Uniform Convergence and Limit Inside Integral When Index Is Not a Natural Number

Question

Suppose we have the quantity $$\lim_{\varepsilon \to 0} \int_{0}^{2 \pi} f\left(\varepsilon e^{ri\theta}\right)d\theta,$$ where $\varepsilon > 0, r>0$ and we want to know if we can move the integral inside. Before this gets marked as a repeat, I have seen questions where we might have the following: $$\lim_{n \to \infty} \int_{0}^{2 \pi} f\left(\frac{1}{n} e^{ri\theta}\right)d\theta.$$ However, in this second example, we can let $g_n(x) := f\left(\frac{1}{n} e^{ri\theta}\right)d\theta$ and then show $g_n$ converges uniformly to some function $g$, so we can bring the limit inside the integral. The difference is, in the first example, we cannot construct a sequence indexed by natural numbers that lists $f(\varepsilon e^{ri\theta})$ for every value of $\varepsilon$. So it doesn't seem to me like we can use the same justification to bring the limit inside the first integral. I would like to know a rigorous argument for bringing the limit inside of the first integral. I'm assuming that we would want to approximate each value of $\varepsilon$ by some rational number and construct a sequence that way, but I'm having trouble making a fully rigorous argument.