Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.

For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim_{n \to \infty} l(g^n|w)/n$.

If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality $$1/2 \le sl_F(w|w) \le 1$$ (where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)

The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.

What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$?

(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl_F(w|w)$ comes from the inequality $scl_G(g) \le sl_G(g|w)(scl_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl_F(w|w)$ if one knows $scl_F(w)>1/2$ (the estimate $scl_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?