Find all $a$ such that $\lim_{x\to\infty}\left( \frac{x+a}{x-a} \right)^x = e$. Saw this problem and I thought I'd take a shot at it:
Find all $a$ such that 
$$\lim_{x\mathop\to\infty}\left( \frac{x+a}{x-a} \right)^x = e.$$
 A: 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 $$\lim_{x \to \infty}\left(1+\frac{a}x\right)^x = e^a,\qquad\lim_{x \to \infty}\left(1-\frac{a}x\right)^x = e^{-a},$$ then you get that $a - (-a) = 1$ so $a = 1/2$.
A: Let 
$$L = \lim_{x\to\infty}\left( \frac{x+a}{x-a} \right)^x. $$
Then 
$$\log L = \lim_{x\to\infty} x (\log (x+a) - \log (x-a)) \\
\overset{\text{LHR}}{=} \lim_{x\to\infty} \frac{1/(x+a)-1/(x-a)}{-1/x^2}\\
=\lim_{x\to\infty} \frac{x^2 (x+a) - x^2 (x-a)}{(x+a)(x-a)}\\
=2a.\\[8pt]
\implies \log L = 2a \\
\implies L = e^{2a}.
$$
So it is true only for $a=1/2$.
A: Put $y=x-a$, then
$\lim_{x\to\infty}\left( \frac{x+a}{x-a} \right)^x =\lim_{y\to\infty}\left(1+\frac{2a}{y} \right)^{y+a}= \lim_{y\to\infty}\left(1+\frac{2a}{y} \right)^{y}= e^{2a}.$
A: Let 
$$L = \lim\limits_{x \to \infty} \left(\frac{x+a}{x-a}\right)^x  =  \lim\limits_{x \to \infty} \left(\frac{(x-a) + 2a}{x-a}\right)^x =  \lim\limits_{x \to \infty} \left(1+\frac{2a}{x-a}\right)^x$$
For $a \le 0$, the value of $\left(\frac{x+a}{x-a}\right)$ is bounded above by 1 for positive $x$, so $L \le 1$.  Thus, we may focus on the case $a > 0$.
Define $u= x - a$.  Then, $x = u+a$, and
$$L = \lim\limits_{u \to \infty} \left(1+\frac{2a}{u}\right)^{u+a}$$
Now, there exists a $u_0$ such that for all $u > 0$, and we have that for every value of $\epsilon > 0$, there exists a $u_0$ so that for all $u > u_0$, 
$$  \left(1+\frac{2a}{u}\right)^{u} < \left(1+\frac{2a}{u}\right)^{u+a} < \left(1+\frac{2a}{u}\right)^{(1+\epsilon)u}$$
(specifically, we can choose $u_0 = \frac{a}{\epsilon}$.  Thus, 
$$e^{2a} \le L \le e^{2a(1+\epsilon)}$$
And since $\epsilon$ can be made a small as we wish, 
$$L = e^{2a}.$$
From this, we see that $a= \frac{1}{2}$ is the only solution.
