Let $x_i\geq 0$ for $i=1,2,\ldots,n$, $\epsilon\geq 0$ and $\epsilon\neq 1$, and consider the function $$f(\epsilon)=\Big(\frac{1}{n}\sum_{i=1}^n\Big[\frac{x_i}{\bar{x}}\Big]^{1-\epsilon}\Big)^{1/(1-\epsilon)},$$ where $\bar{x}=\sum_{i=1}^nx_i\Big/n$.
My question is: How do you prove that $$\lim_{\epsilon\to\infty}f(\epsilon)=\frac{1}{\bar{x}}\min_{i\in\{1,2,\ldots,n\}}x_i?$$ This limit is supposed to be true according to Atkinson, A. B. (1970). On the measurement of inequality. Journal of economic theory, 2(3), 244-263. However, I do not see why this would be the case.
Background: I am interested in this since Atkinson's index (a measure of inequality) is defined as $$A_\epsilon=1-\frac{f(\epsilon)}{\bar{x}}.$$
I suppose that $x_i>0$ for all $i$. Suppose that $\min \{x_i\}=L$. We may suppose that $x_1=x_2=...=x_s=L$, and that $x_j\geq M>L$ if $j\geq s+1$ (if all the $x_k$ are equal the problem is trivial). We may suppose also that $\varepsilon>1$. Then we get that $$s(\frac{L}{\overline{x}})^{1-\varepsilon}\leq (\sum (\frac{x_i}{\overline{x}})^{1-\varepsilon} \leq s(\frac {L}{\overline{x}})^{1-\varepsilon}+(n-s)(\frac{M}{\overline{x}})^{1-\varepsilon} $$ (Note that we have used that for $j\geq s+1$, $x_j\geq M$, hence as $1-\varepsilon<0$, $x_j^{1-\varepsilon}\leq M^{1-\varepsilon}$).
This show that
$$\frac{L}{\overline{x}}(\frac{s}{n})^{1/(1-\varepsilon)}\geq f(\varepsilon)\geq \frac{L}{\overline{x}}(\frac{s}{n})^{1/(1-\varepsilon)} (1+\frac{n-s}{s}(M/L)^{1-\varepsilon)})^{1/(1-\varepsilon)}$$
Now as $M/L>1$, $(M/L)^{1-\varepsilon}\to 0$ and as $1/(1-\varepsilon) \to 0$, it is easy to conclude.