Limit of non-archimedean absolute values is eventually constant Let $(x_n)$ be a Cauchy sequence in a field $K$ with respect to a non-archimedean absolute value $|\cdot|$ arising from a discrete valuation $v$ (meaning $|x|:=c^{v(x)}$ for some $c\in(0,1)$. Suppose $\operatorname{lim}|x_n|\neq 0$. I'd like to understand why the sequence $(v(x_n))$ becomes evetually constant.
I found this statement at the top of page three in these notes by Andrew Sutherland.
The only tool at hand seems to be the non-archimedean triangle inequality, but I'm not sure how to use it. What am I missing?

Your questions on $p$-adic numbers are strange, as if to you it was something completely abstract, but $p$-adic numbers are very concrete. If $x\ne 0$ and $x_n\to x$ then $v(x_n-x)>v(x)$ for $n$ large enough so that $v(x_n)=v(x),|x_n|=|x|$. When $K$ is not complete and $(x_n)$ is Cauchy then $x=\lim x_n$ is in its completion instead of $K$ but it doesn't change anything.
$v(x_n-x)>v(x)\implies v(x)=v(x_n)$ is part of the definition:
If $v(x_n) > v(x)$ then $v(x)=v(x_n+(x-x_n))\ge \min(v(x_n),v(x-x_n))$ a contradiction.
