Prove that $ z_0$ is removable By estimating the coefficients of the Laurent Series, prove that if $z_0$ is an isolated
singularity of $f$, and if $(z−z_0)f(z)\rightarrow 0$ as $z\rightarrow z_0$, then $z_0$ is removable.
My attempt
$\displaystyle f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\ldots+\frac{b_n}{(z-z_0)^n}+\ldots$ 
in a punctured disk $0<|z-z_0|<R_2$.
So, $\displaystyle(z-z_0)f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^{n+1}+b_1+\frac{b_2}{z-z_0}+\ldots+\frac{b_n}{(z-z_0)^{n-1}}+\ldots$
in $0<|z-z_0|<R_2$.
How do I proceed?
 A: You are given that $f$ is analytic in some punctured neighbourhood of $z_0$,
hence $f$ has a Laurent expansion, and the coefficients are given by
$a_n = { 1\over 2 \pi i} \int_{\gamma_R} {f(z) \over (z-z_0)^{n+1}} dz$ for
some circle $\gamma_R$ around $z_0$ of sufficiently small radius $R$.
Since $\lim_{z \to z_0} (z-z_0) f(z) \to 0$, we have $\lim_{z \to z_0} (z-z_0)^k f(z) \to 0$ for $k \ge 1$.
Let $0 < \rho < R$, then if $k \ge 1$, then 
$a_{-k-1} = { 1\over 2 \pi i} \int_{\gamma_\rho} {(z-z_0)^k f(z)} dz$, and
$|a_{-k-1}| \le \rho \sup_{|z-z_0|=\rho} |(z-z_0)^k f(z)| \le R \sup_{|z-z_0|=\rho} |(z-z_0)^k f(z)|$, and hence
we have $a_{-k-1} = 0$ for $k = 1,2,...$.
Hence $f$ has the form $f(z) = {a_{-1} \over z-z_0} +\sum_{n=0}^\infty a_n (z-z_0)^n$, and the convergence is uniform on compact subsets of $B(z_0,R) \setminus \{z_0\}$. It follows that $g(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$, converges uniformly on compact subsets of $B(z_0,R)$, and hence is analytic.
Since $\lim_{z \to z_0} (z-z_0) f(z) = a_{-1}$, we see that
$a_{-1} = 0$, and hence $f=g$ on $B(z_0,R) \setminus \{z_0\}$
and so  $z_0$ is a removable singularity.
A: The assumptions, as well as the work you have done, show that $z_0$ is a pole of order zero, which is equivalent to being a removable singularity.
https://en.wikipedia.org/wiki/Pole_(complex_analysis)
hence you are already done
