Consider this function $f(z) = \frac{1}{1+z}$. We can define $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$, which is zero for this case. Since $f(\frac{1}{t}) \rightarrow \frac{t}{t+1}$ is analytic at point $t=0$, we can deduce that $f(z)$ is analytic at $z=\infty$.
On the other hand, if we consider the type of singularity at $t=0$ of $f(\frac{1}{t})$. There is singularity because of the existence of $\frac{1}{t}$ but the singularity is removable. So the singularity at $z=\infty$ is removable.
The definition is $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$. So how we discuss the singularity and think of it as removable if the value of $f(\infty)$ is taken from the limit?