# 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 $p_n = f(t_n)$.
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