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?
Answer
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.
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