I want to evaluate this limit expression. I have this problem when I explored a corollary of infinity q-binomial theorem.
$ \lim\limits_{{q}\to {1}^{-}} \prod\limits_{k=0}^{+\infty} \dfrac{1-{q}^{a+k}x}{1-{q}^{k}x} \\ x \in {\mathbb{R}} \quad , \quad \left|x\right|<1 \quad , \quad a \in \mathbb{C} $
When $a\in{\mathbb{R}^{+}}$ , it is clear that
$ \lim\limits_{{q}\to {1}^{-}} \prod\limits_{k=0}^{+\infty} \dfrac{1-{q}^{a+k}x}{1-{q}^{k}x}= \lim\limits_{{q}\to {1}^{-}} \prod\limits_{k=0}^{a-1} \dfrac{1}{1-{q}^{k}x}= \dfrac{1}{{(1-x)}^{a}} $
Now I need to extend this conclusion to the broader situation that $a \in \mathbb{C}$ . Here the trouble is that I don't know how to do with it. So I ask for help here.
The idea is to observe numerators and denominators separately and to analyze the logarithm of the function $$\Pi_x(q)=\prod_{n\geq0}(1-xq^n).$$ There: $$-\ln(\Pi_x(q))=-\sum_{n\geq0}\ln(1-xq^n)=\sum_{n\geq0}\sum_{k>0}\frac{x^kq^{nk}}{k}=\sum_{k>0}x^k\frac{\sum_{n\geq0}q^{nk}}{k}=\sum_{k>0}\frac{x^k}{k(1-q^k)}.$$ Now all we need is the difference of these logarithms for $\Pi_x$ and $\Pi_{xq^a}$: $$\sum_{k>0}\frac{x^k(1-q^{ak})}{k(1-q^k)}=\sum_{k>0}\frac{x^k}{k}\left(a+\mathcal{O}\left((1-q^k)\frac{a(a-1)}{2}\right)\right)=$$ $$=\sum_{k>0}\frac{x^k}{k}\left(a+\mathcal{O}\left(k(1-q)\right)\right)=-a\ln(1-x)+\mathcal{O}(1-q)\frac{x}{1-x}\to -a\ln(1-x).$$ What I did when taking the estimated Big-O error out of the sum is justified because the constant in the error is uniformly bounded. The derivation of the insides of the Big-O is given by the derivative (and value) of the function $$q\mapsto\frac{1-q^a}{1-q}$$ at $q=1$. The value is $a$ and the derivative is $\frac{a(a-1)}{2}$ and the Taylor's theorem gives the estimate I used. Note that $a$ is just a constant so it can vanish from the Big-O. Later, The same was done for $$q\mapsto1-q^k:$$ the value is $0$ and the derivative is $-k$.
Finally, taking the result $-a\ln(1-x)$ to the exponent gives you your hypothesized result.