Let $K \subset \mathbb{R}^n$ be a closed convex set. Given a point $u \in K$, the tangent cone of $K$ at $u$ is defined as (or characterized by) $$ T_K(u) := \mathrm{cl}(\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}). $$

My question is about the cases where $K$ is a closed convex **cone**.
The equation right after Eq. (2.1) of this paper says that, if $K$ is a closed convex cone, we have
$$
T_K(u) = \left\{ v - tu \mid v \in K, t \geq 0 \right\}.
$$
I've tried to prove this identity, but I'm having trouble with the last step of the proof. So can anyone help me figure out the right way?

### What I've tried so far:

The inclusion $\left\{ v - tu \mid v \in K, t \geq 0 \right\} \subset \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset T_K(u)$ is easily proven with some algebra, so I will prove the opposite inclusion. It is straightforward to see that $$ \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset \left\{ v - tu \mid v \in K, t \geq 0 \right\} $$ so taking the closures of both sides, we obtain $$ T_K(u) \subset \mathrm{cl}(\left\{ v - tu \mid v \in K, t \geq 0 \right\}). $$ So it suffices to show that the set $\left\{ v - tu \mid v \in K, t \geq 0 \right\}$ is closed, but I can't figure out how I can do it. Given an arbitrary sequence $\{v_n - t_n u\}_{n=1}^\infty$ ($v_n \in K, t_n \geq 0$) that converges to some point, my goal is to prove the limit can be written in the form of $v - tu$ with $v \in K, t \geq 0$. If both $\{v_n\}$ and $\{t_n\}$ converge, this is easy because $K$ is closed. But we can easily construct examples in which neither $\{v_n\}$ nor $\{t_n\}$ doesn't converge, e.g., $v_n = nu$ and $t_n = n$, so we cannot assume the convergence of $\{v_n\}$ or $\{t_n\}$.