Big Gamma $\Gamma$ meets little gamma $\gamma$

As shown in this answer, $\Gamma'(1)=-\gamma$. Thus, $\Gamma\left(1+\frac1x\right)=1-\frac\gamma{x}+O\left(\frac1{x^2}\right)$ and therefore, $$ x\log\left(\Gamma\left(1+\frac1x\right)\right)=-\gamma+O\left(\frac1x\right) $$ and $$ \lim_{x\to\infty}\Gamma\left(1+\frac1x\right)^{\large x}=e^{-\gamma} $$

The second question is essentially the same as $$ \lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e $$ mentioned in this answer if we set $x=\frac1n$, since $n!=\Gamma(1+n)$.

By Stirling's Approximation, $$ n!\sim\sqrt{2\pi n}\,n^ne^{-n} $$ therefore, $$ \begin{align} \lim_{n\to\infty}\frac1n(n!)^{1/n} &=\lim_{n\to\infty}\frac1n\frac ne\lim_{n\to\infty}\sqrt{2\pi n}^{1/n}\\ &=\frac1e \end{align} $$

• First limit: take the logarithm and use that $\gamma=-\psi(1)=-\Gamma'(1)$.

• Second limit: take the logarithm and use Stirling's approximation $\ln\Gamma(1+z)=z(\ln z-1)+O(1)$ as $z\rightarrow+\infty$.

An excellent discussion of this topic can be found in the book The Gamma Function by James Bonnar. In regards to the first question, see Chapter 8 which covers the Weierstrass product form of the Gamma function. In regards to the second question, see Chapter 15 on Stirling's formula. Both of these results are derived in the book.