If limx→∞ e1−e1e−x1+xx=α, then the value of loge⁡α1+loge⁡α equals :

α=limx→∞ e1−e1e−x1+xx1∞ form ∴α=eLWhere L=limx→∞ xe1−e1e−x1+x−1⇒L=limx→∞ e1−ex1e−x1+x−1−ee⇒L=e1−elimx→∞ x1−x1+x⇒L=e1−elimx→∞ xx+1⇒L=e1−e⋅1⇒L=e1−e∴α=ee1−e⇒log⁡α=e1−e∴Required value =e1−e1+e1−e=e

If limx→∞ e1−e1e−x1+xx=α, then the value of loge⁡α1+loge⁡α equals :

α=limx→∞ e1−e1e−x1+xx1∞ form ∴α=eLWhere L=limx→∞ xe1−e1e−x1+x−1⇒L=limx→∞ e1−ex1e−x1+x−1−ee⇒L=e1−elimx→∞ x1−x1+x⇒L=e1−elimx→∞ xx+1⇒L=e1−e⋅1⇒L=e1−e∴α=ee1−e⇒log⁡α=e1−e∴Required value =e1−e1+e1−e=e

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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

e²

e^-1

answer is A.

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