(Asked in [MSE](https://math.stackexchange.com/questions/3760125) but got no response.)


The generating function $\frac{1}{(1-t)^N}=\sum\_k {N+k-1\choose k}t^k=\sum\_k h\_k(1)t^k$ and the Jacobi–Trudi formula $s\_{\lambda/\mu}=\det(h\_{\lambda\_i-i-\mu\_j+j})$ tell me that the value of the skew Schur function at the identity is
$$ s\_{\lambda/\mu}(1\_N)=\det\left({N+\lambda\_i-i-\mu\_j+j-1\choose \lambda\_i-i-\mu\_j+j}\right).\tag{1}\label{1}$$


However, I was reading a paper by Chen and Stanley ([A Formula for the Specialization of Skew Schur Functions](https://doi.org/10.1007/s00026-016-0312-2)) and they state that
$$s\_{\lambda/\mu}(1,q,q^2,\dotsc)=\frac{1}{\prod\_{u\in\lambda/\mu}[N+c(u)]\_q}\det\left(\genfrac[]{0pt}{}{N+\lambda\_i-i}{\lambda\_i-i-\mu\_j+j}\_q\right),\tag{2}\label{2}$$
where $c(u)$ is the content of the box $u$ in the Young diagram of shape $\lambda/\mu$ and the $q$-quantities are $[x]\_q=1-q^x$, $[a]\_q!=[a]\_q[a-1]\_q\cdots$ and $\genfrac[]{0pt}{}a b\_q=\frac{[a]\_q!}{[b]\_q![a-b]\_q!}$.


I am not an expert in this $q$-business, and I am confused by this equation. I have a few closely related questions.


1. Since the left hand side of \eqref{2} is a polynomial in $q$, it should have a limit when $q\to 1$ and this should be the skew Schur at the identity. Is this correct? But what is the number of arguments?
2. The determinant at the right hand side on \eqref{2} has a limit when $q\to 1$, but the prefactor does not. How to take the limit $q\to 1$ of this equation?
3. How to obtain equation \eqref{1} from equation \eqref{2}?



On the value of a skew Schur function at the identity
You might want to read up on the [principal specialization](https://www.symmetricfunctions.com/standardSymmetricFunctions.htm#standardSpecialization).


It specializes a symmetric function into a formal power series.
For example, the symmetric function $s\_1(x) = x\_1+x\_2+ \dotsb$
has principal specialization $s\_1(1,q,q^2,\dotsc) = 1+q+q^2+\dotsb = \frac{1}{1-q}$, so it does not make sense to let $q\to 1$. The expression only exist as a formal power series. However, you always compute the finite version,
$s\_\lambda(1,q,q^2,\dotsc,q^{n})$, which is gonna be a polynomial in $q$.


So, the difference between your Jacob-Trudi formula and the cited q-formula,
is the number of variables (finite vs, infinite).


The principal specialization is in many instances nicer than the finite number of variables specialization.


