Let $f$ be holomorphic on $\Omega - \{z_0\}$. Then $g$ defined by $g(z)=(z-z_0)f(z)$ for $z\in \Omega-\{z_0\}$, $g(z_0)=0$ is holomorphic on $\Omega$. Edit: Also given
$$
\lim_{z\rightarrow z_0} (z-z_0)f(z)=0
$$
It's easy to see that $g$ is holomorphic on $\Omega-\{z_0\}$, so we only need to worry about $g'(z_0)$. By definition, we have
\begin{align*}
\lim_{z\rightarrow z_0} \frac{g(z)-g(z_0)}{z-z_0} &= \lim_{z\rightarrow z_0} \frac{(z-z_0)f(z)-0}{z-z_0} \\
&= \lim_{z\rightarrow z_0} f(z).
\end{align*}
Thus, we need only to show that $\lim_{z\rightarrow z_0}f(z)$ exists. This is where I'm stuck. It makes perfect sense that if $f$ is holomorphic in a neighborhood of $z_0$, and $\lim_{z\rightarrow z_0}(z-z_0)f(z)=0$, then $f$ is continuous at $z_0$, but I'm having trouble showing this.
Thanks so much!
 A: Edit: the question changed.
Now your new assumption makes $z_0$ a removable singularity for $f$: http://en.wikipedia.org/wiki/Removable_singularity
So you can extend $f$ holomorphically over $z_0$.
In particular, $g$ is holomorphic on $\Omega$.
A: It is easy to see that $g$ is continuous.  A function that is holomorphic away from a point and continuous at that point is holomorphic.  See the question Which sets are removable for holomorphic functions? for much more along these lines.
If you look at the proof that $f$ has a removable singularity from julien's link, you'll notice that the trick is to multiply by $z-z_0$ yet again to smooth things out more.  The function $h(z)=(z-z_0)^2f(z)$, $h(z_0)=0$, is not only easily seen to be continuous, but this time using your method you also see that $h'(z_0)=0=\lim\limits_{z\to z_0}h'(z)$, so $h$ is holomorphic.  Since $h$ has a zero of order at least $2$ at $z_0$, one can see that $\dfrac{h(z)}{z-z_0}$ (your function $g$) and $\dfrac{h(z)}{(z-z_0)^2}$ are also holomorphic, e.g., using power series as in the link from julien's answer.
