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It's another weird limit but different of previous I proposed :

Do we have if:

$$f(x)=\frac{1}{\sqrt{x!-x}}\frac{1}{\sqrt{x!!-x^2}}\cdots,x!=\Gamma(x+1),x!!=\Gamma(\Gamma(x+1)+1)$$

$$\lim_{x\to 0} (f(x))^{1/x}=^?e$$

For some approach see Conjecture: $\lim\limits_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots )^{-1/x}\stackrel?=e$ :

Is it true as closed form ?

Update :

The only calculus is :

$$1/2(1+\gamma +\sum_{n=2}^{\infty}(1-\gamma)^{(n-1)}\gamma=1/2(1+\gamma +1-\gamma)=1$$

Credit @Robjohn

Second question :

Have you another sketch or proof ?

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  • $\begingroup$ Another limit equal to $e$ is $$\lim_{x\to 0}\left(x!^{x!!^{x!!!^{\cdots}}}\right)^{\frac{1}{x!-1}}$$ $\endgroup$
    – DesmosTutu
    Feb 12 at 18:04

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