JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumLimit, Continuity and Differentiability420× since 2002If limx→∞((e1−e)(1e−x1+e))x=α\lim_{x \to \infty} \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+e} \right) \right)^x = \alphalimx→∞((1−ee)(e1−1+ex))x=α, then the value of logeα1+logeα\frac{\log_e \alpha}{1 + \log_e \alpha}1+logeαlogeα equals:AOption A. e−1BOption B. e2COption C. e−2DOption D. eCheckStep-by-step hintSkip · show solution
