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

\(\begin{aligned}& \lim _{x \rightarrow \infty}\left(\left(\frac{e}{e-1}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x \\& \lim _{x \rightarrow \infty}\left(\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}\right)^x\right) \\& \alpha=e^{\lim _{x \rightarrow \infty}\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}-1\right)x} \\& \ln (\alpha)=\lim _{x \rightarrow \infty}\left(\frac{e }{(1-e)}-\frac{ex}{(1-e)(1+x)}\right) x \\& \lim _{x \rightarrow \infty} \frac{ex}{(1-e)(1+x)}=\frac{e}{1-e} \\& \text { Now, } \frac{\ln \alpha}{1+\ln \alpha}=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e\end{aligned}\)