How to evaluate $\lim_{z \to i} \frac{z^2+i}{z^4-1}$?

Question

This is a pretty basic question but I've been stuck on it for a while now.

$$\lim_{z \to i} \frac{z^2+i}{z^4-1}$$

My attempt:

$z^4-1=(z^2+i)(z^2-i)-2$ and then dividing the numerator and denominator by $z^2+i$

Using long division, I arrived at the same as above.

How else can I evaluate this limit?

The answer is $-\frac12$.

Answer 1

Unless there is a typo, the limit doesn't exist--or, if you allow it, the limit is $\infty$ (the point at infinity on the Riemann sphere).

The numerator approaches $i^2+i=-1+i$, but the denominator approaches $i^4 -1=1-1=0$.