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JEE MAIN 2025
22-01-2025 SHIFT-2
Question
If $\lim _{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alpha$, then the value of $\frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha}$ equals :
Select the correct option:
A
e
B
$e^{-2}$
C
$e^2$
D
$e^2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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Mathematics
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