If $\mathbb F$ is some field, the group $PSL(2, \mathbb F)$ consists of the mappings $$ x \mapsto \frac{ax + b}{cx + d} $$ with $a,b,c,d \in \mathbb F$ and $ad - bc = 1$. These mappings are defined over the extended field $\mathbb F_{\infty} := \mathbb F \cup \{\infty\}$ (or the so called projective line). As for example written on wikipedia:Möbius transformation we have $f(\infty) = \infty$ if $c = 0$.
But I am asking myself what happens with $f(\infty)$ if $c \ne 0$?
There is another way to view these mappings, consider $SL(2, \mathbb K)$, the group of matrices $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ with $ad - bc = 1$. Then each such matrix could be mapped onto such a mapping as above, and this correspondence is surjective. By computation we find $$ \frac{a\left( \frac{a'x + b'}{c' + d'} \right) + d}{c\left( \frac{a'x + b'}{c' + d'}\right) + d} = \frac{(aa' + c'b)z + (ab' + bd')}{(ca' + dc')z + (cb' + d'd)} $$ and of course $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} = \begin{pmatrix} aa' + bc' & ab' + bd' \\ a'c + dc' & cb' + dd' \end{pmatrix} $$ so this correspondence is a homomorphism. Such a mapping is the identity if $a = d, c = b = 0$, and the condition $ad = 1$ leaves the possiblities $a = b = \pm 1$, hence the kernel of this correspondence is $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. $$ and this is precisely the center in the group $SL(2, \mathbb F)$. Hence $PSL(2, \mathbb F) \cong SL(2, \mathbb F) / Z(SL(2,\mathbb F))$. Now $SL(2, \mathbb F)$ acts on $\mathbb F^2$ in a natural way by matrix multiplication, hence we have an induced action of $PSL(2,\mathbb F)$ on $\mathbb F^2$. Now each $v = (x, y)^T\in \mathbb F^2$ could be identified with $(x/y, 1)^T$ if $y \ne 1$ and $(1,0)^T$ if $y = 0$, and by this we get an identification of $\mathbb F^2$ with $\mathbb F_{\infty}$ where $(1,0)^T$ corresponds to $\infty$. As $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix} $$ I guess we would have $f(\infty) = a/c$ if $c \ne 0$ (which corresponds to the projective point $(a,c)^T \equiv (a/c, 1)^T$). But I am not sure if this is reasonable, so I am asking what is $f(\infty)$ in the general case?
$f(\infty)$ is $\frac{a}{c}$, as you say. Intuitively, plugging in $\infty$ takes the leading terms of the numerator and denominator. Formally it's what you wrote.