If $\lim\inf_{r\to 0}{r}\cdot \max_{|z|=r}|{f(z)}|=0$ then $0$ is removable singularity. $\lim\inf_{r\to 0}{r}\cdot \max_{|z|=r}|{f(z)}|$ show $0$ is removable singularity, given $f$ is analytic in a punctured neighborhood of $z=0$. 
What makes it difficult for me is the fact that the expression above is of $\lim\inf$ form, which has always been very confusing for me, although I did I was exposed to the expression a long time ago. 
Attempt: I know $\inf r_2\max_{|z|=r_2}|{f(z)}|<\inf r_1\max_{|z|=r_1}|{f(z)}|$ if $r_2<r_1$. At least I guess so, for $r$ for which $f$ is analytic in $\{|z|\le r,z\ne 0\}$. That would mean $f$ has to be bounded. Perhaps I am drawing conclusions inadequately. I sure am in need of some guidance here. 
 A: Let $f$ be analytic in the punctured disk $0 < |z| < R$ and 
denote with $M(r) := \max_{|z|=r}|{f(z)}|$ the maximum value of $|f|$ on the circle with radius $r$. Then your condition is
$$ \tag{1}
   \liminf_{r \to 0} \, r \, M(r) = 0
$$
which means that there is a strictly decreasing sequence $(r_n)$
of positive radii with
$\lim_{n \to \infty} r_n = 0$, such that
$$
 \lim_{n \to \infty } \, r_n \, M(r_n) = 0 \, .
$$
The aim is to show that $(1)$ holds with the $\liminf$ replaced
by the limit.
Your argument for $r_2 M(r_2) < r_1 M(r_1)$ for $r_2 < r_1$ is not
immediately apparent because it would require that $f$ is analytic
in the disk $|z| \le r_1$, including the point $z=0$.
But one can proceed as follows:
Let $\varepsilon > 0$. There exists a $m \in \Bbb N$ such that
$r_n \, M(r_n) < \varepsilon $ for $n \ge m$.
For $0 < |z| < r_m$ choose $n$ so large that $r_n < |z| < r_m$.
Applying the maximum modules principle to that annulus shows that
$$
   |z \, f(z) | \le \max \{ r_n \, M(r_n), r_m \, M(r_m) \} < \varepsilon \, .
$$
So we have shown that for every $\varepsilon > 0$ there is a $r > 0$
such that $|z \, f(z)| < \varepsilon$ for $0 < |z| < r$.
In other words:
$$
  \lim_{z \to 0} \, z \, f(z) = 0 \, .
$$ 
This is a well-known condition and implies that $f$ has a removable singularity at
$z = 0$.

Alternatively, one can use the "Hadamard three-circle theorem"
which states that $\log M(r)$ is a convex function of $\log r$.
Using properties of convex functions, it follows from
$$
   \liminf_{r \to 0} \, r \, M(r) = 0
$$
that actually the limit exists (and has the same value):
$$
   \lim_{r \to 0} \, r \, M(r) = 0 \, .
$$
