I am looking for an example of a finitely generated torsion-free nonabelian group that is CSA but not a limit group, that is, not residually free. Recall that a group is CSA if all maximal abelian subgroups are malnormal.
The more elementary the example the better.
Lemma. A two-generated limit group is necessarily free or free abelian.
Proof. Write $G=\langle a, b\rangle$, and suppose $G$ is neither free nor free abelian. Then $G$ is non-abelian and so the set $\{a,b,[a,b]\}$ contains only non-trivial elements. As $G$ is two-generated and non-free, any non-trivial map to a free group must have cyclic image. Hence, $[a,b]$ is always contained in the kernel of such a map. Therefore, $G$ is not fully residually free, and hence not a limit group. QED
Therefore, with only three exceptions ($\mathbb{Z}$, $\mathbb{Z}^2$, $F_2$), any non-trivial torsion-free two-generated CSA group gives you your example. For example, take $G=\langle a, b\mid abab^2\cdots ab^n\rangle$ for large $n$, which is metric small cancellation, hence hyperbolic, hence CSA, and is torsion-free as the relator is not a proper power (this is a curious, non-trivial fact of both one-relator and small cancellation presentations).