Many problems in science and engineering are related to the analytic continuation and in particular infinitesimal analytic continuation to the upper or lower complex plane, i.e., a generic complex function $f(\omega)$ with real $\omega$ is changed to $$F(\omega)=\lim_{\eta\rightarrow0^+}f(\omega\pm i\eta).$$ In many applications, only the part introduced by $\eta$ is necessary, which is $$A(\omega)=\lim_{\eta\rightarrow0^+}[f(\omega+i\eta)-f(\omega-i\eta)].$$

Question: How to symbolically obtain $A(\omega)$?
See examples below: simple Limit does not work. A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

The reason why simple Limit does not work is the following Sokhotski–Plemelj formula that involves DiracDelta $\delta$-function $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ where $\delta$-function is the effect of $\eta$ and a principal value is formally noted. A more general version $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n+1)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ involves the $(n+1)$-th derivative of $\delta$-function. To finally take the limit and express the result, we use such a formula. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

This can always be calculated by hand; I think MMA is a perfect match for this task.

In such a calculation, as the limit is concerned, only the leading order effect of $\eta$ is relevant to us; hence many higher-order terms of $\eta$ can be dropped.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, we have \begin{equation} \begin{split} A(\omega)&=2i\lim_{\eta\rightarrow0^+}\Im f(\omega+i\eta)\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a(\omega+i\eta)+\sqrt{c+(\omega+i\eta)^2}}{e}\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a(\omega+i\eta)+c+\omega^2+i\eta\frac{\omega}{c+\omega^2}}{e}\\ &=2i\lim_{\eta\rightarrow0^+}\frac{ai\eta+i\eta\frac{\omega}{c+\omega^2}}{e}\\ &=0. \end{split} \end{equation} For $f(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=2i\lim_{\eta\rightarrow0^+}\Im f(\omega+i\eta)\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a+b(\omega+i\eta)}{\omega-c+i\eta}\\ &=2i\lim_{\eta\rightarrow0^+}\Im\,\{[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})][a+b(\omega+i\eta)]\}\\ &=2i\lim_{\eta\rightarrow0^+}[-\pi(a+b\omega)\delta(\omega-c)+b\eta\mathcal{P}(\frac{1}{\omega-c})]\\ &=-\pi(a+b\omega)\delta(\omega-c). \end{split} \end{equation} This second example shows that simple Limit does not work, which merely gives 0 as shown below.

f[\[Omega]_] := (a + b \[Omega])/(\[Omega] - c);
Limit[f[\[Omega] + I \[Eta]] - f[\[Omega] - I \[Eta]], \[Eta] -> 0, 
 Direction -> "FromAbove"]

I thought about this for some time, but due to my very limited MMA skill, I don't see how to realize such a calculation. For instance, I think we need to Series expand with respect to $\eta$ and take the leading order, but it is unclear how to maintain and recognize the pattern of denominators in the formulae.