Box dimension as the critical value of the fractal content

Question

Let $M \subseteq \mathbb{R}^n$ be bounded and $N_{\epsilon}(M)$ the minimum number of 'squares' of side $\epsilon$ with center in M necessary to cover $M$. The box dimension of M is then defined as $\dim_B \; M = \lim\limits_{\epsilon \to 0+} \frac{\log \; N_{\epsilon}(M)}{\log \; \epsilon^{-1}}$. The following proposition shows that the s-dimensional fractal content $N_{\epsilon}(M) \epsilon^s$ presents a $0-\infty$-behaviour similar to the s-dimensional Hausdorff measure:

Proposition:

1.$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$ if $s < dim_B M$

2.$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = 0$ if $s > dim_B M$

In argument 1, I show the existence of an element $s_0$ where the above behaviour happens. That is, I prove the proposition replacing $dim_B M$ by $s_0$.

The difference with the Hausdorff dimension definition is that in that setting we defined the dimension as the jumping value in the observed in the $0-\infty$-behaviour. Here, I need to show that $s_0 = dim_B M$.

Until now I only have an informal argument for the case $dim_B M \notin \{0,\infty\}$. So:

Question

How can I formally show that $s_0 = dim_B M$?

Argument 1

For $t > s \ge 0$ write $N_{\epsilon}(M) \epsilon^t = \epsilon^{t-s} N_{\epsilon}(M) \epsilon^s$. If $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s < \infty$ then $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t = 0$. On the other hand, if $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t < \infty$ and $\neq 0$ then $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$. So we can confirm a $0-\infty$-behaviour as shown in the diagram:

References:

Falconer, Fractal Geometry: Mathematical Foundations and Applications, 2nd edition, page 46. Elgar, Measure, Topology and Fractal Geometry, 2nd edition, page 213.

Answer 1

Since $M$ is a bounded subset of $\mathbb R^n$ it automatically holds that $N_\epsilon \le \frac{C}{\epsilon^{-n}}$, hence $d=\dim_B M \le n$.

By the definition of limit for any $\delta>0$ there exists $\epsilon_0>0$ such that for all $\epsilon \in (0,\epsilon_0)$ it holds $d-\delta < \frac{\log N_\epsilon}{\log \epsilon^{-1}} < d+\delta$, hence $\epsilon^{-(d-\delta)} < N_\epsilon < \epsilon^{-(d+\delta)}$. So if $s > d+\delta$ then $N_\epsilon \epsilon^{s} \to 0$ as $\epsilon\to 0$. Similarly, if $0< s < d-\delta$ then $N_\epsilon \epsilon^{s} \to \infty$ as $\epsilon \to 0$. This argument shows that $s_0 \in [d-\delta, d+\delta]$ and by arbitrariness of $\delta>0$ we conclude that $s_0 = \delta$.