Let $K$ be a discrete valuation field where $\nu:K\longrightarrow\mathbb Z$ is a surjective valuation. Let $\lambda\in]0,1[\subseteq\mathbb R $, then the valuation $\nu$ induces a metric $d_\nu$ on $K$ defined as follows: $$d_\nu(x,y):=\lambda^{\nu(x-y)}$$
Now consider a Cauchy sequence $(x_n)$ in $K$ (with respect to the metric $d_{\nu}$); I have to prove that the following limit exists: $$\lim_{n\to\infty}\nu(x_n)$$
Do you have any idea? Why do I need the fact that $(x_n)$ is a Cauchy sequence? Note that the sequence $n\mapsto \nu(x_n)$ has values in $\mathbb Z$.
If $(x_l)_{l \in \mathbb N}$ is a Cauchy sequence w.r.t. $d_v$, then for arbitrary $\epsilon > 0$ we may choose $N \in \mathbb N$ such that $$ \forall n, k > N : d_v(x_k, x_n) = \lambda^{v(x_k - x_n)} < \epsilon. $$ But $$ \lambda^{v(x_k - x_n)} < \epsilon \Leftrightarrow v(x_k - x_n) > \frac{\ln(\epsilon)}{\ln(\lambda)}. $$ As $\epsilon \to 0$, the latter expression converges to infinity. Hence, we must have $v(x_k - x_n) \to \infty$.
Due to the definition of a valuation, either eventually $x_n = x_k = x$ for a fixed $x \in K$ or $\min \{ v(x_k), v(x_n)\} \to \infty$, which would mean $v(x_l) \to \infty, l \to \infty$.