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?
The point is that $1-\frac{1}{n}$ is less than $1$, so raising it to a large power will make it even less-er than $1$. On the other hand, $1+\frac{1}{n}$ is bigger than $1$, so raising it to a large power will make it even bigger than $1$.



There's been some brouhaha in the comments about this answer. I should probably add that $(1-\epsilon(n))^n$ could go to any value less than or equal to $1$, and in particular it could go to $1$, as $n$ increases. It so happens that in this example, it goes to something less than $1$. The reason it goes to something less than $1$ is because we end up raising something sufficiently less than $1$ to a sufficiently high power.
The first definition of $e$ is $$ \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = e^1 $$ which is basically just answering the question of what happens if you take the limit of discrete compounded growth by 100% to continuous growth. Note that $e&gt;2$, i.e. this limit of discrete compounded growth asymptotically approaches a value that is greater than one we would have arrived at with the initial rate of growth. This means that even though we're chipping away at the rate we're growing by with every step of compounding, the aggregate effect is more growth. Also, I just want to note that $e$ is the universal constant from continuous growth by a certain rate - meaning that simply raising $e^{rt}$ will give the effect of continuously growing at a rate $r$ for $t$ units of time.

If we decide instead to see what happens when we take the limit of discrete compounded decay instead of growth, $$ \lim_{n \to \infty} \left( 1- \frac{1}{n} \right)^n = e^{-1} $$ we see that the opposite happens. Going from discrete compounded decay to instantaneous decay lessens the amount to which we are decaying by with every step of compounding, and asymptotically approaches a value $1/e&gt; 0$, the value we would have been at if we did a single step of decay at a rate of 100%.

Don't know if this helps at all.