Suppose $\Sigma$ is a hypersurface in $R^n$, let $x$ be a fixed point on $\Sigma$ and $\nu$ be the unit normal at $x$ to the hypersurface. I would like to know what the following limit would be. $$\lim_{y\rightarrow x, y\in \Sigma} \frac{(x-y)\cdot \nu(x)}{|x-y|^2}.$$
I feel like it is the largest principal curvature at $x$. Is there any rigorious proof for it?
SKETCH of an ANSWER: Let $e_1,\dots,e_n$ be the principal directions at $x$, with corresponding principal curvatures $\kappa_1,\dots,\kappa_n$. Then we can think of $\Sigma$ locally as a graph over the tangent plane at $x$: Indeed, using the coordinates corresponding to the orthonormal basis $e_1,\dots,e_n$, it is the graph of $$f(t_1,\dots,t_n) = \tfrac12\sum \kappa_i t_i^2 + o(\|t\|^2).$$ That is, points $y$ near $x$ are of the form $$y=x+\sum t_ie_i + \left(\tfrac12\sum \kappa_it_i^2 + \dots\right)\nu(x).$$ So it looks like you're looking at the limit $$\lim_{(t_1,\dots,t_n)\to (0,\dots,0)} \frac12\frac{\sum \kappa_i t_i^2}{\sum t_i^2}.$$ This limit will not exist unless $x$ is an umbilic point, i.e., all $\kappa_i$ are equal.