If $$\lim_{x \rightarrow \infty}((\frac{e}{1-e})(\frac{1}{e}-\frac{x}{1+x}))^{x}=\alpha$$ then the value of $$\frac{\log_{e}^{\alpha}}{1+\log_{e}^{\alpha}}$$ equals :

Set $$f(x)=\frac{e}{1-e}\Bigl(\frac{1}{e}-\frac{x}{1+x}\Bigr).$$

Observe that $$\frac{1}{e}-\frac{x}{1+x}=\frac{1+x-ex}{e(1+x)},$$ so $$f(x)=\frac{e}{1-e}\cdot\frac{1+x-ex}{e(1+x)}=\frac{1+x-ex}{(1-e)(1+x)}=\frac{x(e-1)-1}{(e-1)(1+x)}.$$ 

Since $$(e-1)(1+x)=(e-1)x+(e-1)$$, one obtains $$f(x)=1-\frac{e}{(e-1)(1+x)}\,. $$

As $$x\to\infty$$ this expression approaches $$1$$, giving a $$1^\infty$$ form. 

Therefore we take logarithms: $$\ln\alpha=\lim_{x\to\infty}x\ln f(x) =\lim_{x\to\infty}x\ln\!\Bigl(1-\frac{e}{(e-1)(1+x)}\Bigr) =\lim_{x\to\infty}x\Bigl(-\frac{e}{(e-1)(1+x)}+o\bigl(\tfrac1x\bigr)\Bigr) =-\frac{e}{e-1}\,. $$

Finally, $$\frac{\ln\alpha}{1+\ln\alpha} =\frac{-\frac{e}{e-1}}{1-\frac{e}{e-1}} =\frac{-\frac{e}{e-1}}{\frac{e-1-e}{e-1}} =\frac{-e}{-1} =e\ $$ Thus the required value is $$\displaystyle e\,$$.