The divergence as flux density 
Let $S_\epsilon$ be the sphere with center in $(0,0,0)$ and radius $\epsilon$, and the vectorfield $F(x,y,z) = (yx^2, xyz, x^2y^2)$. Let $\hat{N}$ be the unit-normal vector for $S_\epsilon$ pointing outwards. Evalute the limit 
  $$\lim_{\epsilon\to 0^+} \frac{3}{4\pi\epsilon^3}\oint_{S_\epsilon} F \cdot\hat{N}dS$$

I've swapped the original vectorfield in the task with a random one just to clarify if I understand this correctly. I read about this limit in my calculus book which was simply defined as the divergence (as flux density) div F. Does evaluating this limit mean to simply find the divergence, which is very easy for my vectorfield; div F $= x(2y+z)$. Or have i misunderstood the meaning of this limit and how it correlates to divergence itself?
 A: You are right that by the divergence theorem,
$$\lim_{\epsilon\to 0^+} \frac{3}{4\pi\epsilon^3}\oint_{S_\epsilon(x,y,z)} F \cdot\hat{N}dS=\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$$
However, which side of that equation is the definition of the divergence depends on the text.  It sounds like you are given the left-hand side for the definition.  This means you can't just substitute the right-hand side unless the theorem is already proved.
Since the question is asking about a specific point, it could be that you are expected to explicitly calculate the integral on the left-hand side and verify that the indicated limit is $0$ (that is, $x(2y+z)$ when $x=y=z=0$).
A: Usually one defines $\mathrm{div}F:=\partial_x F_1 + \partial_y F_2 +\partial_z F_3$ and then proves the divergence theorem:
$$
\iiint_{B_\epsilon} \mathrm{div} F ~\mathrm{d}V = \iint_{S_\epsilon}  F \cdot N~\mathrm{d}S.
$$
The left intergal is taken over a ball with center $0$ and  radius $\epsilon$, thus its volume is $V(B_\epsilon)= 4\pi\epsilon^3/3$. Consequently
$$
\frac{3}{4 \pi \epsilon^3}\iint_{S_\epsilon}  F \cdot N~\mathrm{d}S = \frac 1{V(B_\epsilon)}\iiint_{B_\epsilon} \mathrm{div} F ~\mathrm{d}V.
$$
Now the right hand side is the average value of the function $\mathrm{div}F$ over the ball $B_\epsilon$. Provided $F$ is reasonably nice (continously differentiable would be enough), these avarages then tend to $\mathrm{div}F(0,0,0)$ as $\epsilon \rightarrow 0$.
