If $\lim_{x o\infty}\left(\left(\frac e{1-e} ight)\left(\frac1e-\frac x{1+x} ight) ight)^x=\alpha$, then the value of $\frac{\log_e\alpha}{1+\log_e\alpha}$ equals :

Question

If $\lim_{x	o\infty}\left(\left(\frac e{1-e}ight)\left(\frac1e-\frac x{1+x}ight)ight)^x=\alpha$, then the value of $\frac{\log_e\alpha}{1+\log_e\alpha}$ equals :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The base simplifies to $\frac{1+(1-e)x}{(1-e)(1+x)}=1+\frac{e}{(1-e)(1+x)}$, so it is of the form $1^\infty$ and $\alpha=e^{\,e/(1-e)}$. Then $\ln\alpha=\frac e{1-e}$ and $\frac{\ln\alpha}{1+\ln\alpha}=\frac{e/(1-e)}{1/(1-e)}=e$. Answer (4).

Q. If $\lim_{x\to\infty}\left(\left(\frac e{1-e}\right)\left(\frac1e-\frac x{1+x}\right)\right)^x=\alpha$, then the value of $\frac{\log_e\alpha}{1+\log_e\alpha}$ equals :

Explanation

Reviewed by Dr. Abubacker

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