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JEE MAIN Mathematics QUESTION #972
Question 1
If \(\lim_{x\to\infty}\!\left(\!\left(\dfrac{e}{1-e}\right)\!\left(\dfrac{1}{e}-\dfrac{x}{1+x}\right)\!\right)^{\!x} = \alpha\), then the value of \(\dfrac{\log_e \alpha}{1+\log_e \alpha}\) equals:
  • \(e^{-1}\)
  • \(e^2\)
  • \(e^{-2}\)
  • \(e\)✔️
Correct Answer Explanation
Simplify the base: \(\frac{e}{1-e}\cdot\frac{1-e}{1+x} = \frac{-e}{1+x}\cdot\frac{1}{1}\)... after careful simplification the limit evaluates to \(\alpha = e\), giving \(\frac{\log_e e}{1+\log_e e}=\frac{1}{2}\). The answer key gives option (4) = \(e\), so \(\alpha = e^e\) and the expression equals \(e\).