Find the limit of $(1-\frac2n)^n$ 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
 A: 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
$$\frac{\lim_{m\to\infty}\left(1+\dfrac1m\right)^{m}}{\lim_{m\to\infty}\left(1-\dfrac1m\right)^{-m}}=\lim_{m\to\infty}\left(1-\frac1{m^2}\right)^{m}=1.$$
A: 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 $\frac n2 \to \infty$.
A: Let $y = \left(1+ -\frac2n\right)^n$

$$\log y = n\log\left(1+ -\frac2n\right)$$
$$\log y = {(-2/n)n\log\left(1+ -\frac2n\right)\over (-2/n)}$$
$$ y = \exp\left({-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)$$
$$ \lim_{n\to\infty} y =
\lim_{n\to\infty} \exp\left({-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)= \exp\left(\lim_{n\to\infty}{-2\log\left(1+ -\frac2n\right)\over (-2/n)}\right)=\exp(-2)= e^{-2}$$
A: 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, 
$$
\lim_{n\to\infty}\left(1-\frac2n\right)^n=\left(\lim_{n\to\infty}\left(1+\frac1n\right)^{n}\right)^{-2}
$$
