for what value of a is the following equation true?
lim ((x+a)/(x-a))^x = e
x->inf
You have differentiated the numerator incorrectly. You want:
$\displaystyle \frac{d}{dx}\left(\ln\left(\frac{x+a}{x-a} \right) \right)=\frac{1}{\frac{x+a}{x-a}}\cdot\frac{-2a}{(x-a)^2}=\frac{2a}{a^2-x^2}$
Now, putting this together with the derivative of the denominator, we have:
$\displaystyle \lim_{x\to\infty}\frac{2ax^2}{x^2-a^2}=1$
which we can write as:
$\displaystyle 2a\lim_{x\to\infty}\frac{1}{1-\frac{a^2}{x^2}}=1$
From here, it is a piece of cake.