$\begin{equation} \lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha, \text { find } \frac{\ln \alpha}{1+\ln \alpha} \end{equation}$ (22 Jan, Shift II, Memory Based)

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YESPYQ · Accepted Answer

Correct answer: b) e. $\begin{aligned} & \lim _{x \rightarrow \infty}\left(\left(\frac{e}{e-1}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x \ & \lim _{x \rightarrow \infty}\left(\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}\right)^x\right) \ & \alpha=e^{\lim _{x \rightarrow \infty}\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}-1\right)x} \ & \ln (\alpha)=\lim _{x \rightarrow \infty}\left(\frac{e }{(1-e)}-\frac{ex}{(1-e)(1+x)}\right) x \ & \lim _{x \rightarrow \infty} \frac{ex}{(1-e)(1+x)}=\frac{e}{1-e} \ & \text { Now, } \frac{\ln \alpha}{1+\ln \alpha}=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e \end{aligned}$

\(\begin{equation} \lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\ri…

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