Why is $ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} = \frac{1}{e^4}$ 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?
 A: The definition of $e$ is
$$\lim_{n \to \infty} \left (1+\frac{1}{n}\right)^n$$
There are many ways in which this may be shown.  One is to use logarithms: let $L$ be the limit; then
$$\log{L} =\lim_{n \to \infty} n \log{\left (1+\frac{1}{n}\right)} = \lim_{n \to \infty} n \frac{1}{n} = 1$$
so that $L = e$.  Further, you can show similarly that
$$\lim_{n \to \infty} \left (1+\frac{x}{n}\right)^n = e^x$$
Your limit is then
$$\left ( \frac{1/e}{e}\right)^2 = \frac{1}{e^4}$$
A: If you already know that 
$$\lim_{n\to\infty}\left(1+\frac x{f(n)}\right)^{f(n)}=e^x\;,\;\;\forall\,x\in\Bbb R$$
and for any function $\,f\,$ s.t. $\,f(n)\xrightarrow[n\to\infty]{}\infty\;$ , then
$$\left(\frac{n-1}{n+1}\right)^{2n+4}=\left[\left(1-\frac2{n+1}\right)^{n+1}\right]^2\left(1-\frac2{n+1}\right)^2\xrightarrow[n\to\infty]{}(e^{-2})^2\cdot1^2=e^{-4}$$
A: $$ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} $$
$$ =\lim_{n \to \infty} \left(\left(1+\frac{(-2)}{n+1} \right)^{\frac{n+1}{-2}}\right)^{\frac{-2(2n+1)}{n+1}}$$
$$ = \left(\lim_{n \to \infty}\left(1+\frac{(-2)}{n+1} \right)^{\frac{n+1}{-2}}\right)^{\lim_{n \to \infty}\left(\frac{-4-\frac2n}{1+\frac1n}\right)}$$
$$=(e)^{-4}\text{ as } n \to \infty, \frac{n+1}{-2}\to-\infty\text{ and } \lim_{m\to\infty}\left(1+\frac1m\right)^m=e$$
