Answer is (B). Of course, we may express $s$ in terms of $r$ using cosine law and take the limit and show the limit is $\cos(70)<1$. But I don't think that this method is intended because if it were they could have just asked for the exact value of the limit. I wonder if there is a geometric argument that allows us to see this immediately.
On
Here is a picture for the geometric limit:
The points $A$, $O$, the length of $AO=1$, and the half-line where $X$ moves on are fixed. We construct $B$ as in the picture, $BO\perp OX$, $AB\|OX$, and the angle in $O$ in $\Delta AOB$ has the measure $20^\circ$. Let $X$ be the point going to the infinity, and for each fixed position of $X$, let $Y$ (depending on $X$) be the intersection of $AX$ and $OB$.
Then we have a sandwich for $XY-OX$, $$ 0\le\boxed{\ XY-OX\ }=\frac{XY^2-OX^2}{XY+OX}=\frac{OY^2}{XY+OX} \le \frac{OB^2}{XY+OX}\to 0\ ,$$ so $$ \begin{aligned} s-r &=AX-OX \\ &=SY +(YX-OX) \\ &\to AB+0 =AB \\ &=\sin 20^\circ\ . \end{aligned} $$
Extend $r$ and make the extension equal to $s$. You will get a small triangle with sides $1, s-r, x$ and angle between sides $1$ and $s-r$ equal to $70°$. Obviously $s-r < 1$ because it's opposite to a smaller angle (we have isosceles triangle with sides $s, s, x$ so the base angles are equal).