Let $\binom{n}{j}_q$ be a $q$-binomial coefficient. I would like to find a simple method to prove that
$$ \lim_{q\to1}\frac{\sum\limits_{j = 0}^{2n} (-1)^j q^{m(j^2+j)} \binom{2n}{j}_q}{\sum\limits_{j = 0}^{2n} (-1)^j q^{(j^2+j)} \binom{2n}{j}_q} = (2m-1)^n. $$
Can anyone give me a hint?
I believe that you limit is not exactly correct. I find for $n=1,..,6$ that the limit gives ${m,2 m-1,m (2 m-1),(2 m-1)^2,m (2 m-1)^2,(2 m-1)^3}$ so for even $n$, the limit is $(2m-1)^{\frac{m}{2}}$ and for odd $n$ the limit is $m(2m-1)^{\frac{n-1}{2}}$. I have only checked with with a compute algebra system up to $n=10$, I have not worked on a proof yet. I believe I can prove as and induction step, that if the limit formula is correct for even $n$, that is $lim=(2m-1)^{\frac{n}{2}}$ then for $n+1, lim=m(2m-1)^{\frac{n}{2}}$
