I've been solving problems from my Complex Analysis course, and I am having trouble solving this one. It says:
Prove that if $f$ is continuous in a neighbourhood of point $a$, then $$\lim_{r\to0}\int_{|z-a|=r}\frac{f(z)}{z-a}dz=2\pi if(a)$$
I know this should be trivial knowing the Cauchy's Formula (because $\text{Ind}_{|z-a|=r}(a)=1$). The problem is that this exercise in my course notes is located before the Cauchy's Formula is introduced, so there must be a way to solve it without using it.
The only two things I see in my course notes that may be useful are the definition of winding number: $$\text{Ind}_\gamma(a)=\frac{1}{2\pi i}\int_\gamma\frac{dz}{z-a},$$ and the fact that, given a path $\gamma$ and a function $f:\text{supp}(\gamma)\to\mathbb C$ that is continuous, then te function $F:\mathbb C\setminus\text{supp}(\gamma)\to\mathbb C$ defined as $$F(a)=\int_\gamma\frac{f(z)}{z-a}dz,\quad a\notin\text{supp}(\gamma)$$ is analytical.
However I don't see how can it be useful. Also, the statement says that $f$ is continuous in a neighbourhood of point $a$, and I guess that may be needed to apply whatever it's used to solve it. I feel like this problem should be so trivial but pretending I don't know Cauchy's Formula makes it seem impossible to me.
How can I solve this not using Cauchy's Formula? Are those two things that I wrote useful in this problem? Any help will be appreciated, thanks in advance.
Hint: Consider $\int_{|z-a|=r} \frac {f(z)-f(a)} {z-a}dz$. Use continuity of $f$ at $a$ to show that this tends to $0$ as $ r \to 0$. Now split this integral into two parts $\int_{|z-a|=r} \frac {f(z)} {z-a}dz$ and $\int_{|z-a|=r} \frac {f(a)} {z-a}dz$. Pull out $f(a)$ from the second integral.