This is my answer to a question that was asked on math.SE, by someone who needed clarification on the limit-integral definition of the divergence operator. There is a lot to talk about, so I thought I would post a complete version of it here.
Let's put it this way. Suppose you have defined the divergence as a differential operator: that is, for a general function $\mathbf f : A \to \mathbb R^n$, where $A$ is a subset of $\mathbb R^n$, and for $\mathbf x_0 \in A \setminus \partial A$ at which $\mathbf f$ is differentiable, let the divergence of $\mathbf f$ at $\mathbf x_0$ be defined as the real number $$\operatorname{div} \mathbf f (\mathbf x_0) := \operatorname{tr} \mathsf D^1\mathbf f(\mathbf x_0) = \sum_{i=1}^n \frac{\partial f_i}{\partial x_i}(\mathbf x_0), \tag1$$ where $\mathsf D^1\mathbf f(\mathbf x_0)$ is the Jacobian matrix of $\mathbf f$ at $\mathbf x_0$ and $\operatorname{tr}$ indicates the trace operator.
Then the following theorem holds:
