Cauchy's formula proof: why can we say $i\int_0^{2\pi} f(z_0+\epsilon e^{i\theta})d\theta \rightarrow2\pi i f(z_0)$ as $\epsilon\to 0$ For $n=1$ the proof of Cauchy's formula 
$$f^{(n)}(z_0)={n!\over2\pi i}\int\limits_\gamma\frac{f(z)}{(z-z_0)^{n+1}}dz$$
defines $g(z)={f(z)\over z-z_0}$ given the closed path $\gamma$ around $z_0$ in the open subset $\Omega\subset\Bbb C$ and a holomorphic function $f:\Omega\mapsto\Bbb C$ with $\epsilon>0$ small enough so that $\overline{B_\epsilon(z_0)}$ is inside $\Omega$ using another theorem and since $\mathcal O=\Omega\backslash B_\epsilon(z_0)$ is an open domain and $g$ is holomorphic on it, $\int\limits_{\mathcal O}g(z)dz=0$
From that we eventually get $\int\limits_\gamma \frac{f(z)}{z-z_0}dz=i\int\limits_0^{2\pi} f(z_0+\epsilon e^{i\theta})d\theta  \xrightarrow[\epsilon\to0]{}2\pi i f(z_0)$ 
Why can we move the limit $\xrightarrow[\epsilon\to0]{}$ inside the integral?
 A: We can use Lebesgue dominated convergence theorem.
A: By the continuity of $f$, as $\epsilon\to 0$ $$f(z_0+\epsilon e^{i\theta}) \to f(z_0)$$ uniformly over $\theta\in [0,2\pi]$.
A: You want to show that: $i\int\limits_0^{2\pi} f(z_0+\epsilon e^{i\theta})d\theta \xrightarrow[\epsilon\to0]{}2\pi i f(z_0)$. That is: 
For any $\delta >0$. There is $\eta>0 $ such that $\left | \int_{0}^{2\pi}f(z_0+\epsilon e^{i\theta })d\theta  -2\pi f(z_0)\right |< \delta  $ whenever $\left | \epsilon \right |\leq \eta $.
But:
$\left | \int_{0}^{2\pi}f(z_0+\epsilon e^{i\theta })d\theta  -2\pi f(z_0)\right |=\left | \int_{0}^{2\pi}(f(z_0+\epsilon e^{i\theta })  - f(z_0))d\theta\right | \leq \int_{0}^{2\pi}\left |f(z_0+\epsilon e^{i\theta })  - f(z_0)  \right |d\theta $
Note that: $\left | z_0+\epsilon e^{i\theta }-z_0 \right |=\epsilon $
Now by continuity of $f$ at $z_0$, We can find $\eta$ such that:
$\left |f(z)  - f(z_0)  \right |\leq \frac{\delta }{2\pi}$ when $\left | z-z_0 \right |\leq \eta $
Can you take it from here?
