Well I continue my previous question with :
Define :
$$f(x)=\left(\left|\int_{0}^{x}\prod_{n=1}^{\infty}\left(1-e^{2-t-n}\right)dt-x-1\right|\right)^{\frac{1}{x}}$$
Does we have :
$$\lim_{x\to 0}f(x)\overset{?}=e$$
Well this problem found by myself is so new to me so I have no idea to attack this problem/conjecture I know some inequalities like Holder or Andersson with a product in the integral . A first step should to show it admits an asymptote to show the existence of a finite limit .
Also I don't know how to use Taylor's series in this case .
How to (dis)prove it ?
Yes, the claim is true. To see this, let $$ F(x) := \int_0^x\prod_{n\geq 1}\left(1-e^{2-t-n}\right)\mathrm{d}t. $$ Of course $F(0)=0$. Moreover, $$F'(x) = \prod_{n\geq 1}\left(1-e^{2-x-n}\right) \implies F'(0)=\prod_{n\geq 1}\left(1-e^{2-n}\right)=0$$ since the $n=2$ term in the infinite product is already zero. Hence (taking for granted that $F$ is twice continuously differentiable in a neighbourhood of $0$) we have $F(x) = O(x^2)$ by Taylor expansion. Now your desired limit is the same as $$ \lim_{x\to 0}\left(1+x+O(x^2)\right)^{1/x} $$ which of course is the standard limit definition for $e$. (The quadratic term vanishes quickly and does not contribute to the limit).
Basically, there's nothing magical happening here; it's just that the integral term vanishes rapidly, so the only contribution to the limit is the $x+1$ that is added outside the integral by hand. The eventual limit comes down to the behaviour of that term only.