According to WolframAlpha, the limit of
$$ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} = \frac{1}{e^4}$$
and I wonder how this result is obtained.
My approach would be to divide both nominator and denominator by $n$, yielding
$$ \lim_{n \to \infty} \left(\frac{1-\frac{1}{n}}{1 + \frac{1}{n}} \right)^{2n+4} $$
As $ \frac{1}{n} \to 0 $ as $ n \to \infty$, what remains is
$$ \lim_{n \to \infty} \left(\frac{1-0}{1 + 0} \right)^{2n+4} = 1 $$
What's wrong with my approach?