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 $A(\omega)$? 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. Also, to finally take the limit and express the result, the following formula is used $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ 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. Note that all these will just appear as convenient symbols, *not* to be evaluated. Two examples are given below, where all parameters are real. Therefore the imaginary part purely generated by $\eta$ is taken, which is simply $iA/2$. $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta+\sqrt{c+2di\eta}}{e}=\lim_{\eta\rightarrow0^+}\frac{(b+d)\eta}{e}=0$$ and $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta}{c+i\eta}=\lim_{\eta\rightarrow0^+}\Im(-i\pi\delta(c)+\mathcal{P}(\frac{1}{c}))(a+bi\eta)=\lim_{\eta\rightarrow0^+}-a\pi\delta(c)+\eta\mathcal{P}(\frac{1}{c})=-a\pi\delta(c).$$ 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.