If =_x[((e)/(1-e))(1e-(x)/(1+x))]^x, then the value of _e1+_e equals:

Question

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

Accepted Answer

Write $\dfrac{x}{1+x}=1-\dfrac1{1+x}$, so $\dfrac1e-\dfrac{x}{1+x}=\dfrac1e-1+\dfrac1{1+x}=-\dfrac{e-1}{e}+\dfrac1{1+x}$.

$$\left(\frac{e}{1-e}\right)\left(-\frac{e-1}{e}+\frac1{1+x}\right)=\left(\frac{e}{1-e}\right)\left(-\frac{e-1}{e}\right)+\left(\frac{e}{1-e}\right)\frac1{1+x}=1-\frac{e}{(e-1)(1+x)}$$

So $\alpha=\displaystyle\lim_{x\to\infty}\left(1-\frac{e}{(e-1)(1+x)}\right)^x=e^{-\frac{e}{e-1}}$ (using $\lim(1-a/x)^x=e^{-a}$).

$$\log_e\alpha=-\frac{e}{e-1}$$

$$\frac{\log_e\alpha}{1+\log_e\alpha}=\frac{-\dfrac{e}{e-1}}{1-\dfrac{e}{e-1}}=\frac{-\dfrac{e}{e-1}}{-\dfrac{1}{e-1}}=e$$

Solutions · 1