# 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.

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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 $p_n = f(t_n)$.

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+1 No ill will...Besides, our posts were/are good! ;) –  amWhy Jan 26 '13 at 0:47
@Guillermo: I have been on the other end of a similar situation, where my answer, arrived at independently, looked very similar to another answer that was posted just before mine. When answers are very close, voters will often go with the earlier answer. If a pattern of abuse forms (which I seriously doubt), flag us again. –  robjohn Jan 26 '13 at 0:49
Why do we write $t$ approaching to $0$ from the right side? –  Anne Feb 9 '13 at 19:40
so that when I write $\|(f(t)-p)/t\|$ I don't have to deal with the sign of $t$. –  William Feb 11 '13 at 3:32