If \(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha\…

If \(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha\), then the value of \(\frac{\log _e \alpha}{1+\log _e \alpha}\) equals :

[JEE Main 2025, 22 Jan (Shift 2)]