Let $R(z)$ be some fixed rational function, and define $Res_R(w)$ to be the residue of $f(z)=\dfrac{1}{z-w}R(z)$ for any $w\in\mathbb{C}$. I would have thought that $Res_R$ would be continuous in $w$, because it seems like the entries of the Laurent series of $\dfrac{1}{z-w}R(z)$ centered at $w$ should be continuous at $w$, but this is not the case.
According to the limit formula for residues, $Res_R(w)=R(w)$ as long as $w$ is not a pole of $R$, but then of course $Res_R(w)\to\infty$ as $w$ approaches any pole of $R$. However $Res_R(w)$ is finite everywhere, thus is not continuous at the poles of $R$. Could someone give an intuitive reason why $Res_R$ is not continuous at the poles of $R$?
If $\Gamma$ is a simple closed positively-oriented contour, and $f$ is analytic on and inside $\Gamma$, then $\dfrac{1}{2\pi i} \oint_\Gamma f(z)\; dz$ is the sum of the residues of the poles of $f$ inside $\Gamma$. Now in your case for a suitably chosen $\Gamma$, $f(z) = R(z)/(z - w)$ will have two poles inside $\Gamma$, one pole (say $p$) of $R(z)$ and the pole $w$ of $1/(z-w)$ (assuming $w \ne p$). As you pointed out, as $w$ approaches $p$ both residues get large, but their sum stays bounded (as $f(z)$ doesn't change much on $\Gamma$). However, when $w = p$ we have only one pole inside $\Gamma$, and its residue is small (again, because $f(z)$ hasn't changed much on $\Gamma$). The quantity that is continuous is the sum of the residues.