Setting:
Let $\mathfrak{U}$ be an ultrafilter on an index set $I$.
Let $G$ be a compact group with identity $e$, and let $\mathbb{T}$ denote the unit circle in the complex plane.
For each $i\in I$, let $\gamma_{i}:G\to \mathbb{T}$ be a continuous group homomorphism, and choose $s_{i},t_{i}\in G$.
Question:
Under what circumstances can I compute the limit
$$\displaystyle\lim_{i\to\mathfrak{U}}\gamma_{i}(s_i^{-1}t_{i})$$
iteratively, by instead considering
$$\displaystyle\lim_{i\to\mathfrak{U}}\displaystyle\lim_{j\to\mathfrak{U}}\gamma_{i}(s_j^{-1}t_{j})?$$
Motivation:
I can easily compute the latter limit as 1, since in my setting I have the additional assumption that $\displaystyle\lim_{j\to\mathfrak{U}}s_j^{-1}t_{j} = e$
Does not hold in general
A simple case... $$ G = \mathbb T = \{z \in \mathbb C \; : \; |z|=1\},\qquad\text{with mutiplication} \\ I = \{1,2,3,\dots\}, \\ \mathfrak U = \text{ any free ultrafilter} \\ \gamma_n(z) = z^n \\ s_n = 1, t_n = e^{i \pi/n} \\ \lim_n \gamma_n(s_n^{-1}t_n) = \lim_n \big(e^{i \pi/n}\big)^n = \lim_n e^{i\pi} = \lim_n (-1) = -1 \\ \lim_n\lim_m \gamma_n(s_m^{-1}t_m) = \lim_n\lim_m \big(e^{i\pi/m}\big)^n =\lim_n\big(\lim_m e^{i\pi/m}\big)^n =\lim_n\big(1\big)^n = \lim_n 1 = 1 $$