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?

When $K$ is a $p$-adic field (rather than just a nonarchimedean field), what you are asking is true, and you have much more tools on hand than just the nonarchimedean inequality.  Let $f: K \rightarrow [0,\infty)$ be the function $f(x) = |x|$.
Let $\omega \in K$ be a uniformizer (if $K = \mathbb Q_p$ you can take $\omega = p$), so that $f(\omega) = \rho$ for some $0 < \rho < 1$.  Since every $0 \neq x \in K$ is equal to $u \omega^n$ for a unique integer $n$ and unique $u \in K$ with $|u| = 1$, it is clear that
$$\operatorname{Image}(f) = \{0\} \cup \{..., \rho^2, \rho, 1, \rho^{-1}, \rho^{-2}, ... \}$$
Now if $\{x_n\} \in K$ is a Cauchy sequence, then $\{f(x_n)\}$ is a convergent sequence in $\operatorname{Image}(f)$.  If we assume that $\{x_n\}$ does not converge to $0$, then $\{f(x_n)\}$ is a convergent sequence in
$$\{..., \rho^2, \rho, 1, \rho^{-1}, \rho^{-2}, ... \}.$$
This last set being discrete (every singleton set is open), the only convergent sequences therein are those which are eventually constant.
