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?