1 to the infinity?

$\lim_{x\to 0}\sqrt[x]{x!} =\exp\left(\lim_{x\to 0}\dfrac{\log_e x!}{x}\right)=\exp\left(\lim_{x\to 0} \dfrac{ \log_e\Gamma (x+1)}{x} \right)$ we apply L-hopital's rule to get $\exp\left(\lim_{x\to 0} \dfrac{\Gamma(x+1)\cdot \psi(x+1)}{\Gamma(x+1)}\right) =e^{\psi(1)}=e^{-\gamma} \approx 0.561$

From the series representation of the Digamma function we have $\psi(s+1)=-\gamma +\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+s}\right)$ setting $s=0$ we have $\psi(1)= -\gamma$

$\begin{aligned} L & = \lim_{x \to 0} (x!)^{\frac 1x} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, }\implies \lim_{x \to a} f(x)^{h(x)} = e^{\lim_{x \to a} h(x)(f(x)-1)} \\ & = \exp \left(\lim_{x \to 0} \frac {x! - 1}x \right) & \small \color{#3D99F6} \text{A } 0/0 \text{ case, L'Hôpital's rule applies} \\ & = \exp \left(\lim_{x \to 0} \frac {{\color{#3D99F6}\Gamma(x+1)} - 1}x \right) & \small \color{#3D99F6} \Gamma (\cdot) \text{ denotes the gamma function} \\ & = \exp \left(\lim_{x \to 0} \frac {\Gamma (x+1)\psi (x+1)}1 \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \exp (0! \psi (1)) & \small \color{#3D99F6} \text{Digamma function }\psi (1) = - \gamma \text{, where } \\ & = e^{-\gamma} & \small \color{#3D99F6} \text{Euler-Mascheroni constant }\gamma \approx 0.5772 \\ & \approx \boxed{0.561} \end{aligned}$

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References:

• L'Hôpital's rule
• Gamma function
• Digamma function
• Euler-Mascheroni constant