JEE Mains · Maths · STD 11 - 12. limits
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 :
- A \(e^{-1}\)
- B \(\mathrm{e}^2\)
- C \(e^{-2}\)
- D e
Answer & Solution
Correct Answer
(D) e
Step-by-step Solution
Detailed explanation
\begin{aligned} & \alpha=\lim _{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}\right)\right)^{\mathrm{x}}\left(1^{\infty} \text { form }\right) \\ & \therefore \alpha=\mathrm{e}^{\mathrm{L}} \\ &…
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