If $f(z)$ is continuous in the neighborhood of the point $z=a$, then $$\lim_{r\to 0^+} \int_{\vert z-a\vert =r} \frac{f(z) dz}{z-a}=2\pi i\,f(a)\tag{1}$$
In my desperate intents of proving the above exercise, I tried to prove a 'particular' case of it, which is when $z=0$.
So if $z=0$ and as $f$ is continuous on a neighborhood of zero, given $\epsilon>0$, there exists $\delta>0$ such that $|f(re^{i \theta})-f(0)|\le \epsilon$ for $r\in[0,\delta)$ and $\theta\in\mathbb R$. Hence,
$\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| <2\pi\epsilon \Rightarrow \limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| \le2\pi\epsilon.$
If $\epsilon\to0$, then $\limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right|=0 \Rightarrow \lim_{r\to0}\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta=0.$
Is this proof correct?
How can I make the proof for $(1)$ following the 'particular proof'? Can someone help me?
Your idea is nice. You can extended it to the general case by a change of variables as follows: \begin{align} \int_{|z-a|=r}\frac{f(z)}{z-a}\,dz&=\int^{2\pi}_0 \frac{f(a+re^{i\theta})}{re^{i\theta}}ire^{i\theta}\,d\theta\\ &=\int_{|z|=r}\frac{f(a+z)}{z}\,dz \end{align} Since the function $g(z)=f(z+a)$ is continuous around $z=0$, we use what you have proved and get: \begin{align} \lim_{r\to 0^+}\int_{|z|=r}\frac{g(z)}{z}\,dz=2\pi ig(0)=2\pi i f(a) \end{align} And therefore the claim follows.