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

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?

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## 3 Answers

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

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

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so both can be transformed into the same limit, which is then solvable with L'Hospitals rule – nlognfan Apr 22 '13 at 11:01

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

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

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I reached the form lim(n->inf,1/(1-1/n)^n). But why is lim(n->inf,(1-1/n)^n)=e^(-1)? that is not obvious – nlognfan Apr 22 '13 at 10:14
Because $\lim_{n\rightarrow\infty}\left( 1+r/n\right )^n=e^r$ for any real $r$. – Matt L. Apr 22 '13 at 10:23
this answer is also good, but I had to pick one, sorry – nlognfan Apr 22 '13 at 11:01
No problem, fair enough :) – Matt L. Apr 22 '13 at 11:19