I want to understand these two limits:
$\lim_{\delta \to 0} N_{\delta}(F)\delta^{s} = \infty$ when $s < \dim_{M}(F)$
$\lim_{\delta \to 0} N_{\delta}(F)\delta^{s} = 0$ when $s > \dim_{M}(F)$
where $\dim_{M}(F) = \lim_{\delta \to 0} \dfrac{\log N_{\delta}(F)}{-\log \delta}$ is the Minkowski dimension, or box-counting dimension of a bounded non-empty subset $F$ of $\mathbb{R}^n$.
I tried to do this, for the first case (an analogue calculations for the second case, both lead to nowhere):
$s < \lim_{\delta \to 0} \dfrac{\log N_{\delta}(F)}{-\log \delta} = \lim_{\delta \to 0} \log_{\delta} \dfrac{1}{N_{\delta}(F)} \implies \delta^s > \lim_{\delta \to 0} \dfrac{1}{N_{\delta}(F)} \implies \\ \implies N_{\delta}(F)\delta^s > N_{\delta}(F)\lim_{\delta \to 0} \dfrac{1}{N_{\delta}(F)} \implies \\ \implies \lim_{\delta \to 0} N_{\delta}(F)\delta^{s} \ge \lim_{\delta \to 0} \left(N_{\delta}(F)\lim_{\delta \to 0} \dfrac{1}{N_{\delta}(F)}\right)=1$
Then I have $\lim_{\delta \to 0} N_{\delta}(F)\delta^s\ge 1$, but why would this limit be infinity?
Probably I am not doing my calculations right...
Consider $\lim_{\delta \to 0} N_\delta(F) \delta^{\dim_M (F) + \epsilon}$. By a property of exponents, this equals $\lim_{\delta \to 0} N_\delta(F) \delta^{\dim_M (F)} \delta^{\epsilon}$. When the individual limits are finite, this can be split into a product of limits: $\left(\lim_{\delta \to 0} N_\delta(F) \delta^{\dim_M (F)}\right) \left(\lim_{\delta \to 0}\delta^{\epsilon}\right)$. Now assuming $\epsilon > 0$, $\lim_{\delta \to 0}\delta^{\epsilon} = 0$.
To show that the other limit in the product is finite: $\lim_{\delta \to 0} N_\delta(F) \delta^{\dim_M (F)} = K \in \mathbb{R}$ iff $\lim_{\delta \to 0} \log N_\delta(F) + \dim_M(F) \log \delta = \log K$ iff $\dim_M(F) = \lim_{\delta \to 0} \frac{\log N_\delta(F) - \log K}{-\log \delta} = \lim_{\delta \to 0} \frac{\log N_\delta(F)}{-\log \delta}$.