The calculation of the box dimension for the Cantor middle third set $C$ typically goes as such
In the $n$th step of constructing $C$, there are $2^n$ line segments of length $\frac{1}{3^n}$ each remaining.
So, if we take $\epsilon_n = \frac{1}{3^n}$, then we get $N(\epsilon_n) = 2^n$, and thus \begin{equation} D_B = - \lim_{\epsilon\to 0} \frac{\log N(\epsilon)}{\log \epsilon} = \lim_{n \to\infty} \frac{\log 2^n}{\log 3^n} = \frac{\log 2}{\log 3} \end{equation}
My question arose when I decided to generalize this by taking $\epsilon_n = \frac{1}{3^{n+\delta(n)}}$ instead, for some $\delta(n) \in\mathbb{N}$. Then, we would get $N(\epsilon_n) = 2^n 3^{\delta(n)}$ (still an integer), and so \begin{equation} D_B = \lim_{n \to\infty} \frac{\log 2^n 3^{\delta(n)}}{\log 3^{n+\delta(n)}} = \frac{\log 2}{\log 3} \lim_{n \to\infty} \frac{n}{n+\delta(n)} + \lim_{n \to\infty} \frac{\delta(n)}{n+\delta(n)} \end{equation}
In this case, the limit depends on how $\delta(n)$ varies with $n$. But shouldn't the box dimension be independent of this? If so, where am I going wrong?