Let $f$ be a continuous function on $D(0,1)$ and $\mathbb{C}$-differentiable at $0$.
I want to prove that $$\lim_{r \rightarrow 0}\frac{1}{r^2}\int_{C(0,r)}f(z)dz=0$$
My idea is to use polar coordinates
we'll have : $\frac{1}{r^2}\int_{C(0,r)}f(z)dz=\frac{i}{r}\int_0^{2\pi}f(r\exp(i\theta))\exp(i \theta)d\theta$
Just note that$\frac i r \int_0^{2\pi} f(re^{i\theta})e^{i\theta}d\theta=i\int_0^{2\pi} [f(re^{i\theta})-f(0)] /[re^{i\theta}] e^{2i\theta}d\theta$ (because $\int_0^{2\pi} f(0)e^{i\theta}d\theta=0$). As $ r \to 0$ this converges to $i\int_0^{2\pi} f'(0)e^{2i\theta}d\theta=0$. I will leave it to you to justify taking the limit inside the integral.