Does this $\frac{0}{0}$ 2-dimensional limit exist? This is a updated question closely related to the one I posted several days ago in math.SE. (I've put on math.SE, but there is no answer so far.)
Thanks to Christian Blatter's answer to that question, the limit (there are 9 limits here indeed.)
$$
\lim_{y\to\xi}\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot n(y)}{|\xi-y|^5},\quad 1\leq i,j\leq 3,\tag{1}
$$
does not exist in general. Here $S\subset{\mathbb R}^3$ is a surface which has a continuously varying normal vector, $\xi=(\xi_1,\xi_2,\xi_3)\in S$, $y=(y_1,y_2,y_3)\in S$, $n(y)$ is the unit normal vector at point $y$. Here $(\xi-y)\cdot n(y)$ is the dot product.
The key point in the counterexample is that the quotient is of order $\frac{1}{|\xi-y|}$. I am interested in the following "updated" limit:
$$
\lim_{y\to\xi}\,[\psi_j(\xi)-\psi_j(y)]\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot n(y)}{|\xi-y|^5},\quad 1\leq i\leq 3,\tag{2}
$$
where the Einstein summation convention is applied for $j$ here and $\psi_j:S\to{\mathbb R}$ is assumed to be $C^{\infty}$. Or without normalization, consider the limit
$$
\lim_{y\to\xi}\,[\psi_j(\xi)-\psi_j(y)]\frac{(\xi_i-y_i)(\xi_j-y_j)({{\xi}-y})\cdot \frac{\partial y}{\partial\alpha}\times\frac{\partial y}{\partial\beta}}{|\xi-y|^5},\quad 1\leq i\leq 3,\tag{3}
$$
where $y(\alpha,\beta)=(y_i(\alpha,\beta))_{1\leq i\leq3}$ is a parameterization of $S$.
Here is the new question:

Does the updated limit (2) or (3) exist? 

Intuitively, the $[\psi_j(\xi)-\psi_j(y)]$ term may compensate the order of the numerator. In the trivial case where $S$ is a plane, the quotient is $0$. But I don't have a strategy for the general case (e.g., when $S$ is a unit ball). 
 A: In general, these limits won't exist, though they might exist for special choices of the $\psi_j$.  You can see this as follows:  By translation, you might as well assume that $\xi=0$.  Moreover, by rotation, you might as well assume that $n(0)=(0,0,1)$, and then the surface can be parametrized near $0\in\mathbb{R}^3$ in the form 
$$
y = \bigl(u,v,h(u,v)\bigr)
$$
where $h(u,v) = a\ u^2 + b\ v^2 + O(3)$ for some real constants $a$ and $b$.  Then $y\cdot n(y) = - a\ u^2 - b\ v^2 + O(3)$ and $|y| = \sqrt{(u^2 + v^2)} + O(2)$.  
Without loss of generality, you can assume that $\psi$ vanishes at $0\in S$, so $y\cdot\psi(y) = r\ u^2 + 2s\ uv + t\ v^2 + O(3)$ for some real numbers $r, s, t$, which could be any real numbers, depending on your choice of $\psi$.  So now, you are basically asking whether the limits
$$
\lim_{(u,v)\to(0,0)} \frac{u\ (r\ u^2 + 2s\ uv + t\ v^2)(a\ u^2 + b\ v^2)}{(u^2 + v^2)^{5/2}} 
$$
and 
$$
\lim_{(u,v)\to(0,0)} \frac{v\ (r\ u^2 + 2s\ uv + t\ v^2)(a\ u^2 + b\ v^2)}{(u^2 + v^2)^{5/2}} 
$$
exist, whatever the values of $a,b,r, s,t$, and it is clear that, for most such values, these limits will not exist.
