Possible new definition of Gamma (Euler-Mascheroni Constant): $\lim_{x \to 0} (-\ln ( \sqrt[x]{x!} )) = \gamma$ I think I've discovered a new definition for the Euler-Mascheroni Constant (Gamma)
I can't find it online anywhere, has anyone seen it before?
$$\lim_{x \to 0} (-\ln ( \sqrt[x]{x!} )) = \gamma$$
 A: Write
$$
\ln ( \sqrt[x]{x!} ) =
\ln ({x!}^{1/x} ) =
\frac{\ln ({x!})}{x} =
\frac{\ln (\Gamma(x+1))}{x} =
\frac{f(x)-f(0)}{x-0}
$$
for
$$
f(x)=
\ln (\Gamma(x+1))
$$
Then
$$
\lim_{x \to 0} \ln ( \sqrt[x]{x!}) = f'(0)
$$
Now
$$
f'(x)= \frac{\Gamma'(x+1)}{\Gamma(x+1)}
$$
and so
$$
f'(0) = \frac{\Gamma'(1)}{\Gamma(1)} = \Gamma'(1) = -\gamma
$$
The last equality is the only difficult part, but it is a standard fact.
A: The Limit
The recurrence $\Gamma(x+1)=x\Gamma(x)$ gives us that
$$
\begin{align}
\left(\frac{\Gamma(x+n+1)}{\Gamma(n+1)}\right)^{1/x}
&=\Gamma(x+1)^{1/x}\left(\frac{(x+1)(x+2)\cdots(x+n)}{n!}\right)^{1/x}\\
&=\Gamma(x+1)^{1/x}\left(1+\frac x1\right)^{1/x}\left(1+\frac x2\right)^{1/x}\cdots\left(1+\frac xn\right)^{1/x}\tag{1}
\end{align}
$$
As shown below, the log-convexity of $\Gamma(x)$ guarantees that
$$
(n+x)^x<\frac{\Gamma(x+n+1)}{\Gamma(n+1)}<(n+1)^x\tag{2}
$$
Combining $(1)$ and $(2)$ gives
$$
1\lt\lim_{x\to0}\Gamma(x+1)^{1/x}e^{H_n-\log(n)}\lt 1+\frac1n\tag{3}
$$
Because $n$ is arbitrary, the Squeeze Theorem yields
$$
\lim_{x\to0}\Gamma(x+1)^{1/x}=e^{-\gamma}\tag{4}
$$
which is a restatement of the question.

Log-Convexity of $\,\boldsymbol{\Gamma(x)}$
Since $\Gamma(x)$ is log-convex, we have
$$
\begin{align}
\Gamma(x+n+1)
&\le\Gamma(n+1)^{1-x}\Gamma(n+2)^x\\
&=\Gamma(n+1)^{1-x}[(n+1)\Gamma(n+1)]^x\\
&=\Gamma(n+1)(n+1)^x\tag{5}
\end{align}
$$
and
$$
\begin{align}
\Gamma(n+1)
&\le\Gamma(x+n)^x\Gamma(x+n+1)^{1-x}\\
&=\left[\frac{\Gamma(x+n+1)}{n+x}\right]^x\Gamma(x+n+1)^{1-x}\\
&=\Gamma(x+n+1)(n+x)^{-x}\tag{6}
\end{align}
$$
Combining $(5)$ and $(6)$ yields $(2)$.
A: More in general we have
$${a}^{-1}b\,\lim _{x\rightarrow 0}(-\ln  \left(  \left(  \left( a\,x \right) !
 \right) ^{{\frac {1}{b\,x}}} \right)) =\gamma
$$
