Show that $\lim\limits_{n\to\infty}\left\langle N,\frac{p_{n}-p}{|p_{n}-p|}\right\rangle =0$ on a regular surface I am trying to do the following question, because I think it is very interesting as it formalises the idea that the tangent space is practically indistinguishable from the surface if one looks in a very small neighbourhood of a point. The question is from "differential Geometry of Curves and Surfaces" by Tapp. 
My first idea was to show that $\lim\limits_{n\to\infty}\frac{p_{n}-p}{|p_{n}-p|}=:v\in T_{p}S$ but I don't see why/if this limit exists.
I would be very greatful for some hints or even a solution.
Thanks a lot in advance!:) 
Definition of regular surface:

I tried to use $p_{n}-p=\sigma(x_{n})-\sigma(x))=d_{p}\sigma(x_{n}-x)+o(|x_{n}-x|)$ to rewrite $\langle N,\frac{p_{n}-p}{|p_{n}-p|}\rangle$ 

 A: It may very well be that $\lim_{n \to \infty} \frac{p_n-p}{|p_n-p|}$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_{n_i}$ of $p_n$.
First, taking a parametrization $\varphi$ near $p$, we have that $p_{n_i}$ is in that chart for $i$ sufficiently big. Letting $q_{n_i}=\varphi^{-1}(p_{n_i})$, we know that
\begin{align}
\frac{p_{n_i}-p}{|p_{n_i}-p|}&=\frac{\varphi(q_{n_i})-\varphi(q)-d\varphi(q_{n_i}-q)+d\varphi(q_{n_i}-q)}{|q_{n_i}-q|}\cdot \frac{|p_{n_i}-p|}{|q_{n_i}-q|} \\
&=\frac{\varphi(q_{n_i})-\varphi(q)-d\varphi(q_{n_i}-q)}{|q_{n_i}-q|} +d\varphi\left(\frac{q_{n_i}-q}{|q_{n_i}-q|}\right)\cdot \frac{|p_{n_i}-p|}{|q_{n_i}-q|}.
\end{align}
By passing to subsequences yet again (twice), this converges to 
$K\cdot d\varphi(h)$ for some $h \in \mathbb{R}^2$, $K \in \mathbb{R}$. Therefore, the limit of this subsequence $\frac{p_{n_{i_j}}-p}{|p_{n_{i_j}}-p|}$ of $p_{n_i}$ belongs to $T_pS$ and it follows that
$$\lim_{j \to \infty} \left\langle N, \frac{p_{n_{i_j}}-p}{|p_{n_{i_j}}-p|}\right\rangle=0.$$
Letting $a_n=\left\langle N, \frac{p_n-p}{|p_n-p|}\right\rangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
