The definition of $f(z)$ being analytic at point $\infty$ 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?
 A: As M Turgeon points out, the whole idea of a removable singularity is that we can "fix" the function at that point so it is equal to the limit, and is then analytic there.
The same idea applies, for example, to the function $f(z)=\dfrac{\sin(z)}{z}$. As it is defined, here, there is a singularity at $z=0$. However, it is removable, since $$\lim_{z\to 0}\frac{\sin z}{z}=1,$$ so we may simply extend the definition of $f$ so that $f(0)=1$, giving us an entire function.
Addendum: Robert Mastragostino brings up a good point, as well. Note that "removable singularity" doesn't (necessarily) mean that we are replacing a given function value with another--indeed, in the example I gave above, $f$ is initially undefined at $z=0$.
An isolated singularity of a function $f$ is a point $z=a$ such that for some $r>0$, we have that $f(z)$ is defined and analytic on $\{z:0<|z-a|<r\}$, but not on $\{z:|z-a|<r\}$. Such a singularity is removable iff there is a function $g$ that agrees with $f$ on $\mathrm{dom}(f)\smallsetminus\{a\}$--note that that doesn't indicate that $a\in\mathrm{dom}(f)$--such that $g(z)$ is defined and analytic on $\{z:|z-a|<r\}$. This may occur by replacing $f(a)$ with $\lim_{z\to a} f(z)$. This may also occur as in the example I gave above, not with replacement, but with extension.
If we are "removing" a singularity at $\infty$, we will generally be extending, rather than replacing a pre-existing function value. A notable exception is with linear fractional transformations $T(z)=\dfrac{az+b}{cz+d}$, where $ad-bc\neq 0$. In such a case, we tend to explicitly define $T(\infty)=\frac{a}{c}$ at the outset.
