Prove that derivative is tangent to a set This is my question.
Let $p$ be a limit point of a set $ A \subseteq \mathbb{R}^k$. Let $f: (-\delta, \delta) \rightarrow \mathbb{R}^k $ be a function which is differentiable in $0$, $\delta > 0$, for $t>0$, $f(t) \in A$ and $f(0)=p$. Thesis: $f^{'}(0) $ is tangent to $A$ in point $p$.
Well, the definition of a tangent vector $v$ is: $v = t\lim_{n \to \infty} \frac{p_n-p}{||p_n-p||} $ (where $p_n \to p$) and I think this could be done by definition, but how? Please, help me. 

If you call $v = f'(0)$ then you have
$$
v = \lim_{t \to 0^+} \frac{f(t)-p}{t} = \lim_{t \to 0^+} \frac{f(t)-p}{\|f(t)-p\|}\Bigl\|\frac{f(t)-p}{t}\Bigr\| = |v|\lim_{t \to 0} \frac{f(t)-p}{\|f(t)-p\|}
$$
You can choose any sequence $t_n$ that tends to $0$ and then $p_n = f(t_n)$.
