This seems to be an easy problem. First $r\neq 0$. I simply find the power series of the denominator centered around $c=e^{i\pi/2r}$ to be $1+c^{2r}+2rc^{2r-1}(z-c)+r(2r-1)c^{2r-2}(z-c)^2+...$ and divide nominator and denominator by $z-c$ to obtain $$\frac{z-c}{z^{2r}+1}=\frac{1}{\frac{1+c^{2r}}{z-c}+2rc^{2r-1}+r(2r-1)c^{2r-1}(z-c)+...}.$$ So, $$\lim_{z \to c}\frac{z-c}{z^{2r}+1}=\frac{1}{\lim \frac{1+c^{2r}}{z-c}+2rc^{2r-1}}$$
Apparently, we will require that $1+c^{2r}=0$. But now there is a quite strange thing that happens: on the one hand $1+c^{2r}=1+e^{i \pi/(2r)2r}=0$--so $$\lim_{z \to c}\frac{z-c}{z^{2r}+1}=\frac{1}{2rc^{2r-1}}$$ and this is non-zero for any nonzero $z$
But on the other hand $1+c^{2r}=0$ is not true for $r=1/4$! $1+e^{i \pi/(2*0.25)2*0.25}=1+(e^{i*2 \pi})^{\pi/2}=1+1^{\pi/2}=2$.
Indeed, the answer is (C). The solution simply commented on "there is a branch issue and $r=1/4$ is a counter-example". But by my computation above the limit is simply -2 when $r=1/4$ whereas the true value is 0. Can someone explain this phenomeon?
What did I do wrong?