Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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

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

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

share|improve this answer

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

share|improve this answer
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.