Given a sequence $\{\boldsymbol{x}^{k}\}_{k\in\mathbb{N}} \subset \mathbb{R}^{n}$ such that it converges to $\boldsymbol{x}$, I need to find vectors $\boldsymbol{v}$ and $\boldsymbol{a}$ and sequences $\{\boldsymbol{v}^{k}\}, \{\boldsymbol{a}^{k}\}\subset \mathbb{R}^{n} $ and $\{t_k\}_{k\in\mathbb{N}}\subset \mathbb{R}^{*}_{+} = \{t \in\mathbb{R}: t>0\}$ such that $$\boldsymbol{x}^{k} = \boldsymbol{x} + t_k \boldsymbol{v}^{k} + \dfrac{t_k^2}{2}\boldsymbol{a}^{k},$$ $\boldsymbol{v}^{k}\rightarrow \boldsymbol{v}$, $\boldsymbol{a}^{k}\rightarrow \boldsymbol{a}$ and $t_k \rightarrow 0$. Note that this resembles the determination of the velocity and acceleration of a particle that passes through $\boldsymbol{x}^{k}$ over the time at the position $\boldsymbol{x}^{*}$ . Is this possible?
Motivation from optimization: Let us consider the optimization problem \begin{equation*} \begin{array}{c c} \text{minimize}_{x\in\mathbb{R}^n} & f(x) \\ \text{subject to} & \begin{aligned} x \in \Omega.\\ \end{aligned} \end{array} \end{equation*} The first order tangent cone at $x$, say $T(x)$ is given by the set of vectors $d$ such that $x+t_k d^{k} \in \Omega$ with $d^{k}\rightarrow d$ and $t_k \rightarrow 0.$ Hence, a natural definition for a second order tangent set would be the set of pairs $(v,a)$ such that $$x+t_k v^{k}+ \dfrac{t_k^{2}}{2} a^{k} \in \Omega$$ such that $v^{k} \rightarrow v$, $a^{k} \rightarrow a$ and $t_{k}\rightarrow 0.$ Now, this definition gives rise to other questions like
Does this notion of second order tangent set is enough to model a quadratic feasible set?
Solving this problem, would help me to understand better how this question could be answered.