Rigorous way to find the limit of this difference? This is a question from an old released exam.

By the triangle inequality, $s-r<1$, so I eliminate answers D and E. Intuitively, since the lower angle between $1$ and $r$ is fixed at $110^\circ$, $s$ will always be a little longer than $r$, so I eliminate A and C to find B as the correct answer.
This is pretty informal, is there a more rigorous way one could prove the limit?
 A: From the cosine law, 
$$s^2=1+r^2-2r\cos\alpha,$$  
where $\alpha=110^\circ$.
We can rewrite this as 
$$s-r=\frac{1-2r \cos\alpha}{s+r}$$ 
and remember that $\cos\alpha \lt 0$.
As you say, $0 \lt s-r \lt 1$, so 
$$\frac{1-2r \cos\alpha}{2r+1} \lt s-r \lt \frac{1-2r \cos\alpha}{2r}.$$ 
The left inequality is bounded away from $0$ and the right from $1$, so the answer is B.  In fact, we can say 
$$\frac{1/r-2 \cos\alpha}{2+1/r} \lt s-r \lt \frac{1/r-2 \cos\alpha}{2}$$ 
so the limit is $-\cos\alpha=-\cos 110^\circ$.
A: A qualitative answer.
When $s$ and $r$ tend both to infinity, keeping the said elements fixed, the sides $s$ and $r$ tend to be parallel, so their difference is the projection of the third side on their common direction, given by
$$s-r\to|1\cdot\cos\alpha|$$
A: By the cosine law,
$c = \cos 110 = \frac{1+r^2-s^2}{2r}$,
or $2 r c = 1 + r^2 - s^2$.
Let $d = |c|$, so $d > 0$.
Since $s^2 = r^2 +2rd+1$,
$s = r\sqrt{1 + 2d/r + 1/r^2}
=r (1+d/r + O(1/r^2))
= r + d + O(1/r)$,
so $ s-r \to d = -\cos 110$.
A slightly modified look:
$s^2 = r^2+2rd+d^2 + 1-d^2
= (r+d)^2+1-d^2$,
so $s = \sqrt{(r+d)^2+1-d^2}
=(r+d)\sqrt{1 + (1-d^2)/(r+d)^2}
= (r+d)(1 + O(1/(r+d)^2))
= r+d + O(1/(r+d))
$.
