Splitting a black box rational function

Question

Suppose I can evaluate a black box rational function $f(\vec{x})$. For the purposes of the explanation, let me take an example I can actually write down \begin{equation} f(\vec{x})=\frac{x_1+x_2+x_3+x_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)}. \end{equation} I know some theoretical properties of this black-box function, e.g. \begin{equation} \lim_{x_1\rightarrow -x_2} (x_1+x_2)f(\vec{x})\neq 0, \quad \lim_{x_3\rightarrow -x_4} (x_3+x_4)f(\vec{x})\neq 0, \quad \lim_{x_1\rightarrow -x_2} (x_1+x_2)\lim_{x_3\rightarrow -x_4} (x_3+x_4) f(\vec{x})=0. \end{equation} These can be verified for the specific form I provided above. I am interested in defining $f_1$ and $f_2$ such that $f=f_1+f_2$ and \begin{align} \lim_{x_1\rightarrow -x_2} (x_1+x_2)f_1(\vec{x})\neq 0, \quad \lim_{x_3\rightarrow -x_4} (x_3+x_4)f_1(\vec{x})= 0, \\ \lim_{x_1\rightarrow -x_2} (x_1+x_2)f_2(\vec{x})= 0, \quad \lim_{x_3\rightarrow -x_4} (x_3+x_4)f_1(\vec{x})\neq 0. \end{align} One way to do this is to define \begin{equation} f_1(\vec{x})=f(\vec{x})-\frac{1}{x_3+x_4}\lim_{x_3\rightarrow -x_4}(x_3+x_4)f(\vec{x}), \quad f_2(\vec{x})=\frac{1}{x_3+x_4}\lim_{x_3\rightarrow -x_4}(x_3+x_4)f(\vec{x}), \end{equation} but this requires to take a limit, which I would rather avoid. Is there any other way I can achieve this separation (for example, by multiplying $f$ by two other known rational function that sum up to one)?