Assuming that $g(z)=\frac{f(z)}{z-z_0}$ is continuous at $z_0$, prove that $\int_\gamma \frac{f(z)}{z-z_0}dz=0$. Suppose that $f$ is analytic in some domain D containing the unit circle $\gamma$ and that $z_0$ is a point in D not on $\gamma
$. Assuming that $g(z)=\frac{f(z)}{z-z_0}$ is continuous at $z_0$, prove that $$\int_\gamma  \frac{f(z)}{z-z_0}dz=0.$$
This is where I am so far, we know that from Cauchy's Integral Formula $$\int_\gamma  \frac{f(z)}{z-z_0}dz=2\pi if(z_0)$$ This is equal to $0$ if $f(z_0)=0$. To show that this is equal to $0$, this must have to do something with $g(z)$ being continuous at $z_0$. I know that $g(z)$ is analytic if $f(z_0)=0$, however I cannot show that continuity of $g(z)$ implies analyticity of $g(z)$.
Any comments will be helpful.
 A: You know that $\lim_{z\to z_0}\frac{f(z)}{z-z_0}$ exist. Also, as $f$ is analytic, $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists. Thus, the limit $\lim_{z\to z_0}\frac{f(z_0)}{z-z_0}$ exists. But this can only be true when $f(z_0) = 0$.
A: Since $\lim_{z\to z_0}\frac{f(z)}{z-z_0}$ exists, $f(z_0)=0$. Otherwise, we would have $\lim_{z\to z_0}f(z)\neq0$ and $\lim_{z\to z_0}z-z_0=0$, which would imply that the limit $\lim_{z\to z_0}\frac{f(z)}{z-z_0}$ does not exist (in $\mathbb C$). So$$\int_\gamma\frac{f(z)}{z-z_0}\mathrm dz=2\pi if(z_0)=0.$$
A: $g$ has an isolated singularity at $z_0$. Since $g$ is continuous at $z_0$, Riemann says:
$g$ has aremovable singularity at $z_0$. Hence there is an analytic function $h$ on $D$ such that $h=g$ on $D \setminus \{z_0\}$.
Consequence: $\int_\gamma  \frac{f(z)}{z-z_0} dz=\int_\gamma  g(z) dz=\int_\gamma  h(z) dz=0$.
A: Since $g(z)=\frac{f(z)}{z-z_0}$ is given to be continuous at $z_0$. Then 
\begin{eqnarray*}
g(z_0)&=&\lim_{z \to z_0}\frac{f(z)}{z-z_0}\\
&=&\frac{\lim_{z \to z_0}f(z)}{\lim_{z \to z_0} (z-z_0)} \\
\lim_{z \to z_0}f(z)&=&g(z_0) \lim_{z \to z_0}(z-z_0)\\
f(z_0)&=&0
\end{eqnarray*}
