Suppose that $f$ is holomorphic on $U\setminus\{a\}$ and $\lim_{z\to a}(z-a)f(z)=0$ . Let $$ f(z)=\sum_{k=-\infty}^{\infty}c_k (z-a)^k $$ be the Laurent series of $f$ centered at $a$ . We will show that $c_k=0$ for $k<0$ , so that $f$ can be holomorphically extended to all of $U$ by defining $f(a)=c_0$ .

For any non-negative integer $n$ , the residue of $(z-a)^n f(z)$ at $a$ is $$ \operatorname{Res}((z-a)^n f(z),a)=\frac{1}{2\pi i} \lim_{\delta\to 0^+}\oint_{|z-a|=\delta}(z-a)^n f(z)\mathrm{d}z. $$ This is equal to zero, because \begin{eqnarray*} \left|\oint_{|z-a|=\delta}(z-a)^n f(z)\mathrm{d}z\right| &\le&2\pi\delta\max_{|z-a|=\delta}|(z-a)^n f(z)|\\ &=&2\pi\delta^n\max_{|z-a|=\delta}|(z-a)f(z)| \end{eqnarray*}which, by our assumption, goes to zero as $\delta\to 0$ . Since the residue of $(z-a)^n f(z)$ at $a$ is also equal to $c_{-n-1}$ , the coefficients of all negative powers of $z$ in the Laurent series vanish.

Conversely, if $a$ is a removable singularity of $f$ , then $f$ can be expanded in a power series centered at $a$ , so that $$ \lim_{z\to a}(z-a)f(z)=0 $$ because the constant term in the power series of $(z-a)f(z)$ is zero.

A corollary of this theorem is the following: if $f$ is bounded near $a$ , then $$ |(z-a)f(z)|\le|z-a|M $$ for some $M>0$ . This implies that $(z-a)f(z)\to 0$ as $z\to a$ , so $a$ is a removable singularity of $f$ .