\end{aligned}$$ Hence we get $|H_n|=\Theta_{\omega}(d_n)$. As the function $f_X$ is sub-linear, we also have, $$\begin{aligned}
\label{C} 
    f_X(|H_n|)=o_{\omega}(|H_n|)=o_{\omega}(d_n).
    \end{aligned}$$ Finally we get that $A$ is contained in the union of other two sides in $Con^{\omega}(X,e,d)$, as $\alpha_n=o_{\omega}(d_n)$ (combining $\color{blue}{(\ref*{A}),(\ref*{B}),(\ref*{C})}$). This contradicts out assumption that the triangle $\triangle_{\infty}$ is simple. Hence $Con^{\omega}(X,e,d)$ is an $\mathbb{R}$-tree. That contradicts our assumption that $X$ is not hyperbolic.

Lacunary hyperbolic groups {#lacunarydef}
==========================

A metric space $X$ is called lacunary hyperbolic if one of the asymptotic cones of $X$ is an $\mathbb{R}$-tree. A finitely generated group is called lacunary hyperbolic if it has a finite generating set and the corresponding Cayley graph is lacunary hyperbolic.

Note that lacunary hyperbolicity does not depend on finite generating set. It follows from Theorem \[OOS\] below.

Let $\alpha: G\rightarrow H$ be a group homomorphism and $G=\langle A\rangle ,H=\langle B \rangle$. The injectivity radius $r_A(\alpha)$ is the radius of largest ball centered at identity of $G$ in the Cayley graph of $G$ with respect to $A$ on which the restriction of $\alpha$ is injective.

In the above setting, $r_A(\alpha)$ can be $\infty$, for example when $\alpha$ is injective.

[@OOS07 Theorem 1.1 ]\[OOS\] Suppose $G$ be a finitely generated group. Then following 