MCQMediumJEE Main 2025 · 22 January, Shift 2Limits

Mathematics Question from JEE Main 2025 · 22 January, Shift 2

If limx(e1e(1ex1+x))x=α,\lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, then the value of logeα1+logeα\frac{\log_e \alpha}{1 + \log_e \alpha} equals:

  • A

    e2e^{-2}

  • B

    e1e^{-1}

  • C

    ee

  • D

    e2e^2

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