Show that when $\theta \rightarrow\infty, C(u_1, u_2) \rightarrow \min\{u1, u2\}$, where $C(u_1,u_2)$ is Frank's copula

Question

I was told that this could be solved only using limits. However, I don't understand how a limit could converge to the function $\min\{u_1,u_2\}$. Also, I can't get any coherent result from taking the limit.

Answer 1

Looking it up on Wikipedia, Frank's Copula is $$ C(u,v) = -\frac{1}{\theta}\log\left(1 + \frac{(e^{-\theta u}-1)(e^{-\theta v}-1)}{e^{-\theta}-1}\right)$$ (do please learn Mathjax and type the formula so the question is self-contained in the future).

Say $u < v.$ Then we can write what's inside the log as $$ \frac{e^{-\theta}-1 +(e^{-\theta u}-1)(e^{-\theta v}-1) }{e^{-\theta}-1} =\frac{e^{-\theta u}+e^{-\theta v} -e^{-\theta}-e^{-\theta(u+v)}}{1-e^{-\theta}} \\= e^{-\theta u}\frac{1+e^{-(v-u)\theta}-e^{-(1-u)\theta} + e^{-\theta v}}{1-e^{-\theta}}$$ so we have $$ C(u,v) = u -\frac{1}{\theta}\log\left(\frac{1+e^{-(v-u)\theta}-e^{-(1-u)\theta} + e^{-\theta v}}{1-e^{-\theta}}\right).$$ Since $u < v$ and $u < 1$ and $v>0,$ when we take the limit as $\theta\to \infty$ of what's inside the log above, it goes to one. So the limit of that term goes to zero and we're left with $u = \min(u,v).$