I would like to show the equaivalence of the following two optimization problems. I am able to do it but my way is a bit sketchy and I am looking for a better way. Namely, for $n\in\mathbb{N}$ and
$$r(t,n)=pc_0(n)e^{-nf_0(t)}+(1-p)c_1(n)e^{-nf_1(t)}$$ where $0<p<1$ and, $$\lim_{n\rightarrow \infty}\frac{1}{n}\log c_0(n)=\lim_{n\rightarrow \infty}\frac{1}{n}\log c_1(n)=0$$
I need to show that
$$\arg \lim_{n\rightarrow \infty}\min_t r(n,t)=\arg \min_t \max \{f_0(t),f_1(t)\}$$
Additional information: We have also $f_1(t)=f_0(t)-t$, $f_0$ and $f_1$ are non-negative, continous, bounded and defined on some interval $[a,b]$, $a<0$ and $b>0$, and $f_0$ is an increasing and $f_1$ is a decreasing function.
I think without additional information (except probably for $f_0\geq 0$ and $f_1\geq 0$ ) it should be possible to show the equivalence. With the additional information we also have
$$\arg \lim_{n\rightarrow \infty}\min_t r(n,t)=\arg \min_t \max \{f_0(t),f_1(t)\}=\arg \max_t \min \{f_0(t),f_1(t)\}=0$$