$$ \begin{aligned} \alpha &= \lim_{x \to \infty} \left[\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right]^x. \end{aligned} $$

Begin by rewriting the expression inside the limit. Notice that

$$ \frac{x}{1+x} = 1 - \frac{1}{1+x}, $$

so

$$ \frac{1}{e} - \frac{x}{1+x} = \frac{1}{e} - \left(1 - \frac{1}{1+x}\right) = \frac{1}{e} - 1 + \frac{1}{1+x}. $$

Express the constant part as

$$ 1 - \frac{1}{e} = \frac{e-1}{e}, $$

to obtain

$$ \frac{1}{e} - \frac{x}{1+x} = -\frac{e-1}{e} + \frac{1}{1+x}. $$

Multiplying by the prefactor, we have

$$ \left(\frac{e}{1-e}\right)\left(\frac{1}{e} - \frac{x}{1+x}\right) = \frac{e}{1-e}\left[-\frac{e-1}{e} + \frac{1}{1+x}\right]. $$

Split the expression:

$$ \frac{e}{1-e}\cdot\left(-\frac{e-1}{e}\right) + \frac{e}{1-e}\cdot\frac{1}{1+x} = -\frac{e-1}{1-e} + \frac{e}{(1-e)(1+x)}. $$

Since

$$ 1-e = -(e-1), $$

it follows that

$$ -\frac{e-1}{1-e} = -\frac{e-1}{-(e-1)} = 1. $$

Hence, the expression simplifies to

$$ 1 + \frac{e}{(1-e)(1+x)}. $$

Again, replacing $1-e$ by $-(e-1)$,

$$ 1 + \frac{e}{-(e-1)(1+x)} = 1 - \frac{e}{(e-1)(1+x)}. $$

Thus, the limit becomes

$$ \alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)(1+x)}\right)^x. $$

For large $x$, note that

$$ 1+x \sim x, $$

so we approximate

$$ \alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)x}\right)^x. $$

Recall the limit

$$ \lim_{x \to \infty} \left(1 - \frac{a}{x}\right)^x = e^{-a}, $$

with

$$ a = \frac{e}{e-1}. $$

Therefore,

$$ \alpha = e^{-\frac{e}{e-1}}. $$

Taking the natural logarithm,

$$ \ln \alpha = -\frac{e}{e-1}. $$

Now, compute

$$ \frac{\ln \alpha}{1 + \ln \alpha} = \frac{-\frac{e}{e-1}}{1 - \frac{e}{e-1}}. $$

Express the denominator with a common denominator:

$$ 1 - \frac{e}{e-1} = \frac{e-1}{e-1} - \frac{e}{e-1} = -\frac{1}{e-1}. $$

Thus,

$$ \frac{-\frac{e}{e-1}}{-\frac{1}{e-1}} = \frac{e}{1} = e. $$

The final result is

$$ \frac{\ln\alpha}{1+\ln\alpha} = e. $$