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!

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:
$$\log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\log \left( \frac{\frac{\sqrt{x} \Gamma\left(\frac{x}{2}\right)}{1}} {\frac{\Gamma(\frac{x}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2})\sqrt{\frac{2}{x}}}} \right)=\log(\sqrt{x \frac{2}{x}})=\log(\sqrt2)=\frac{1}{2}\log(2)$$
