On the boundary of a region Assume $U$ is a region with $C^1$ boundary, can we claim this:
If $x\in\partial U$ then for each $\epsilon >0$ there exists $\delta >0$ such that when $|x-y|<\delta$ and $y\in\bar{U}$ imply $n(x)\cdot \frac{y-x}{|y-x|}\leq\epsilon$. $n(x)$ is the unit outward normal vector at $x$.
Thanks in advance!
 A: The assertion can be written as
$$
\lim_{y\in U, y\to x}\big(n(x)\cdot\frac{y-x}{|y-x|}\Big)\le0.
$$
Equivalently, for any limit vector $u=\lim\frac{y-x}{|y-x|}$ with $y\in U$, $y\to x$, we have $n(x)\cdot u\le 0$. But as $n(x)$ is normal outward, the last inequality means that $u$ is not outward. Now this is a differential topology matter and we can forget the normal. 
Consider a local diffeomorphism $\varphi=(\varphi_1,\dots,\varphi_n):V\to B$ from an open nbhd of $x$ in $\mathbb R^n$ onto another $W$ of the origin, such that
$$
\varphi(x)=0,\quad W\cap \overline U=\{y\in W:\varphi_n(y)\ge0\}
$$ 
(in other words, a chart of the manifold with boundary $\overline U$).
Then for $y\in W\cap \overline U$ we have (definition of differential):
$$
0\le \varphi_n(y)=\varphi_n(x)+\nabla\varphi_n(x)\cdot(y-x)+o(|y-x|).
$$
We deduce
$$
0\le\nabla\varphi_n(x)\cdot\frac{y-x}{|y-x|}+\frac{o(|y-x|)}{|y-x|},
$$
and take limits to get 
$$
0\le\nabla\varphi_n(x)\cdot u.
$$
To understand this inequality, note that the vector
$$
d_x\varphi(u)=(...,\nabla\varphi_i(x)\cdot u,...)
$$
is outward if and only if $\nabla\varphi_n(x)\cdot u<0$, hence it is not. But being outward is preserved by charts, hence $u$ is not outward. This completes the argument.
