"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 :"

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"

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

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

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

"

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 :

Solution

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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 :

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