If lim _{x rightarrow infty}left(left(frac{e} | vedclass

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(A) Given $\alpha = \lim _{x \rightarrow \infty} \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) \right)^x$,which is in

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$e$

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(A) Given $\alpha = \lim _{x \rightarrow \infty} \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) \right)^x$,which is in the $1^{\infty}$ form.We know that $\lim _{x \rightarrow \infty} f(x)^{g(x)} = e^{\lim _{x \rightarrow \infty} g(x)(f(x)-1)}$.Let $L = \lim _{x \rightarrow \infty} x \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) - 1 \right)$.Simplifying the expression inside the limit:$\left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) = \frac{1}{1-e} - \frac{ex}{(1-e)(1+x)} = \frac{1+x-ex}{(1-e)(1+x)}$.Now,$L = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex}{(1-e)(1+x)} - 1 \right) = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex - (1-e)(1+x)}{(1-e)(1+x)} \right)$.$L = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex - (1+x-e-ex)}{(1-e)(1+x)} \right) = \lim _{x \rightarrow \infty} x \left( \frac{e}{(1-e)(1+x)} \right) = \frac{e}{1-e}$.Thus,$\alpha = e^{\frac{e}{1-e}}$,so $\log _e \alpha = \frac{e}{1-e}$.The required value is $\frac{\log _e \alpha}{1+\log _e \alpha} = \frac{\frac{e}{1-e}}{1 + \frac{e}{1-e}} = \frac{e}{1-e+e} = e$.

If $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$,then the value of $\frac{\log _e \alpha}{1+\log _e \alpha}$ equals :

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