I am new to ultralimits and I am trying to find out what the asymptotic cone $\operatorname{Cone}_{\omega}(X)$ of $X:=\operatorname{Cay}(\langle\mathbb{Z}^2\vert(1,0),(0,1)\rangle)$ is. And how to show it..
I can see that points that have finite distance are equal in the limit.
Further question: Why is the asymptotic cone of a $\delta$-hyperbolic space a real tree?
I suppose that $\mathrm{Cone}_{\omega}(X)$ corresponds to the ultralimit $\left(X,\frac{1}{n} \cdot \mathrm{dist}, (0,0) \right)$ as $n \to + \infty$.
Imagine that you are looking $X$ from a far away point of view. Then, the squares become smaller and smaller, and finally $X$ seems to fill in the whole plane. Roughly, we want to say that $\mathrm{Cone}_{\omega}(X)$ is simply $\mathbb{R}^2$. More formally, notice that the sequence $(nx,ny)$, viewed as a point of $\mathrm{Cone}_{\omega}(X)$, is naturally associated to $(x,y) \in \mathbb{R}^2$.
For your second question, it is classical to prove that the ultralimit of $\delta_n$-hyperbolic spaces is $\delta$-hyperbolic if $\delta_n \to \delta$. Now, notice that $\left(X,\frac{1}{n} \cdot \mathrm{dist} \right)$ is $\delta/n$-hyperbolic.