JEE Math Practice Question
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 :
- A.$e^{-2}$
- B.$\mathrm{e}^2$
- C.$e$
- D.$e^{-1}$
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