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

How would one compute $\lim_{\delta \rightarrow 0, k\rightarrow\infty} (1+\delta)^{ak}$, where $a$ is some positive constant?

I am finding a lower-bound of the Hausdorff Dimension on a Cantor-like set and this expression appeared in my formula.

Here's what I have, even though I'm not sure if I can use L'Hopital in this case (where $k, \delta$ are approaching $\infty, 0$, respectively.)

$\lim (1+\delta)^{ak}= \lim e^{ak\log(1+\delta)}=\lim e^{ak\log(1+\delta)}=\lim e^\frac{a\log(1+\delta)}{\frac{1}{k}}=\lim e^\frac{-ak^2}{1+\delta}=0,$ which I find troubling since the base is always greater than 1.

Would this change much if the limit as k tends to infinity is the liminf?

share|cite|improve this question
It depends how $\delta$ and $k$ approach their respective limits. If $\delta=\frac ck$ then the limit is $e^{ac}$. – Hagen von Eitzen Sep 29 '12 at 22:36
It is undefined, depends on how $\delta$ and $k$ are behaving with respect to each other. The path $\delta=1/n$, $k=n$ gives a different result than $\delta=1/n$, $k=2n$. Almost anything can happen. – André Nicolas Sep 29 '12 at 22:36
up vote 4 down vote accepted

It’s undefined, because the limit depends entirely on how $k\to\infty$ and $\delta\to 0$. For example:

$$\lim_{k\to\infty}\lim_{\delta\to 0}(1+\delta)^{ak}=\lim_{k\to\infty}1^{ak}=1$$

$$\lim_{\delta\to 0^+}\lim_{k\to\infty}(1+\delta)^{ak}=\lim_{\delta\to 0^+}\infty=\infty$$

$$\lim_{\delta\to 0^-}\lim_{k\to\infty}(1+\delta)^{ak}=\lim_{\delta\to 0^-}0=0$$

$$\lim_{\delta\to 0^+}(1+\delta)^{a/\delta}=\lim_{k\to\infty}\left(1+\frac1{bk}\right)^{ak}=e^{a/b}$$

share|cite|improve this answer
Thanks, I forgot that fact. – The Substitute Sep 30 '12 at 3:17

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.