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

I am looking for a proofs of the following limits:

$$\lim_{x \to \infty} \Gamma \left(1+\frac{1}{x} \right)^x = e^{-\gamma}.$$ I find this limit interesting as it relates the gamma function $\Gamma$ with the other gamma $\gamma$ which is the Euler-Mascheroni constant.

The second limit whose proof I am interested in is $$\lim_{x \to 0} x \Gamma \left(1+\frac{1}{x} \right)^x = e^{-1}.$$

• Do you know the product representation of $\Gamma$? – Daniel Fischer Aug 1 '13 at 12:36
• The first one is just an exponentiated form of the fact that $\Gamma'(1) = -\gamma$, which is well known (just write $x = 1/h$ with $h \rightarrow 0$ and use the limit definition of the derivative for $\log \Gamma(x)$). – KCd Aug 1 '13 at 12:38
• my edit @user60930 – what'sup Aug 1 '13 at 12:39
• The second one (which is meant to be $x \rightarrow 0^+$, I presume) is a quick consequence of the logarithmic form of Stirling's formula. Write down Stirling's asymptotic formula for $\Gamma(t+1)$ as $t \rightarrow \infty$, take the logarithm of both sides, divide by $t$, and then write $t$ as $1/x$. – KCd Aug 1 '13 at 12:43
• @what'sup Several suggestions: 1) do not use \Large 2) do not put all formulas between double dollars, some of them look much better in the text 3) if you change, say $x$ to $\mathrm{x}$, do it everywhere 4) avoid minor edits and try to adress all issues of the post. – Start wearing purple Aug 1 '13 at 13:10

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 $z\rightarrow+\infty$.