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

1
e
2
e-2
3
e2
4
e-1

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