I asked this question
and I tried my own method which I am not sure if it is correct or wrong.
let $L=\lim_{n\to\infty}\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$ $$\ln(L)=\lim_{n\to\infty}\left(\sum_{k=0}^n \ln(2n+2k)-\sum_{k=0}^n \ln(2n+2k+1)\right)$$ let $a=\frac{1}{n}$
then $$\ln(L)=\lim_{n\to\infty}\left(\sum_{k=0}^n\ln(n)+\sum_{k=0}^n \ln(2n+2ak)-\sum_{k=0}^n\ln(n)-\sum_{k=0}^n \ln(2n+2a+a)\right)$$ $$\ln(L)=\lim_{a\to0}\left(\sum_{k=0}^n \ln(2+2ka)-\sum_{k=0}^n \ln(2+2ka+a)\right)$$
$$\ln(L)=\lim_{a\to0}-a\left(\sum_{k=0}^n\frac{- \ln(2+2ka)+ \ln(2+2ka+a)}{a}\right)$$
$m:=\lim_{a\to0}\frac{- \ln(2+2ka)+ \ln(2+2ka+a)}{a}$is the defintion of derivative
then $\ln(L)=-\lim_{a\to0}a\sum_{k=0}^n\frac{1}{2+2ka} $ which is Riemann sum so $$\ln(L)=\int_0^1\frac{-1}{2+2x}dx$$ $$\ln(L)=-0.5\ln(2)$$ $$\ln(L)=\ln(\frac{1}{\sqrt{2}})$$ $$L=\frac{1}{\sqrt{2}}$$