How to prove $\mathbf{M}/r$ is a continuously differentiable vector field? I had previously asked a question here regarding the applicability of divergence theorem.
$\mathbf{M'}$ is a continuous vector field in volume $V'$ (which is compact and has a piecewise smooth boundary $S$). Consider the following integral:
$$\int_V \nabla . \left( \dfrac{\mathbf{M}}{r} \right) dV$$
with $(r=0) \in V$.
The answer says that, since the integrand is infinite at $(r=0)$, if we redefine our integral as:
$$\lim \limits_{\delta \to 0} \int_{V-\delta} \nabla . \left( \dfrac{\mathbf{M}}{r} \right) dV$$
where $\delta$ is a volume surrounding the point $(r=0)$
then we can apply $GDT$.
But that is only condition. How can we ensure $\dfrac{\mathbf{M}}{r}$ is a continuously differentiable vector field?

Edit:
Comment : "Is $\mathbf{M}$ continuously differentiable?"
Maybe... But break it down into two cases: $(i)$ $\mathbf{M}$ is continuously differentiable   $(ii)$ $\mathbf{M}$ is continuous

Please show whether $\dfrac{\mathbf{M}}{r}$ is a continuously differentiable vector field in the two cases.
 A: We have $r = \|\mathbf{x}\|$ and the Euclidean norm is a continuous function from $\mathbb{R}^3$ into $\mathbb{R}$ since by the reverse triangle inequality
$$|\, \|\mathbf{x}\| - \|\mathbf{x}_0\|\,| \leqslant \|\mathbf{x} - \mathbf{x}_0\|.$$
Hence, $\mathbf{x} \mapsto \frac{1}{r}$ is continuous in any region that excludes the origin.  You can show this with an $\epsilon - \delta$ argument, or just quote the theorem that continuity of $f$ with domain in $\mathbb{R}$ implies continuity of $1/f$ on any set where $f(x) \neq 0$.
Since $\mathbf{M}$ is continuous, so too is $\mathbf{x} \mapsto \frac{\mathbf{M}(\mathbf{x})}{r}$, since with $r_0 = \mathbf{x}_0$ we have
$$\left\|\frac{\mathbf{M}(\mathbf{x})}{r} -  \frac{\mathbf{M}(\mathbf{x}_0)}{r_0}\right\| \leqslant \frac{1}{r}\|\mathbf{M}(\mathbf{x})-\mathbf{M}(\mathbf{x}_0)\|+ \|\mathbf{M}(\mathbf{x}_0)\|\left|\frac{1}{r} - \frac{1}{r_0} \right|,$$
and $\mathbf{M}(\mathbf{x}_0)$ and $1/r$ are bounded in any compact neighborhood of $\mathbf{x}_0$.
For continuous differentiability you need $\mathbf{M}$ to be continuously differentiable unless some removable discontinuity arises. It remains to show  that $\mathbf{x} \mapsto \frac{1}{r}$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above. 
