and a direction vector $v$ in $V$, there exists a continuous curve $c\colon[0,+\infty)\to C$ such that $c(0)=p$ and $$\left.\frac{dc(t)}{d t}\right|_{t=0+}=v$$ if, and only if, $v$ is in the sequential closure of the tangent cone of $C$ at $p$.

To prove the “if” part, let $v_1,v_2,\dots$ be a sequence of vectors in the tangent cone to $C$ at $p$ that converges to $v$, $v_i\to v$. For $i=1,2,\dots$, let $\varepsilon_i>0$ be such that $p+\varepsilon_iv_i$ is contained in $C$, and also $\varepsilon_i\to0$ as $i\to +\infty$. Take a sequence $t_1,t_2,\dots$ of positive numbers such that $0<t_{i+1}<t_i{\leqslant}\varepsilon_i$. Define $p_j$ to be equal to $p+t_jv_j$. Note that $p_j\to p$ and $t_j\to 0$ as $j\to+\infty$. To define the curve $c$, first set $c(t_j)=p_j$ and then on the intervals $(t_{j+1},t_j)$ do a linear interpolation between those values . Also set $c(0)=p$; this renders $c$ continuous on $[0,t_1]$. For simplicity, we let $c(t)=p_1$ for $t{\geqslant}t_1$. Then $c$ is continuous, and $c(t)$ is in $C$ for all $t>0$ because of the convexity of $C$ and the fact that $p$ and $p_i$, $i=1,2,\dots$, are all contained in $C$. Also, we have $$\lim_{i\to+\infty}\frac{p_i-p}{t_i}=\lim_{i\to+\infty}v_j=v.$$ Since $c$ is a linear interpolation between these points, $$\lim_{t\to0+}\frac{c(t)-p}{t}=\lim_{i\to+\infty}\frac{p_i-p}{t_i}=v.$$ Thus the curve $c$ exists.

To prove the converse 