Q:

Solving a PDE problem

I have an analytical problem: I need to prove the following
$$\lim_{\varepsilon\rightarrow0}\frac{i}{2\pi}\int_{\partial B_{x_i}(\varepsilon)}\frac{\partial u}{\partial z}dz=0 ;$$
where $u$ is a continuous function satisfies the following condition
$$ |z|\frac{\partial u}{\partial z}\longrightarrow 0, \quad z\rightarrow 0$$ 
and $x_i$ it's just a point.
I obtain this integral in the following way: let $v:S\rightarrow \mathbb{R}$ a function, defined over a compact surface $S$ without boundary, of the form $v=\beta_i log|z|+u$ where $z$ is the local holomorphic coordinate around the point $x_i$. Consider now $d*dv$, where $*$ is the Hodge star; $$i.e. *dv=-i\frac{\partial v}{\partial z}dz+i\frac{\partial v}{\partial \overline{z}}d\overline{z}.$$
Let $B_{x_i}(\varepsilon)$ a disk of radios $\varepsilon$ around $x_i$ and consider $S_{\varepsilon}=S- B_{x_i}(\varepsilon)$.
CLAIM: $$-\frac{1}{2\pi}\int\int_{S}d*dv=\beta_i ;$$
to prove this we can use the Green formula and we have
$$-\frac{1}{2\pi}\int\int_{S_{\varepsilon}}d*dv=\frac{1}{2\pi}\int_{\partial B_{x_i}(\varepsilon)}*dv$$
so $$\frac{1}{2\pi}\int_{\partial B_{x_i}(\varepsilon)}*dv=\frac{\beta_i}{2\pi}\int_{\partial B_{x_i}(\varepsilon)}*dlog|z|+\frac{1}{2\pi}\int_{\partial B_{x_i}(\varepsilon)}*du.$$
The first integral is $\beta_i$, so if I prove that the second integral converge to $0$ when $\varepsilon$ go to $0$ I have done.
Rmk: $$*du=-i\frac{\partial u}{\partial z}dz+i\frac{\partial u}{\partial \overline{z}}d\overline{z}.$$
I have in mind an idea to solve it, but I have some problems to formalize it correctly. Could anyone help me? Thanks you a lot!

A:

The statement to be proved is false as stated (or, if it's true, then it holds for every differentiable $u$). Either the condition 
$$ 
|z|\frac{\partial u}{\partial z}\longrightarrow 0, \quad z\rightarrow 0
$$ 
is not exactly what we want, or we are meant to assume that $x_i$ corresponds to $z= 0$ in this context.  I will assume the latter. (NOTE: OP meant the latter, per comments below).
Hint: For any $C>0$, we may state that for sufficiently small $\epsilon$,
$$
\left|\int_{\partial B(\epsilon)} \frac{\partial u}{\partial z}\,dz\right| \leq
\int_{\partial B(\epsilon)} \frac {C}{|z|} \,dz = |\partial B(\epsilon)| \cdot C \cdot \frac 1\epsilon
$$
Where $|\partial B(\epsilon)|$ denotes the "surface area" of the ball.

