# Saw this problem and I thought I'd take a shot at it: lim_{x->oo}((x+a)/(x-a))^x = e.

Find all $a$ such that $\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x+a}{x-a}\right)}^{x}=e$
Saw this problem and I thought I'd take a shot at it:
Find all $a$ such that
$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x+a}{x-a}\right)}^{x}=e.$
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Coleman Ali
Note that
$\frac{x+a}{x-a}=\frac{1+\frac{a}{x}}{1-\frac{a}{x}}.$
So if you can show (or simply recognize from theorem) that
$\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{a}{x}\right)}^{x}={e}^{a},\phantom{\rule{2em}{0ex}}\underset{x\to \mathrm{\infty }}{lim}{\left(1-\frac{a}{x}\right)}^{x}={e}^{-a},$
then you get that $a-\left(-a\right)=1$ so $a=1/2$

Nash Frank
Let
$L=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x+a}{x-a}\right)}^{x}.$
Then
$\mathrm{log}L=\underset{x\to \mathrm{\infty }}{lim}x\left(\mathrm{log}\left(x+a\right)-\mathrm{log}\left(x-a\right)\right)\phantom{\rule{0ex}{0ex}}\stackrel{\text{LHR}}{=}\underset{x\to \mathrm{\infty }}{lim}\frac{1/\left(x+a\right)-1/\left(x-a\right)}{-1/{x}^{2}}\phantom{\rule{0ex}{0ex}}=\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{2}\left(x+a\right)-{x}^{2}\left(x-a\right)}{\left(x+a\right)\left(x-a\right)}\phantom{\rule{0ex}{0ex}}=2a.\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}L=2a\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}L={e}^{2a}.$
So it is true only for $a=1/2$

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