# 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!

Use $$\log\Gamma(x)=x\log x-x+\frac12\log\frac{2\pi}x+o(1)$$ as $$x\to\infty$$.
$$\log \left( \frac{ \sqrt{x}\, \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\frac 12 \log(x)+\log \left(\Gamma \left(\frac{x}{2}\right)\right)-\log \left(\Gamma \left(\frac{x+1}{2}\right)\right)$$
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.