If , then the value of equals : [2025]

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Solution

Q.
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 : [2025]

1 $e^{- 2}$

2 $e^{- 1}$

3 e

4 $e^{2}$

Ans.
(3)

$α = \lim_{x \to \infty} ((\frac{e}{1 - e}) {(\frac{1}{e} - \frac{x}{1 + x}))}^{x}$ ( $1^{\infty}$ form)

$∴ α = e^{L}, where L = \lim_{x \to \infty} x ((\frac{e}{1 - e}) \times (\frac{1}{e} - \frac{x}{1 + x}) - 1)$

$\Rightarrow L = \lim_{x \to \infty} (\frac{e}{1 - e}) x (\frac{1}{e} - \frac{x}{1 + x} - (\frac{1 - e}{e}))$

$\Rightarrow L = \frac{e}{1 - e} \lim_{x \to \infty} x (1 - \frac{x}{1 + x}) \Rightarrow L = \frac{e}{1 - e} \lim_{x \to \infty} \frac{x}{1 + x}$

$\Rightarrow L = \frac{e}{1 - e} \cdot 1 \Rightarrow L = \frac{e}{1 - e}$

$∴ α = e^{\frac{e}{1 - e}} \Rightarrow \log_{e} α = \frac{e}{1 - e}$

$∴$ Required value $= \frac{\frac{e}{1 - e}}{1 + \frac{e}{1 - e}} = e$ .

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