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.

Perhaps think about the binomial expansions of $\left(1 + \frac{1}{n}\right)^n$ and $\left(1 - \frac{1}{n}\right)^n$. The first two terms are $1 + n \frac{1}{n}$ and $1 - n \frac{1}{n}$ respectively. And after that the terms in $\left(1 + \frac{1}{n}\right)^n$ are all positive, whereas the terms in $\left(1 - \frac{1}{n}\right)^n$ alternate. So the difference between the two limits is going to be at least 2.

The true issue is not why changing the sign has such an impact, it is why adding such a small quantity as $\dfrac1n$ drastically changes the result.

$$1^n\to1\text{ vs. }\left(1+\frac1n\right)^n\to e$$

(and very similarly $\left(1-\frac1n\right)^n\to e^{-1}$.)

The reason is that the tiny quantity gets multiplied over and over so that it becomes a finite quantity,

$$\left(1+\frac1n\right)\left(1+\frac1n\right)\left(1+\frac1n\right)\cdots=1+\frac1n+\frac1n+\frac1n+\cdots>2$$ as there are $n$ terms $\dfrac1n$ (and yet others). The "tininess" of the terms is well compensated by the amount of terms.

Also notice that the "asymmetry" shown by $e-1\ne 1-e^{-1}$ is just due to the non-linearity of the exponential.

Actually you have the stronger true statement that $$ \lim_{x\to0}(1+x)^{1/x}=e, $$ of which the initial limit you stated is a special case, approaching $0$ through the sequence of values $x=\frac1n$ for $n\in\Bbb N_{>0}$. But if you approach $0$ through the sequence of values $x=-\frac1n$ for $n\in\Bbb N_{>1}$, the same limit gives you $$ \lim_{n\to\infty}\left(1-\frac1n\right)^{-n}=e. $$ Now it is a simple matter to see that the sequence of inverses $\left(1-\frac1n\right)^n$ tends to the inverse value $e^{-1}$.

It should be noted that while the first limit above is more general than the limits for $n\to\infty$, it is also less elementary to define, since it involves powers of positive real numbers with arbitrary real exponents. Introducing such powers requires studying exponential functions in the first place, which is why the limit statement with integer exponents is often preferred. But the more general limit statement is true, and can serve to give intuition for the relation between the two limits in your question.

Here's a useful generalization of the limit definition of $e$ from the OP:

Given

$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

Raise both sides to the power of $x$:

$$e^x = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$$

This is trivially true when $x = 0$, as both sides evaluate to 1

Assume $x \ne 0$ and let $m = nx$, i.e., $n = \frac{m}{x}$

As $n\to\infty, \, m\to\infty$

$$e^x = \lim_{m\to\infty}\left(1+\frac{x}{m}\right)^{m}$$

[Note the similarity between this and the first limit in Marc van Leeuwen's answer].

In particular, for $x = -1$

$$e^{-1} = \lim_{m\to\infty}\left(1+\frac{-1}{m}\right)^{m}$$

or

$$e^{-1} = \lim_{m\to\infty}\left(1-\frac{1}{m}\right)^{m}$$

As mathmandan notes in the comments, my derivation is flawed when $x < 0$, since then $n\to\infty \implies m\to -\infty$ :oops:

I'll try to justify my result for negative $x$ without relying on the fact that $e^x$ is an entire function and that there is only a single infinity in the (extended) complex plane.

For any finite $u, v \ge 0$, we have

$$e^u = \lim_{n\to\infty}\left(1+\frac{u}{n}\right)^{n}$$

and

$$e^v = \lim_{n\to\infty}\left(1+\frac{v}{n}\right)^{n}$$

Therefore,

$$e^{u-v} = \lim_{n\to\infty}\left(\frac{1+\frac{u}{n}}{1+\frac{v}{n}}\right)^{n}$$

Let $m = n + v$. For any (finite) $v$ as $n\to\infty, \, m\to\infty$.

$$\begin{align}\\ \frac{1+\frac{u}{n}}{1+\frac{v}{n}} & = \frac{n + u}{n + v}\\ & = \frac{m + u - v}{m}\\ & = 1 + \frac{u - v}{m}\\ \end{align}$$

Thus $$\begin{align}\\ e^{u-v} & = \lim_{n\to\infty}\left(1+\frac{u - v}{m}\right)^{n}\\ & = \lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^{m-v}\\ & = \lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^m \lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^{-v}\\ & = \lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^m\\ \end{align}$$

since

$$\lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^{-v} = 1$$

In other words,

$$e^{u-v} = \lim_{m\to\infty}\left(1+\frac{u - v}{m}\right)^m$$

is valid for any finite $u, v \ge 0$. And since we can write any finite $x$ as $u-v$ with $u, v \ge 0$, we have shown that

$$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$$

is valid for any finite $x$, so

$$e^{-x} = \lim_{n\to\infty}\left(1+\frac{-x}{n}\right)^{n}$$
And hence $$e^{-x} = \lim_{n\to\infty}\left(1-\frac{x}{n}\right)^{n}$$