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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?

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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

1 Answer 1

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:

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

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

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

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

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Thanks, I forgot that fact. –  The Substitute Sep 30 '12 at 3:17

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