Recall that the translation length of an isometry $\gamma : X \to X$ is $|\gamma| := \inf\{ d(x,\gamma.x) : x \in X\}$, and that an isometry is said to be semi-simple if this infimum is achieved.
Bridson and Haefliger's first exercise in chapter II.6 of Metric Spaces of Non-Positive Curvature on semi-simple isometries reads: Let $(X,d)$ be a metric space and let $\gamma$ be an isometry of $X$. Show that $$\lim_{n\to\infty}\frac 1nd(x,\gamma^n.x)$$ exists for all $x \in X$. Show that $\lim_{n\to\infty}\frac1nd(x,\gamma^n.x)$ is independent of $x$, and if $\gamma$ is semi-simple, then $|\gamma| = \lim_{n\to\infty}\frac1nd(x,\gamma^n.x)$.
The only assertion I am struggling to prove is the final one: repeated application of the triangle inequality shows that for suitable choice of $x$, $\frac1nd(x,\gamma^n.x) \leq |\gamma|$, so the limit is certainly bounded above by $|\gamma|$, but how do we show that $\lim_{n\to\infty} \frac 1nd(x,\gamma^n.x) < |\gamma|$ is not possible?
Without assuming $CAT(0)$, there are counterexamples to this assertion. So I am guessing that there is some hypothesis missing.
Here's a simple counterexample.
Take the infinite ladder $$X = \bigl(\mathbb{R} \times \{-1,1\}\bigr) \cup \bigl(\mathbb{Z} \times [-1,1]\bigr) \subset \mathbb{R}^2 $$ and let $X$ be given its natural path metric, obtained by infimizing lengths of paths in $X$.
Let $\gamma : X \to X$ be defined by $$\gamma(x,y) = (x+1,-y) $$ In this case $|\gamma|=3$ but $\lim_{x \to \infty} \frac{1}{n} d(x,\gamma^n \cdot x) = 1$.