are equivalent:\
$\mathbf{a.)}$ $G$ is lacunary hyperbolic group.\
$\mathbf{b.)}$ There exists a scaling sequence $d=(d_n)$, such that $Con^{\omega}(G,d)$ is an $\mathbb{R}$-tree for all non-principal ultrafilter $\omega$.\
$\mathbf{c.)}$ $G$ is the direct limit of a sequence of finitely generated hyperboolic groups and epimorphisms;\
$$\begin{aligned}
\label{lacunarylim} 
    G_1\overset{\alpha_1}{\longrightarrow} G_{2}\overset{\alpha_{2}}{\longrightarrow}\cdots G_i\overset{\alpha_i}{\longrightarrow} G_{i+1}\overset{\alpha_{i+1}}{\longrightarrow}G_{i+2}\overset{\alpha_{i+2}}{\longrightarrow}\cdots 
    \end{aligned}$$ where $G_i$ is generated by a finite set $\langle S_i\rangle $ and $\alpha_i(S_i)=S_{i+1}$. Also $G_i$’s are $\delta_i$ hyperbolic where $\delta_i$=$o(r_{S_i}(\alpha_i))$ (where $r_{S_i}(\alpha_i)$=injective radius of $\alpha_i$ w.r.t. $S_i$ ).

The following theorem gives us a characterization of lacunary hyperbolic group in terms of the hyperbolicity function of the corresponding Cayley graph.

\[lhgdef\] Let G be a finitely generated group with generating set $S$. Then $G$ is lacunary hyperbolic if and only if the hyperbolicity function $f_{G}$ of the corresponding Cayley graph $\Gamma(G,S)$ satisfies $\liminf_{t\rightarrow\infty}\frac{f_{G}(t)}{t}=0$.

Suppose $G$ be lacunary hyperbolic group. Consider any geodesic triangle of perimeter less than or equal to $r_{S_i}(\alpha_i)$, then we use part $c.)$ of Theorem ${\color{blue}\ref*{OOS}}$ and get that $f_G(r_{S_i}(\alpha_i))=\mathcal{O}(\delta_i)=o(r_{S_i}(\alpha_i))$, as $\delta_i=o(r_{S_i}(\alpha_i))$. Hence, $$\lim_{n\rightarrow\infty}\frac{f_G(r_{S_i}(\alpha_i))}{r_{S_i}(\alpha_i)}=0\Rightarrow \liminf_{n\rightarrow\infty}\frac{f_G(t)}{t}=0.$$\
\
Suppose the hyperbolicity function $f_G$ satisfies the following condition: $$\liminf_{t\rightarrow\infty}\frac{f_G(t)}{t}=0$$ Hence there exists a non-decreasing scaling sequence, say $k=(k_n)$ such that: $$\begin{aligned}
\label{S} 
    \lim_{n\rightarrow\infty}\frac{f_G(d_n)}{d_n}=0.
    \end{aligned}$$ Fix the sequence $d=\{d_n\}$. Then we have, 