# 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

## 3 Answers

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.