Let: $$A=\Gamma\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)$$
$$B=\Gamma\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)$$ I spent a few days trying to work on the $\gamma$; (Euler's constant) I got to this point. It is still look complicated. I need help to simplfy the limit further. $$\lim_{n \to \infty}\frac{(1-A)(1-B)}{(2-A-B)(A-B)}=\frac{\gamma(\gamma-1)}{2\gamma-1}\tag1$$
Can anyone help me to simplfy this limit $(1)?$
Hint
I suppose that you need to use (and reuse) the Taylor expansion (have a look at formula 35 here) $$\Gamma \left(\frac{1}{n}\right)=n-\gamma +\frac{6 \gamma ^2+\pi ^2}{12 n}+O\left(\frac{1}{n^2}\right)$$ $$\frac{1}{\Gamma \left(\frac{1}{n}\right)}=\frac{1}{n}+\frac{\gamma }{n^2}+O\left(\frac{1}{n^3}\right)$$ and then $$\Gamma(1+\epsilon)= 1-\gamma \epsilon +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) \epsilon ^2+O\left(\epsilon ^3\right)$$
It works (a bit tedious).