MCQ
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 :
- Ae
- B$e ^{-2}$
- C$e^2$
- D$e^{-1}$
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