Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Is it true that : $\lim_{ x\to\infty } \left( 1+\frac{f \left( x \right) }{x} \right) ^x = \exp \left( \lim_{ x\to\infty } f \left( x \right) \right)$ ?

Assumption is that limit of $\lim_{ x\to\infty } f(x)$ exists.

share|cite|improve this question
Yes. Proof: what can you say about $\log(1+f(x)/x)$ when $x\to\infty$? – Did Apr 10 '12 at 21:00
I was wondering who was that sad man who downvoted both answers... – Norbert Apr 10 '12 at 22:45
up vote 2 down vote accepted

\begin{eqnarray} \lim_{x\to\infty}\left(1+\frac{f(x)}x\right)^x &=&\lim_{x\to\infty}\exp \left(x\log \left(1+\frac{f(x)}x \right) \right)=\exp \left(\lim_{x\to\infty}x\log \left(1+\frac{f(x)}x\right)\right)\\ \ \\ &=&\exp \left(\lim_{x\to\infty}x \left(\frac{f(x)}x+o\left(\frac1{x^2}\right)\right)\right) =\exp \left(\lim_{x\to\infty}f(x)+o\left(\frac1{x}\right)\right)\\ \ \\ &=&\exp \left(\lim_{x\to\infty}f(x)\right). \end{eqnarray}

The limit exchange in the second equality is justified by the fact that the exponential is continuous and that the expression inside converges.

share|cite|improve this answer
Not sure why this was downvoted. This is essentially correct, but I guess some people don't like seeing the second equality before seeing a proof of the convergence of the expression inside. – Aryabhata Apr 11 '12 at 5:27
Could be. Actually, it's probably much better to address the limit inside first. What I don't like is when people downvote without saying why. – Martin Argerami Apr 11 '12 at 5:40
Can't help it. People aren't obligated to explain their upvotes or downvotes. In this case, the person downvoting probably lost an opportunity to learn something from your answer. – Aryabhata Apr 11 '12 at 5:45
One more possible reason for the downvote: You haven't mentioned that $\frac{f(x)}{x} \to 0$, in order to justify the third equality. – Aryabhata Apr 11 '12 at 7:21

Since there exist $\lim\limits_{x\to 0}f(x)=l$ we have that $f$ is bounded in the neighborhood of $\infty$. So $\lim\limits_{x\to\infty}\frac{f(x)}{x}=0$ and we can write $$ \lim\limits_{x\to\infty}\left(1+\frac{f(x)}{x}\right)^x= \lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{f(x)}= \lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{\lim\limits_{x\to\infty}f(x)}= e^l $$

share|cite|improve this answer
Is the second limit in second equality justified by the fact that (1+f(x)/x)^(x/f(x)) converges ? – Qbik Apr 10 '12 at 21:16
I didn't downvote this, but perhaps this was downvoted because you don't deal with the case: $f(x) = 0$ infinitely often? – Aryabhata Apr 11 '12 at 5:22

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.