$$L=\lim_{x\to 1^{-}} \prod_{n=0}^{\infty} \left (\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$$ I found this limit as below $$L=\exp\left[ \lim_{x\to 1^{-}}\sum_{n=1}^{\infty}x^n \ln \left (\frac{1+x^{n+1}}{1+x^n}\right)\right]=\exp\left[ \lim_{h\to 0}\sum_{n=1}^{\infty}(1-h)^n \ln \left (\frac{1+(1-h)^{n+1}}{1+(1-h)^n}\right)\right]$$ Using binomial approximation: $(1+z)^\nu \approx (1+\nu z)$ where $|z|$ is as small as we please. Then $$L=\exp\left[ \lim_{h\to 0}\sum_{n=0}^{\infty}(1-h)^n \ln \left (\frac{2-h(n+1)}{2-hn}\right)\right]=\exp\left[ \lim_{h\to 0}\sum_{n=0}^{\infty}(1-h)^n \ln \left (\frac{1-h(n+1)/2}{1-hn/2}\right)\right]$$ $$L=\exp \left[\lim_{h\to 0} \sum_{n=0}^{\infty}(1-h)^n \ln[(1-(n+1)h/2)(1+hn/2)]\right]$$ $$L=\exp \left[\lim_{h\to 0} \sum_{n=0}^{\infty}(1-h)^n \ln[1-h/2-h^2(n^2+n)/4]\right]=$$ $$L=\exp \left[\lim_{h\to 0} \sum_{n=0}^{\infty}(1-h)^n \ln[1-h/2]\right]=\exp \left[\lim_{h\to 0} (-h/2)\sum_{n=0}^{\infty}(1-h)^n\right]$$ $$\implies L=\exp \left[\lim_{h\to 0} (-h/2)\frac{1}{1-1+h}\right].$$ Finally, $$L=e^{-1/2}.$$ I would like to know if this answer is correct? Somehow numerical results do not seem to support this result!
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