Christopher Bishop defines the expected value of the complete-data log likelihood function (i.e. assuming that we are given both the observable data X as well as the latent data Z) as follows:
$$ \mathbb{E}_\textbf{Z}[\ln p(\textbf{X},\textbf{Z} \mid \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})] = \sum_{n=1}^N \sum_{k=1}^K \gamma(z_{nk})\{\ln \pi_k + \ln \mathcal{N}(\textbf{x}_n \mid \ \boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)\} \tag 1 $$
where $\gamma(z_{nk})$ is defined as:
$$ \frac{\pi_k \mathcal{N}(\textbf{x}_n \mid \ \boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)}{\sum_{j=1}^K \pi_j \mathcal{N}(\textbf{x}_n \mid \ \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)} \tag 2 $$
The idea, as described, is to consider a Gaussian Mixture Model in which the covariance matrices of the mixture components are given by $\epsilon \textbf{I}$, where $\epsilon$ is a variance parameter that is shared by all of the components, such that:
$$ p(\textbf x \mid \boldsymbol \mu_k, \boldsymbol \Sigma_k) = \frac{1}{(2 \pi \epsilon)^\frac{M}{2}} \exp\big\{{-\frac{1}{2 \epsilon} \|\textbf x - \boldsymbol \mu_k\|^2}\big\} \tag 3 $$
and so, $\gamma(z_{nk})$ is now defined as:
$$ \frac{\pi_k \exp\{ - \| \textbf x_n - \boldsymbol \mu_k\|^2 / 2 \epsilon\}}{\sum_{j=1}^K \pi_j \exp\{ - \| \textbf x_n - \boldsymbol \mu_j\|^2 / 2 \epsilon\}} \tag 4 $$
The argument now is the following:
if we consider the limit $\epsilon \to 0$, we see that in the denominator the term for which $\| \textbf x_n - \boldsymbol \mu_j\|^2$ is smallest, will go to zero most slowly, and hence the responsibilities $\gamma(z_{nk})$ for the data point $\textbf x_n$ all go to zero except for term j, for which the responsibility $\gamma(z_{nk})$ will go to unity. Thus, in this limit, we obtain a hard assignment of data points to clusters, just as in the $K$-means algorithm, so that $\gamma(z_{nk}) \to r_{nk}$
where $r_{nk}$ is defined as:
\begin{equation*} f(n) = \begin{cases} 1 & \text{if } k = \text{arg } \text{min}_j \|\textbf x_n - \boldsymbol \mu_j\|^2\\ 0 & \text{otherwise}\\ \tag 5 \end{cases} \end{equation*}
My question is how does the above argument hold? Namely, what does it mean for a term to go to zero $\textbf{most slowly}$ ? And how does taking the limit $\epsilon \to 0$ in eqn $4$ result in a binary responsibility?