Limits Continuity And Differentiability

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