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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I asked recently in this question how to use the definition of

$$e:= \lim_{x\to\infty}\left( 1+\frac1x \right)^x$$

to show that

$$\lim_{x\to -\infty}\left( 1+\frac1x \right)^x = e.$$

A helpful answer said that

$$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$

and $-(\eta + 1) \xrightarrow[]{\eta \to -\infty} \infty$.

It struck me as a brilliant way to solve the problem, and I wondered how one might come up with it. Is the rearrangement done in this answer an instance of a more general strategy, or is it more or less just fiddling until one gets the right form?

share|cite|improve this question
The strategy is to change $x \rightarrow -x$, and try to compute the limit with the limit you already have. – Secret Math Oct 14 '13 at 17:20
up vote 1 down vote accepted

In general, if $h(x)=1+O(x^{-2})$ then $\lim_{x\to\infty} h(x)^x = 1$.

Now, let $h(x)=\left(1-\frac{1}{x^2}\right)=\left(1-\frac{1}{x}\right)\left(1+\frac 1x\right)$.

Since $h(x)^x\to 1$ as $x\to\infty$ and $(1+1/x)^x\to e$ as $x\to\infty$ we have that $\left(1-\frac 1x\right)^x\to \frac{1}{e}$ as $x\to\infty$.

But then $\left(1-\frac1x\right)^{-x}\to e$ as $x\to\infty$. Replacing $x$ with $-x$, we see that:

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

So all you have to prove is the first claim.

share|cite|improve this answer
That is definitely food for thought. Thank you as always for your helpful posts. – Eric Auld Oct 14 '13 at 17:42
Just reread this. That's a beautiful answer! – Eric Auld Oct 17 '13 at 3:39

Hint : Given limit is of the form $1^\infty$, It's a indeterminate form. You can have a look of this to find the limit in such cases.

Why is $1^{\infty}$ considered to be an indeterminate form

share|cite|improve this answer
I'm not sure this is relevant, since it is an explanation of indeterminate forms, and recommends taking the logarithm, which is not how the above problem was solved. Thanks anyway – Eric Auld Oct 14 '13 at 17:41

Your Answer


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.