For what $a$ is $\lim_{x\to\infty}(\frac{x+a}{x-a})^x=e$? I am trying to figure out for which $a$ the following equation is true:
$$\lim_{x\to\infty}\left(\frac{x+a}{x-a}\right)^x=e$$
It seems like an application of L'Hospital could work perhaps, but I'm not really sure what to do with the exponential. Any hints or suggestions would be appreciated!
 A: Hint: $$\left(\frac{x+a}{x-a}\right)^x = \left(\frac{1+(a/x)}{1-(a/x)}\right)^x = \frac{(1+(a/x))^x}{(1-(a/x))^x} \to \frac{e^a}{e^{-a}} = e^{2a}.$$
A: Hint:
Let $L = \lim _{ x\rightarrow \infty  }{ (\frac { x+a }{ x-a }  } )^x=e$. 
Take ln's of both sides: 
$\ln L = \lim _{ x\rightarrow \infty  }{ x \ln\frac{(x+a)}{(x-a)}} = \lim _{ x\rightarrow \infty  }{ \frac{\ln\frac{(x+a)}{(x-a)}}{ x^{-1}}}$, which is of the form $\frac{0}{0}$ 
Since $\frac{d}{dx} \ln\frac{(x+a)}{(x-a)} 
= \frac{d}{dx} [\ln(x+a) - \ln(x-a)] = \frac{1}{(x+a)} - \frac{1}{(x-a)} = \frac{-2a}{(x^2 - a^2)}$, 
applying L'Hopital's Rule yields 
$\ln L = \lim _{ x\rightarrow \infty  }{ [\frac{\frac{-2a}{(x^2 - a^2)}}{ (-x^{-2}) }}]
= \lim _{ x\rightarrow \infty  }{ \frac{2ax^2} { (x^2 - a^2)}} 
= \lim _{ x\rightarrow \infty  }{\frac{2a} { (1 - a^2/x^2) }}
= 2a$. 
Hence, $L = e^{(2a)}$. 
Since we want the limit to equal $e$, we conclude that 
$2a = 1 \therefore a = 1/2$. 
A: The following is one possible approach (the part denoted by $\stackrel{\text{H}}{=}$ is where you use L'Hospital): if $L=\lim_{x\to\infty}\left(\frac{x+a}{x-a}\right)^x$, then $L$ has the indeterminate form $1^\infty$. Thus:
\begin{align}
\ln L &= \lim_{x\to\infty}\ln\left(\frac{x+a}{x-a}\right)^x\\[1em]
&= \lim_{x\to\infty}x\ln\left(\frac{x+a}{x-a}\right)\\[1em]
&= \lim_{x\to\infty}\frac{\ln(x+a)-\ln(x-a)}{1/x}\\[1em]
&\stackrel{\text{H}}{=}\lim_{x\to\infty}\frac{\frac{1}{x+a}-\frac{1}{x-a}}{-1/x^2}\\[1em]
&= \lim_{x\to\infty}\left[\frac{(x-a)-(x+a)}{(x+a)(x-a)}\cdot\frac{-x^2}{1}\right]\\[1em]
&= \lim_{x\to\infty}\frac{2ax^2}{x^2-a^2}\\[1em]
&= \lim_{x\to\infty}\frac{2a}{1-a^2/x^2}\\[1em]
&= 2a.
\end{align}
Thus, we have that $\ln L=2a$, so $L=e^{2a}$. From the original equation, we want to have $L=e^1$; thus,
$$
L=e^1\Longrightarrow 2a=1\Longrightarrow a=\frac{1}{2},
$$
and you're done.
A: suppose $$f(a) = \lim_{x \to \infty}\left(\frac{x-a}{x+a}\right)^x.$$  we can see the following:  $f(0) = 1,\, f(-a)f(a) = 1, \,f(1)= e^2.$ therefore it is enough to consider $a > 0.$
changing the variable $x = au,$ we have $$f(a) = \lim_{x \to \infty}\left(\frac{x-a}{x+a}\right)^x = \lim_{u \to \infty}\left(\frac{au-a}{au+a}\right)^{au}=\lim_{u \to \infty}\left(\frac{u-1}{u+1}\right)^{au} =\left(f(1)\right)^a.$$ 
