This question is a follow up to [Are $G$-limits of a slender group $G$ also slender?](https://mathoverflow.net/questions/253686/are-g-limits-of-a-slender-group-g-in-the-space-of-marked-groups-also-slender) The MO user @Ycor gave an excellent answer to this question, in which they show that $\mathbb{Z}\wr\mathbb{Z}$ embeds in a limit of $\mathbb{Z}^2 \rtimes \mathbb{Z}$, answering the question negatively.


Nonetheless $\mathbb{Z}\wr\mathbb{Z}$ is still *small*, i.e. it does not contain a non-abelian free group. Examples of f.g. small group with a non-small limit are groups containing a copy of every finite group. This includes the group $\mathrm{Sym}\_0(\mathbb{Z})\rtimes\mathbb{Z}$, which is not clearly slender.


So my new question is:


1. Are $G$-limits of a slender group $G$ in the space of marked groups small?
2. What if $G$ is also amenable? (I know little about amenability.)


My motivation is to understand the actions of the limits of slender groups on trees.



Are $G$-limits of a slender group $G$ in the space of marked groups small?
Thanks to Ycor's answer and an other [answer due to Anton Klyatchko](https://mathoverflow.net/questions/146812/minimal-number-of-generators-of-subgroups-of-noetherian-groups), the answer is **no, limits of slender groups are not small.**


In that Anton Klyatchkos's answer, [Obraztsov's embedding theorem](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=2990&option_lang=eng) is used to produce 2-generated group $H$ in which the isomorphisms classes of its subgroups consists of infinite cyclic and all odd ordered finite subgroups. As explained in the excellent answer to [Are $G$-limits of a slender group $G$ in the space of marked groups also slender?](https://mathoverflow.net/questions/253686/are-g-limits-of-a-slender-group-g-in-the-space-of-marked-groups-also-slender) such a group has a limit containing a free limit. Indeed since free groups are residually $p$-groups for any prime $p$, there is a sequence of finite groups converging to the free group within the class of free groups with odd order.


All known slender (or noetherian) groups are either virtually polycyclic or "monstruous", i.e. weird and perfect. If $G$ is polycyclic then as explained [again here](https://mathoverflow.net/questions/253686/are-g-limits-of-a-slender-group-g-in-the-space-of-marked-groups-also-slender) it may have a non slender limit, however since polycyclic groups are *Equationally Noetherian*, every limit $L$ of $G$ must fully residually $G$ and therefore can be discriminated via maps $L \to G$. In particular if $G$ has derived length $n$ then so must $L$. It therefore follows that $L$ must also be small.


