The cross ratio $(\infty,z_2,z_3,z_4)=\dfrac{(z_3-\infty)(z_4-z_2)}{(z_3-z_2)(z_4-\infty)} = \dfrac{z_4-z_2}{z_3-z_2}$ As stated in wikipedia (https://en.wikipedia.org/wiki/Cross-ratio#:~:text=Kingdon%20Clifford.%5B4%5D-,Definition,-%5Bedit%5D) and in my textbook the cross ratio $(\infty,z_2,z_3,z_4)=\dfrac{(z_3-\infty)(z_4-z_2)}{(z_3-z_2)(z_4-\infty)} = \dfrac{z_4-z_2}{z_3-z_2}$. What I don't get is how the infinities get cancelled out. I am pretty sure that $\infty / \infty$ is undefined, because $1/\infty = 0$, while $\infty0$ is undefined (am I wrong?).

The general definition of the cross ratio is
$$
 (z_1, z_2, z_3, z_4) = T(z_1)
$$
where  $T$ is the unique Möbius transformation which maps $(z_2, z_3, z_4)$ to $(1, 0, \infty)$, respectively. If all $z_j$ are finite then this is equal to
$$
 (z_1, z_2, z_3, z_4) = \frac{(z_3-z_1)(z_4-z_2)}{(z_3-z_2)(z_4-z_1)} \, .
$$
If one of the $z_j$ is equal to $\infty$ then one can compute the cross ratio as the limit for $z_j \to \infty$, for example
$$
(\infty, z_2, z_3, z_4) = \lim_{z_1 \to \infty}\frac{(z_3-z_1)(z_4-z_2)}{(z_3-z_2)(z_4-z_1)} = \frac{z_4-z_2}{z_3-z_2} 
$$
or
$$
(z_1, \infty, z_3, z_4) = \lim_{z_2 \to \infty}\frac{(z_3-z_1)(z_4-z_2)}{(z_3-z_2)(z_4-z_1)} = \frac{z_3-z_1}{z_4-z_1} \, .
$$
