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

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e^-2

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Detailed Solution

$\begin{array}{l} α = lim_{x \to \infty} {((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))}^{x} (1^{\infty} form) \\ ∴ α = e^{L} \end{array}$
Where $L = lim_{x \to \infty} x ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}) - 1)$
$\begin{array}{l} \Rightarrow L = lim_{x \to \infty} (\frac{e}{1 - e}) x (\frac{1}{e} - \frac{x}{1 + x} - (\frac{1 - e}{e})) \\ \Rightarrow L = \frac{e}{1 - e} lim_{x \to \infty} x (1 - \frac{x}{1 + x}) \\ \Rightarrow L = \frac{e}{1 - e} lim_{x \to \infty} \frac{x}{x + 1} \\ \Rightarrow L = \frac{e}{1 - e} \cdot 1 \\ \Rightarrow L = \frac{e}{1 - e} \\ ∴ α = e^{\frac{e}{1 - e}} \Rightarrow \log α = \frac{e}{1 - e} \end{array}$
$∴$ Required value $= \frac{\frac{e}{1 - e}}{1 + \frac{e}{1 - e}} = e$