= Solution
For every fixed real $a$, the base is positive once $x>|a|$. Taking its logarithm gives
$$
\log\left[\left(\frac{x+a}{x-a}\right)^x\right]
=x\left[\log\left(1+\frac ax\right)-\log\left(1-\frac ax\right)\right].
$$
The <Taylor series> of the logarithm makes the bracket $2a/x+O(x^{-3})$, so this logarithm tends to $2a$. Continuity of the exponential gives
$$
\boxed{\lim_{x\to\infty}\left(\frac{x+a}{x-a}\right)^x=e^{2a}}.
$$
For $a=0$ the base is identically one, consistently giving the same answer.
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