If there's a line $L\subset \mathbb{R}^n$ and a sequence $t_k\to a$ where $f(t_k)\in L$, then $L$ is the tangent line at $f(a)$ 
Let $f:I\to\mathbb{R}^n$ be a differentiable path, with $f'(a)\neq 0$
  for some $a\in I$. If there is a line $L\subset \mathbb{R}^n$ and a
  sequence of distinct numbers such that $t_k\to a$ such that $f(t_k)\in
 L$, then the line $L$ is the tangent line to $f$ at the point $f(a)$

First, the tangent line $L$ at the point $a$ is just the line that has 'slope' $f'(a)$ and passes at the point $f(a)$, right?
So I need to prove that
$$t_k\to a \implies f(t_k)\in L \implies \lim_{t\to 0}\frac{f(a+t)-f(a)}{t} = L$$
?
Intuitively, I see that $t_k\to a$ implying $f(t_k)\in L$ means that, no matter which side we approximate, if we're getting closer and closer to $a$, then $f(t_k)$ is in $L$. I guess the other definition of limit would help me better:
$$L = \lim_{t\to a}\frac{f(t)-f(a)}{t-a}$$
But how to show that 
$$t_k\to a \implies f(t_k)\in L \implies \lim_{t\to 0}\frac{f(a+t)-f(a)}{t} = L$$
implies in the limit definition above?
 A: Note that you cannot write $\lim_{t \to 0} \frac{f(a + t) - f(a)}{t} = L$ because the left hand side is a vector while the right hand side is a line (which is a subset). In general, a line $L$ in $\mathbb{R}^n$ can be described as the set
$$ L = \{ p_0 + tv \, | \, s \in (-\infty, \infty) \} $$
where $p_0 \in \mathbb{R}^n$ is some point on the line and $v \neq 0$ is the direction vector of $L$. By assumption, you know that there exists $s_k$ such that
$$ f(t_k) = p_0 + s_k v $$
(this is what it means for $f(t_k)$ to lie on the line $L$). Now, $f$ is continuous so we must have
$$ f(t_k) = p_0 + s_k v \to f(a) $$
which means that $s_k v \to f(a) - p_0$. Using $v \neq 0$, show that this actually implies that $s_k \to s$ for some $s \in \mathbb{R}$ so we have $sv = f(a) - p_0$ which shows that $f(a) \in L$. At this point, we can replace $p_0$ with $f(a)$ and assume that $f(t_k) = f(a) + s_k v$ for some $s_k \in \mathbb{R}$ with $s_k \to 0$.
Finally,
$$ f'(a) = \lim_{k \to \infty} \frac{f(t_k) - f(a)}{t_k - a} = \lim_{k \to \infty} \left( \frac{s_k}{t_k - a} \right)v.$$
Show like before that this implies that $\lim_{k \to \infty} \frac{s_k}{t_k - a}$ exists and deduce that $v$ and $f'(a)$ are parallel which shows that $L$ is the tangent line to $f$ at the point $f(a)$.
A: We have by hypothesis* that
$$f(t_k)=f(a)+v_k$$
for some $v_k $ belonging to a $1$-dimensional subspace $L'$ of $\mathbb{R}^n$ (the line $L$ translated to the origin). It follows that
$$\frac{f(t_k)-f(a)}{t_k-a}=\frac{v_k}{t_k-a}.$$
Since the limit of the LHS exists, the limit of the RHS also exists, say it is $v$. Since $L'$ is closed, $v$ must belong to $L'$. Therefore, $f'(a)=v$, where $v \in L'$, which is what we wanted to prove.
*This follows since a line is closed, and therefore $f(a)$ belongs to the line.
