Well I give an answer here and I want to know if it's true for a special example
Answer :
Not an answer just some speculation about gamma function and Bernstein's polynomial which is of independent interest here but could help .
Using fallaciously the inverse function of $f(x)$:
$$f\left(x\right)=e^{x^{4}-\ln\left(\frac{1}{y}\right)}!,x!=\Gamma(x+1),0<y<1$$
Then using Bernstein form it seems we have $\forall x>0$:
$$g(x)=\lim_{n\to \infty}\sum_{k=0}^{n}\frac{f\left(\frac{k}{n}\right)n!}{k!\left(n-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(1+x^{\frac{1}{6}}\right)\right)^{\left(n-k\right)}=y!$$
End of the answer :
Now the special example :
Let $f(x)$ as above and $y=1/2$ have we for $0<x\leq 1$ :
$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{f\left(\frac{k}{n}\right)n!}{k!\left(n-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(1+x^{\frac{1}{6}}\right)\right)^{\left(n-k\right)}=\sqrt{\frac{\pi}{4}}$$
Is this special example true ?
Remark : I have no idea to tackle it rigourosly .
Addendum :
It's seems not always true using Askey-Kasper inequality for $x\simeq 0^+$ as we have :
Let :
$$f\left(x,y\right)=\frac{x!}{y!\left(x-y\right)!},h\left(x\right)=\sum_{n=0}^{p}\frac{\sum_{s=0}^{n}f\left(n+a,n-s\right)f\left(n+b,s\right)\left(\frac{x-1}{2}\right)^{s}\left(\frac{x+1}{2}\right)^{n-s}}{\frac{\left(n+b\right)!}{n!\left(b\right)!}},a>0,b>0,0<x\leq 1$$
$$m\left(x\right)=\sum_{k=0}^{P}\frac{r\left(\frac{k}{P}\right)P!}{k!\left(P-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(x^{\frac{1}{6}}+1\right)\right)^{\left(P-k\right)},r\left(x\right)=h\left(e^{x^{4}-\ln\left(10\right)}\right)$$
We have some problem around zero $m'(0)=\infty$
FYI :
taking two different value one lower than 1 and one greater than one in $m(x)$ and calculating $m(0)$ then the Askey-Gasper is true at some other point .