Intuitive explanation for why $\left(1-\frac1n\right)^n \to \frac1e$ I am aware that $e$, the base of natural logarithms, can be defined as:
$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$
Recently, I found out that
$$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n = e^{-1}$$
How does that work? Surely the minus sign makes no difference, as when $n$ is large, $\frac{1}{n}$ is very small?
I'm not asking for just any rigorous method of proving this. I've been told one: as $n$ goes to infinity, $\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n = 1$, so the latter limit must be the reciprocal of $e$. However, I still don't understand why changing such a tiny component of the limit changes the output so drastically. Does anyone have a remotely intuitive explanation of this concept?

Let me try. Consider $$A=\left(1+\frac{a}{n}\right)^n$$ Take logarithms $$\log(A)=n\log(1+\frac a n)$$ Now, when $x$ is small, by Taylor, $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ Replace $x$ by $\frac{a}{n}$. This makes $$\log(A)=n \Big(\frac{a}{n}-\frac{a^2}{2 n^2}+O\left(\frac{1}{n^3}\right)\big)=a-\frac{a^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ Now, $$A=e^{\log(A)}=e^a-\frac{a^2 e^a}{2 n}+O\left(\frac{1}{n^2}\right)$$ Now, play with $a$. 
Hoping that this makes things clearer to you.
