Let $z_0\in \Bbb C$ and $R>0$. Let $f:B(z_0,R)\to\Bbb C$ be a complex function such that $f=u+iv$ (where $B(z_0,R)$ is the open disc centered at $z_0$ with radius $R$). If $u$ and $v$ have continuous partial derivatives then prove that $$\lim\limits_{r\to 0}\frac{1}{r^2}\int\limits_{C_r}{f(z)dz}=2\pi i\frac{\partial f}{\partial \bar{z}}(z_0),$$ where $0<r<R$ and $C_r$ is the circle $|z-z_0|=r$ oriented counter-clockwise.
I have proved it using Green's theorem. I wanted to know if there is any proof of this without using the Green's theorem.
Take $z = z_0 + r e^{i \theta}$ with $\theta \in [0,2\pi]$, since the function has continuous derivatives we can approximate
$$ f(z) \approx f(z_0) + re^{i\theta}\frac{\partial f (z_0)}{\partial z} + re^{-i\theta} \frac{\partial f (z_0)}{\partial \bar{z}} $$ Using $dz = i r e^{i\theta} d\theta$ in the integral above we see that only the third term in the r.h.s. above survives integration, leading to the desired result.