I am struggling to understand the definition of tangent vector.
$\textbf{Definition.}$ Let $\bar{x}\in Q \subseteq R^n$. A vector $v\in R^n$ is tangent to $Q$ at $\bar{x}\in Q$ if there exists sequences $\{x_k\}\subseteq Q$ and $\{t_k\} \subset R_+$ such that $$x_k\rightarrow \bar{x},\ t_k\downarrow 0,\ \frac{1}{t_k}(x_k-\bar{x})\rightarrow v.$$
$\textbf{Question.}$ This is the first time I've seen such a definition for a "tangent" vector, as I always thought that a tangent vector means a vector that touches a curve at exactly one point. But that aside, what really confuses me about this definition is that it seems to me that any vector $v$ can be tangent since as long as $$x_k\rightarrow \bar{x},\ t_k\downarrow 0,$$ we have $$x_k\rightarrow \bar{x},\ t_kv\downarrow 0$$ and so $$x_k \rightarrow \bar{x} + t_kv.$$
$\textbf{Edit.}$ Does the definition imply that $v$ is a tangent vector if $\exists \epsilon>0$ s.t. $\bar{x}+\epsilon v \in Q$?
Your argument is almost perfectly correct, but at the last step, you've constructed a sequence $x_k$ that converges to $x$ as required, adn whose differences from $x$ all pooint in the right direction, etc.
But you've missed the important requirement on the $x_i$s, namely, that each $x_i$ be an element of $Q$.
Consider the case where $Q$ is the x-axis of the plane, and $v$ is the vertical vector $(0, 1)$, $t_k = 1/k$, and work through the details of what you've said to see why you have NOT proved that $v$ is tangent to $Q$.