Calculate limit of \Gamma function Show that
$$\lim _{x \to \infty} \log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right) = \frac{1}{2} \log(2),$$ where $\Gamma$ is the Gamma function.
I reduced this problem to calculate the limit:
$$\lim_{x \to \infty}  \frac{B\left(\frac{x}{2},\frac{x}{2}\right)}{B\left(\frac{x+1}{2},\frac{x+1}{2}\right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.
Any help will be very appreciated, thanks!
 A: Hint
Use $$\log\Gamma(x)=x\log x-x+\frac12\log\frac{2\pi}x+o(1)$$ as $x\to\infty$.
A: $$\log \left( \frac{ \sqrt{x}\, \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\frac 12 \log(x)+\log \left(\Gamma \left(\frac{x}{2}\right)\right)-\log \left(\Gamma \left(\frac{x+1}{2}\right)\right)$$
Now, using Stirling approximation, as Kemono Chen commented,
$$\log\left(\Gamma \left(t\right)\right)=t (\log (t)-1)+\frac{1}{2} \left(-\log \left({t}\right)+\log (2 \pi
   )\right)+\frac{1}{12 t}+O\left(\frac{1}{t^3}\right)$$ Use this formula, simplify and continue with Taylor expansion to get 
$$\log \left( \frac{ \sqrt{x}\, \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\frac{\log (2)}{2}+\frac{1}{4 x}+O\left(\frac{1}{x^3}\right)$$ which shows the limit and how it is approached.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\lim _{x \to \infty}
\ln\pars{\root{x}\Gamma\pars{x/2} \over
\Gamma\pars{\bracks{x + 1}/2}}} =
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\Gamma\pars{x} \over \Gamma\pars{x + 1/2}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\bracks{x - 1}! \over \bracks{x - 1/2}!}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\,{\root{2\pi}\bracks{x - 1}^{x - 1/2}
\expo{-\pars{x - 1}} \over
\root{2\pi}\bracks{x - 1/2}^{x}\expo{-\pars{x - 1/2}}}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2} + \lim _{x \to \infty}
\ln\pars{\root{x}\,{x^{x - 1/2}\bracks{1 - 1/x}^{x - 1/2}
\root{\expo{}} \over
x^{x}\bracks{1 - \pars{1/2}/x}^{x}}}
\\[5mm] = &\
{1 \over 2}\,\ln\pars{2}\ +\
\underbrace{\ln\pars{{\expo{-1}\root{\expo{}} \over\expo{-1/2}}}}
_{\ds{=\ 0}}\ =\ \bbx{{1 \over 2}\,\ln\pars{2}}\approx 0.3466
\end{align}
A: Old topic, but there is a much easier way to prove it.
First of all, the following relationship exists between Beta and Gamma functions:
$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
So we have:
$$\Gamma \left( \frac{x+1}{2}\right) = \Gamma \left( \frac{x}{2}+\frac{1}{2}\right) = \frac{\Gamma(\frac{x}{2})\Gamma(\frac{1}{2})}{B(\frac{x}{2},\frac{1}{2})}$$
On the other hand, Stirling's approximation gives the equation below for large x:
$$B(\frac{x}{2},\frac{1}{2})=\Gamma(\frac{1}{2})(\frac{x}{2})^{(-\frac{1}{2})}=\Gamma(\frac{1}{2})\sqrt{\frac{2}{x}}$$
By substituting all in the original formula, we have:
$$\log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\log \left( \frac{\frac{\sqrt{x} \Gamma\left(\frac{x}{2}\right)}{1}} {\frac{\Gamma(\frac{x}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2})\sqrt{\frac{2}{x}}}} \right)=\log(\sqrt{x \frac{2}{x}})=\log(\sqrt2)=\frac{1}{2}\log(2)$$
