TITLE: what if take limit to negative infinity in the definition of e as a limit
QUESTION [5 upvotes]: By definition $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e.$$
But what about a similar limit where $n$ tends to negative infinity, i.e, $$\lim_{n\to -\infty}\left(1+\frac{1}{n}\right)^n?$$
Why does that equal to $e$ too? Could someone give a simple proof?

REPLY [5 votes]: The following argument is based solely on the standard limit $\lim_{n\to\infty}\left(1+{1\over n}\right)^n=e$.
For $|x|<1$ Bernoulli's inequality guarantees $(1+x)^n\geq1+nx$. Therefore
$$1-{1\over n}\leq \left(1-{1\over n^2}\right)^n\leq1\qquad(n\geq2)\ ,$$
and it follows that $\lim_{n\to\infty}\left(1-{1\over n^2}\right)^n=1$. From this we conclude that
$$\left(1-{1\over n}\right)^n={\left(1-{1\over n^2}\right)^n\over \left(1+{1\over n}\right)^n}\to{1\over e}\qquad(n\to\infty)\ .$$