How to find the limit $z\rightarrow i$? How to calculate the limit of $\frac{z^{10}+1}{z^6+1}$ as $z\rightarrow i$?
I tried to take the limit but the denominator becomes zero. Does this mean that the limit is infinite?

To avoid L'Hopital rememeber that
$$\begin{align}x^{10}+1=&(x^2+1)(x^8-x^6+x^4-x^2+1)\\
x^6+1=&(x^2+1)(x^4-x^2+1)\end{align}$$
So the fraction simplifies to
$${x^8-x^6+x^4-x^2+1\over x^4-x^2+1}\to {5\over 3}$$
