I am trying to find the limit of $$(1-\frac2n)^n$$ I know how $e$ is defined and I am sure the prove will involve substituting a term with $e$ at some point. But I do not really know where to start. I tried rewriting the term, simplifying it, using the binomial theorem, but all that does not seem to work out that well. Where do I start?
Edit: As $n$ goes towards infinity
The familiar limit is with $\dfrac1n$. Then try the change of variable
$$-\frac2n=\frac1m$$ which gives
$$\lim_{n\to\infty}\left(1-\frac2n\right)^n=\lim_{m\to-\infty}\left(1+\frac1m\right)^{-2m}=\left(\lim_{m\to-\infty}\left(1+\frac1m\right)^{m}\right)^{-2}=e^{-2}.$$
There is a little flaw in the above derivation, because the limit is to $-\infty$. You can fix that by using
$$\frac{\lim_{m\to\infty}\left(1+\dfrac1m\right)^{m}}{\lim_{m\to\infty}\left(1-\dfrac1m\right)^{-m}}=\lim_{m\to\infty}\left(1-\frac1{m^2}\right)^{m}=1.$$
Note that $$ \left(1-\frac2n\right)^n = \left(1-\frac{1}{n/2}\right)^n = \left(\left(1-\frac{1}{n/2}\right)^{n/2}\right)^2 $$ Now take the limit as $\frac n2 \to \infty$.
Let $y = \left(1+ -\frac2n\right)^n$
$$\log y = n\log\left(1+ -\frac2n\right)$$
$$\log y = {(-2/n)n\log\left(1+ -\frac2n\right)\over (-2/n)}$$
$$ y = \exp\left({-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)$$
$$ \lim_{n\to\infty} y = \lim_{n\to\infty} \exp\left({-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)= \exp\left(\lim_{n\to\infty}{-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)=\exp(-2)= e^{-2}$$
Consider the product $$ \left(1+x\right)^2\left(1-2x\right)=1-3x^2-2x^3 $$ which leads to $$ \left(1+\frac1n\right)^{2n}\left(1-\frac2n\right)^n =\left(1-\frac3{n^2}-\frac2{n^3}\right)^n $$ By Bernoulli, one gets for the right side $$ 1\ge\left(1-\frac3{n^2}-\frac2{n^3}\right)^n\ge1-\frac3{n}-\frac2{n^2} $$ so by squeeze theorem its limit is $1$. By the quotient rule of limits, $$ \lim_{n\to\infty}\left(1-\frac2n\right)^n=\left(\lim_{n\to\infty}\left(1+\frac1n\right)^{n}\right)^{-2} $$
