Taking limit for $\left(\frac{1}{\delta} \log \left(1 + \frac{(e^{\delta u_1} -1 )(e^{\delta u_2} - 1)}{e^{\delta} -1}\right)\right)$

Question

Find $$\lim_{\delta \to 0} \left(\frac{1}{\delta} \log \left(1 + \frac{(e^{\delta u_1} -1 )(e^{\delta u_2} - 1)}{e^{\delta} -1}\right)\right).$$

I was trying to prove this limit equals $u_1 u_2$. Not sure how to start but thought about using L'hopital's rule. Any hints will be helpful.

Context: This is frank's copulas and appears when I try to show independence.

Answer 1

The interesting part is the following:

Can we expand $\log \big(1 + \frac{(e^{\delta u_1}-1)(e^{\delta u_2}-1)}{e^\delta-1}\big)$ to show that it is equal to

$ \delta u_1 u_2 - \frac{1}{2} \delta^2 ((u_1 - 1) u_1(u_2 - 1) u_2) + \frac{1}{12} \delta^3 u_1 (2 u_1^2 - 3 u_1 + 1) u_2 (2 u_2^2 - 3 u_2 + 1) - \frac{1}{24} \delta^4 ((u_1 - 1) u_1 (u_2 - 1) u_2 (u_1^2 (6 u_2^2 - 6 u_2 + 1) + u_1 (-6 u_2^2 + 6 u_2 - 1) + (u_2 - 1) u_2)) + \frac{1}{720} \delta^5 u_1 (2 u_1^2 - 3 u_1 + 1) u_2 (2 u_2^2 - 3 u_2 + 1) (u_1^2 (36 u_2^2 - 36 u_2 + 3) + u_1 (-36 u_2^2 + 36 u_2 - 3) + 3 u_2^2 - 3 u_2 - 1) + o(\delta^6)$

If we can calculate the coefficient of $\delta$ in $\frac{(e^{\delta u_1}-1)(e^{\delta u_2}-1)}{e^\delta-1}$, that will give us the answer directly (Since upon dividing by $\delta$, all the $o(\delta^2)$ terms, (i.e., the terms containing $\delta^2$ and higher powers of $\delta$) will $\to 0$ as $\delta \to 0$).

How do we find the coefficient of $\delta$?

Let $\frac{(e^{\delta u_1}-1)(e^{\delta u_2}-1)}{e^\delta-1} = a_0 + a_1 \delta + a_2 \delta^2 + \ldots$

Or, $(e^{\delta u_1}-1)(e^{\delta u_2}-1) = (a_0 + a_1 \delta + a_2 \delta^2 + \ldots)(e^\delta-1)$

Or, $(1 + \delta u_1 + \ldots -1)(1 + \delta u_2 + \ldots -1) = (a_0 + a_1 \delta + a_2 \delta^2 + \ldots)(1 + \delta + \ldots -1)$

Comparing the coefficients of $\delta$ and $\delta^2$ from both sides, we find $a_0 = 0$ and $a_1 = u_1u_2$