I wanted to calculate a simple example for the integral representation of the divergence $$\vec\nabla\cdot\vec{A}=\lim\limits_{\Delta V \rightarrow 0}{\frac{1}{\Delta V}}\iint_{\partial(\Delta V)} \vec{A}\cdot d\vec{F}$$ with $\Delta V$ being an infinitesimally small volume.
Thus, I assumed a radial vector field $\vec{A}(\vec{r})=A(r)\vec{e}_r$ and the $\Delta V$ around $0$. The RHS gives you $\lim\limits_{r \rightarrow 0}{\frac{3}{r}}A(r)$, the LHS yields $\left(\frac{2}{r}+\partial_r\right)A(r)$, thus we have $$\lim\limits_{r \rightarrow 0}{\partial_r A(r)}=\lim\limits_{r \rightarrow 0}{\frac{A(r)}{r}}.$$ How can I see that both sides are equal?
One can do the following: $$ \lim\limits_{r \rightarrow 0}{\partial_r A(r)}=\lim\limits_{r \rightarrow 0}{\left(\frac{A(r)}{r}-\frac{A(0)}{r}+\frac{A(0)}{r}\right)}\\ \lim\limits_{r \rightarrow 0}{\partial_r A(r)}=\lim\limits_{r \rightarrow 0}{\left(\partial_r A(r)+\frac{A(0)}{r}\right)} $$ This tells us that $A(0)$ has to be zero. This makes sense one should remember that we started in local coordinates and thus a finite value of the vector field in the origin would end up with an indefinite direction in the coordinate center.