what if take limit to negative infinity in the definition of e as a limit

Question

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?

Answer 1

$$\lim_{n\to\pm \infty}\left(1+\frac{1}{n} \right)^n = \lim_{n\to \pm \infty}\exp (n\log(1+\frac 1 n))=\lim_{x\to 0}\exp(\dfrac{\log(1+x)}{x})=\exp(1)=e $$

Where we make the substitution $x=\frac 1 n$. The limit with the logarithm exists, so we get the answer.

Answer 2

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

Answer 3

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 $$\left(1-{1\over n}\right)^n={\left(1-{1\over n^2}\right)^n\over \left(1+{1\over n}\right)^n}\to{1\over e}\qquad(n\to\infty)\ .$$