Limit of $\lim \limits_{x \to\infty} (\frac{x}{1-x})^{2x}$ I've been struggling with this term for a while now:
$$\lim \limits_{x \to\infty} (\frac{x}{1-x})^{2x}$$
I know it has to do something with $\lim \limits_{x \to\infty} (1+\frac{n}{x})^x=e^n$ but didn't come further than this though: 
$$\lim \limits_{x \to\infty}  (1+ \frac{2x-1}{1-x})^{2x} = \lim \limits_{x \to\infty} (1+ \frac{2x-1}{1-x})^{1-x})^{\frac{2x}{1-x}} \overset{?}{=} e^{2x-1 {\lim \limits_{x \to\infty} \frac{2x}{1-x}}}  $$ 

Since $\frac{x}{1-x}\lt0$ for $x\gt1$, this limit does not exist unless $x\in\mathbb{Z}$ is specified (so that the exponent is an even integer). However,
$$
\begin{align}
\lim_{x\to\infty}\left(\frac{x}{x-1}\right)^{2x}
&=\lim_{x\to\infty}\left(1+\frac1{x-1}\right)^{2x-2}\lim_{x\to\infty}\left(1+\frac1{x-1}\right)^2\\
&=e^2\cdot1
\end{align}
$$
