A basic application of the Gauss-Bonnet theorem for polygons. I'm studying for an exam and having trouble to apply this theorem.
What the exercise says...

Let the regular geodesic polygon $P_n(x)$ with $n$ sides and center on the Poincaré's Disc, where $x$ is the distance of its vertices to the center. All its interior angles have the same measure $\alpha_n(x)$. 
Show that
i) $n\alpha_n(x)=(n-2)\pi-A_n(x)$, where $A_n(x)$ is the area of $P_n(x)$.
  ii) $\alpha_n:]0,\infty[ \rightarrow ]0, \frac{(n-2)\pi}{n}[$ is a decreasing function
  $$
\lim_{x\rightarrow0}{\alpha_n(x)}=\frac{(n-2)\pi}{n}
$$
  $$
\lim_{x\rightarrow +\infty}{\alpha_n(x)}=0
$$
  iii)If $n \geq 5$ exists $x$ such that $\alpha_n(x)=\frac{2 \pi}{n}$.

My attempt
i) This is ok! I apply the Gauss Bonnet Theorem and have this result.
ii) Here I have some insecurity if is clear... 
Note that fixed $n$, $-A_n(x)$ is a decreasing function of $x$ (because its measure the area). So, $\alpha_n(x)=\frac{\pi(n-2)-A_n(x)}{n}$ and $\frac{\pi(n-2)}{n}$ is fixed, we have that $\alpha_n(x)$ is a decreasing function.
I don't know how show the second limit!
iii) I have not idea how to do this
Thank you!
 A: (iii) is just an application of Intermediate value theorem. The function $\alpha_n$ is continuous because $A_n$ (the area) is. Since $n\ge 5$, the number $2\pi/n$ is strictly between the two limits stated in (ii).
One way to deal with the second limit in (ii) is to use the fact that the area of an ideal triangle (geodesic triangle whose sides meet at infinity) is $\pi$. This can be established by direct integration. By the invariance of area under Möbius transformations, we can choose the positions of the three vertices. It also helps to switch to the halfplane model. If the triangle has vertices $\pm 1,\infty$, then its area is
$$
\int_{-1}^1 \int_{\sqrt{1-x^2}}^\infty \frac{dy}{y^2} = \int_{-1}^1 \frac{dx}{\sqrt{1-x^2}} =\sin^{-1}x\,\bigg|_{-1}^1 = \pi
$$
The union $\bigcup_{x>0}P_n(x)$ is an ideal $n$-gon, which can be triangulated into $(n-2)$ ideal triangles.  Hence, its area is $(n-2)\pi$. It follows that $\lim_{x\to\infty} A_n(x)=(n-2)\pi$.
The above may be unnecessary if your form of the Gauss-Bonnet theorem allows vertices to lie on the ideal boundary (with angle understood as zero).  
