Proof that the divergence is the limit of the ratio of the surface integral and the volume

Question

Let us assume that I have a series of compact smooth sets $A_n$ that converges to the set with a single point $\{x\}$. For any given $\epsilon$ there is an index in the sequence from which on all the sets are in the open neighbourhood of $x$ of size $\epsilon$. I am currently working on a proof that the divergence in a point for the function $F$, which is continuously differentiable, is the limit of the the divergence on the surface of $A_n$ divided by the volume of $A_n$. I tried to model $\lim_{n \to \infty} \lambda^d (A_n)^{-1} \int_{A_n} div F(x) d \lambda^d (x)$ as a difference quotient, but that does not seem to work out because then I would have infinitly-dimensional function (taking a set and yielding a scalar). Does anybody have a hint how I could continue?