Tangent line to a curve statement I am having problems understanding some parts of the proof of some statement related to tangent line to a curve. I'll copy the exact statement and proof and then my doubts.
Statement
If $\mathcal C$ is a curve that admits a parametrization $\gamma:[a,b] \to \mathbb R^3$ with $\gamma \in  C[a,b]$, inyective, differentiable at $t_0 \in [a,b]$ such that $\gamma'(t_0):=(x'(t_0),y'(t_0),z'(t_0)) \neq (0,0,0)$, it follows that $\mathcal C$ has a tangent line at the point $P_0=\gamma(t_0)$ and that this line has the direction of the vector $\gamma'(t_0)$. It is easy to show that $t_n \to t_0$.
Proof
Let $P_n \in \mathcal C$ with $P_n \to P_0$. We denote $S_{P_n,P_0}$ to the secant line through $P_n$ and $P_0$. We have that for every $n$ there is a unique $t_n \in [a,b]$ such that $P_n=\gamma(t_n)$. Analogously, there is a unique point $t_0 \in [a,b]$ with $\gamma(t_0)=P_0$.
Let's show that $S_{P_n,P_0}$ tends to $L_{P_0}$, the line containing the point $P_0$ with direction $\gamma'(t_0)$. In effect, $S_{P_n,P_0}$ has the direction of the vector $$\gamma(t_n)-\gamma(t_0)$$.
It also generates the same line the vector $$\tau_n :=\dfrac{1}{t_n-t_0}(\gamma(t_n)-\gamma(t_0)),$$
which converges to $\gamma'(t_0)$
As $\gamma'(t_0) \neq (0,0,0)$, we have $\dfrac{\tau_n}{||\tau_n||} \to \dfrac{\gamma'(t_0)}{||\gamma'(t_0)||}$, which means the angle between  $S_{P_n,P_0}$ and $L_{P_0}$ tends to $0$ when $n \to \infty$. By definition of tangent line, $L_{P_0}$ is the tangent line of $\mathcal C$ at the point $P_0$.
Questions
Now, my basic question is: why is it that $\dfrac{\tau_n}{||\tau_n||} \to \dfrac{\gamma'(t_0)}{||\gamma'(t_0)||}$ implies the angle between $S_{P_n,P_0}$ and $L_{P_0}$ tends to $0$? Is it because  $\dfrac{\tau_n}{||\tau_n||}$ and $\dfrac{\gamma'(t_0)}{||\gamma'(t_0)||}$ are the slopes of $S_{P_n,P_0}$ and $L_{P_0}$ respectively? Why is it necessary to divide by the norms of this vectors?, I mean, couldn't we arrived to the same conclusion just by the fact that $\tau_n \to \gamma'(t_0)$?
I hope my doubts are clear and I would appreciate if someone could clarify this for me.
 A: The missing piece of the argument follows from this statement:
Claim. Let $v,w$ be two unit vectors that subtend an angle $\theta$. Then $v\to w$ if and only if $\theta \to 0$.
Proof.  Recall that $\cos\theta = v\cdot w$. 
($\Rightarrow$): If $v\to w$, then $\cos\theta \to w\cdot w=\|w\|^2=1$. Thus $\theta \to \{0,\pi\}$. We can rule out the case $\theta=\pi$, since then $v\to -w$. Thus $\theta \to 0$. 
($\Leftarrow$): If $\theta\to 0$, then $v\cdot w\to 1$. Hence we compute
$$
\|v-w\|^2=\|v\|^2-2v\cdot w+\|w\|^2\to 1-2+1=0,
$$
which means $v\to w$.

Given this statement, we can answer both of your questions. The first question is answered directly by the claim. To answer the second question, note that the claim only applies to unit vectors. Thus we have to divide by their lengths.
Here's a counterexample of what happens when you don't divide by the length:
$$
v=\langle t,0,0\rangle,\qquad w=\langle 0,-t,0\rangle.
$$
Note that $v\cdot w=0$, so the vectors are always perpendicular (so their angle is always $\pi/2$). However, as $t\to 0$, both vectors tend to $\langle 0,0,0\rangle$.
