Show that $|z|_0=\lim_{\epsilon\to0}\sum_{i=1}^n\frac{\log\left(1+z_i\epsilon^{-1}\right)}{\log\left(1+\epsilon^{-1}\right)}$? I found this property of the zero norm $|z|_0$ where $z\geqslant0$ a vector of $\mathbb{R}^n$:
$$|z|_0=\lim_{\epsilon\to0}\sum_{i=1}^n\frac{\log\left(1+z_i\epsilon^{-1}\right)}{\log\left(1+\epsilon^{-1}\right)}.$$
The zero norm of a vector $z=[z_1,\ldots,z_n]^\top$, $|z|_0$, is the number of non-zero elements in $z$.
How to prove this property?
 A: If $z_i = 0$, then clearly
$$\frac{\log (1 + z_i\epsilon^{-1})}{\log (1 + \epsilon^{-1})} = \frac{\log 1}{\log (1 + \epsilon^{-1})} = 0.$$
For $z_i \neq 0$, note
$$\frac{\log (1 + z_i\epsilon^{-1})}{\log (1 + \epsilon^{-1})} = \frac{\log \epsilon^{-1} + \log (\epsilon + z_i)}{\log \epsilon^{-1} + \log (1 + \epsilon)} = \frac{1 - \frac{\log (\epsilon + z_i)}{\log \epsilon}}{1 - \frac{\log (1 + \epsilon)}{\log \epsilon}}.$$
In this form, the limit as $\epsilon \to 0$ is not hard to determine.
A: Suppose $t>0$; then
$$
\lim_{\varepsilon\to0^+}
\frac{\log(1+t\varepsilon^{-1})}{\log(1+\varepsilon^{-1})}=1
$$
(the limit doesn't make sense for $\varepsilon\to0^{-}$). Just apply l'Hôpital to the $\infty/\infty$ form
$$
\lim_{\varepsilon\to0^+}
\frac{\log(1+t\varepsilon^{-1})}{\log(1+\varepsilon^{-1})}=
\lim_{\varepsilon\to0^+}
\frac{\log(\varepsilon+t)-\log\varepsilon}
     {\log(1+\varepsilon)-\log\varepsilon}=
\lim_{\varepsilon\to0^+}
\frac{\dfrac{1}{\varepsilon+t}-\dfrac{1}{\varepsilon}}
     {\dfrac{1}{1+\varepsilon}-\dfrac{1}{\varepsilon}}
$$
