right),$$
with $t=\sqrt{2}n^{1/6}(x-\sqrt{2n})$
one must find an expression for $K_N(x,y)$, and evaluate the limit
$$K(X,Y):=\lim_{N\to\infty} \frac{1}{2^{1/2}N^{1/6}} K_N\left(\sqrt{2N}+\frac{X}{2^{1/2}N^{1/6}},\sqrt{2N}+\frac{Y}{2^{1/2}N^{1/6}}\right)$$
to prove that $$K(X,Y)=\frac{Ai(X)Ai'(Y) - Ai(Y)Ai'(X)}{X-Y}.$$

So far I can easily show that the norm of the $N-1$ polynomial is  $2^{-N-1}\Gamma(N)\sqrt{\pi}$ and thus that
  $$ K_N(x,y) = \frac{e^{-x^2/2}e^{-y^2/2}}{2^N\Gamma(N)\sqrt{\pi}}\left(\frac{H_N(x)H_{N-1}(y)-H_{N-1}(X)H_N(Y)}{x-y}\right)$$
  which can then be related to the asymptotic extension. However I have no idea of how to relate $H_n$ and $H_{n-1}$ in a way that yields a derivative of the Airy function... even in the limiting case $N\to\infty$. Any hint appreciated.

 A: 
I found a way to to use Taylor Series. I post it here for further reference.

Using the formula for the Kernel directly, with renormalized and re-entered variables $\widetilde{X}:=\sqrt{2N}+2^{-1/2}N^{-1/6} X$ et $\widetilde{Y}:=\sqrt{2N}+2^{-1/2}N^{-1/6} Y$, we have
\begin{align}
        K(X,Y)  
        &= \lim_{N\to\infty}\frac{e^{-\widetilde{X}^2\!/2}e^{-\widetilde{Y}^2\!/2}}{2^{N+\frac{1}{2}}\pi^{\frac{1}{2}}N^{\frac{1}{6}}(N-1)!}
           \frac{H_N(\widetilde{X})H_{N-1}(\widetilde{Y})-H_{N-1}(\widetilde{X})H_{N}(\widetilde{Y})}{\widetilde{X}-\widetilde{Y}}.\notag\\
        &\sim \lim_{N\to\infty}\frac{\pi^2}{N^{\frac{1}{12}}(N-1)^{\frac{1}{12}}}\sqrt{\frac{N!}{(N-1)!}}  
              \frac{\text{Ai}(X)\Psi_{N-1}(\widetilde{Y}) - \Psi_{N-1}(\widetilde{X})\text{Ai}(Y)}{X-Y}. \notag\\
        &:=\lim_{N\to\infty} f(N) \frac{\text{Ai}(X)\Psi_{N-1}(\widetilde{Y}) - \Psi_{N-1}(\widetilde{X})\text{Ai}(Y)}{X-Y}
\end{align}
where $f(N)$ is some function of $N$. To simplify further calculation, we have used
\begin{align}
    \Psi_{N-1}(\widetilde{X_i})&=\text{Ai}\left(\sqrt{2}(N-1)^{1/6}\left(\sqrt{2N}-\sqrt{2(N-1)}+\frac{X_i}{\sqrt{2}N^{1/6}}\right)\right):=\text{Ai}(\alpha X_i+\beta).
\end{align}
with $\alpha=[(N-1)/N]^{1/6}$, $\beta:=2(N-1)^{1/6}(\sqrt{N}-\sqrt{N-1})$ and $X_i=X,Y$.
Now, let's expand the Airy functions  $\Psi(X_i)$ about points $X_{i_0}$:
    \begin{align}
        \text{Ai}(\alpha X+\beta)=& \text{Ai}(\alpha X_0+\beta)+\alpha (X-X_0) \text{Ai}'(\alpha X_0 + \beta) + \mathcal{O}\bigl((X-X_0)^2\bigr)\\
        \text{Ai}(\alpha Y+\beta)=& \text{Ai}(\alpha Y_0+\beta)+\alpha (Y-Y_0) \text{Ai}'(\alpha Y_0 + \beta) + \mathcal{O}\bigl((Y-Y_0)^2\bigr),
    \end{align}
where we suppose that higher order term are not contributing for now. Substituting the $\Psi(X_i)$ in $K(X,Y)$ yields
\begin{align}
        K(X,Y)  
        \approx \lim_{N\to\infty} \alpha f(N) \frac{(Y-Y_0)\text{Ai}(X)\text{Ai}'(\alpha Y_0+\beta) - (X-X_0)\text{Ai}'(\alpha X_0+\beta)\text{Ai}(Y)}{X-Y}.    
\end{align}
Since we wish to eliminate the dependencies in $(X-X_0),(Y-Y_0)$ as well as the $\alpha f(N)$ coefficient simultaneously, we select $X_{i_0}=X_i-\alpha^{-1}f(N)^{-1}$ as the point of expansion. Hence
\begin{align}
        K(X,Y)  
        \approx \lim_{N\to\infty} \frac{\text{Ai}(X)\text{Ai}'\bigl(\alpha Y - f(N)^{-1}+\beta\bigr) - \text{Ai}'\bigl(\alpha X - f(N)^{-1}+\beta\bigr)\text{Ai}(Y)}{X-Y}.  
\end{align}
Using Stirling approximation, it is easily shown that 
\begin{align}
    \frac{1}{f(N)}
    = \frac{N^{\frac{1}{12}}(N-1)^{\frac{1}{12}}}{\pi^2}\sqrt{\frac{(N-1)!}{N!}} 
    \sim \frac{e^{\frac{1}{2}}}{\pi^2} \frac{ (N-1)^{\frac{N}{2}-\frac{1}{6}} }{ N^{\frac{N}{2}+\frac{1}{6}}}
\end{align}
which in turn allows us to evaluate the limit operations
    \begin{align}
        &\lim_{N\to\infty} \alpha = \lim_{N\to\infty} \left(\frac{N-1}{N}\right)^{1/6} = 1 \\
        &\lim_{N\to\infty} \beta = \lim_{N\to\infty} 2(N-1)^{\frac{1}{6}}\bigl(\sqrt{N}-\sqrt{N-1}\bigr) = 0 \\
        &\lim_{N\to\infty} \frac{1}{f(N)} = 0 
    \end{align}
Since $\text{Ai}'\bigl(\alpha Y - f(N)^{-1}+\beta\bigr) $ is continuous in the limit of $N\to\infty$, order of evaluation does not matter, and we can evaluate the argument first. Thus 
\begin{align}
        K(X,Y)  = \frac{\text{Ai}(X)\text{Ai}'(Y) - \text{Ai}'( X)\text{Ai}(Y)}{X-Y}.   
\end{align}
We now double check that the TS approximation is justified. One can readily verify that superior order terms scale as $\mathcal{O}(f(N)^{-d})$, with $d$ the order. Hence our approximation is not unreasonable. 
$\mathcal{QED}.$
