Calculating limits of areas of circle 
If $r$ is the radius of circle then I calculated $f(t)$ and $g(t)$ to be 
$$f(t)=\frac{1}{2}r^2t-\frac{1}{2}r^2\sin t=\frac{1}{2}r^2(t-\sin t)$$
$$AB^2=4r^2\sin^2 \frac{t}{2}= 2AC^2+2AC^2\cos(\pi-t)$$
$$g(t)=\frac{1}{2}AC^2\sin (\pi-t)-\frac{1}{2}r^2(t-\sin t )= \frac{r^2\sin^2 \frac{t}{2}}{1-\cos t}-\frac{1}{2}r^2(t-\sin t)$$
$$\frac{f(t)}{g(t)} = \frac{1}{\frac{2\sin^2 \frac{t}{2}\sin t}{(1-\cos t)(t-\sin t)}-1}$$
Thus $$\ \lim_{x\rightarrow 0 } \frac{f(t)}{g(t)} = 0$$
Is this working correct? Intuitively we can see that as $t$ approaches $0$, line segment $AB$ becomes smaller and thus $f(t)$ approaches $0$ and $g(t)$ approaches the area $ABC$  giving a limit of $0$ which is my answer. 
 A: Alternatively:
$$\begin{align}
f(t)=&\frac{1}{2}r^2t-\frac{1}{2}r^2\sin{t}=\frac{1}{2}r^2(t-\sin t). \\
AC=&r\tan{\frac{t}{2}}. \\
S_{ABC}=&\frac{1}{2}AC^2\sin{(\pi-t)}=\frac{1}{2}r^2\tan^2{\frac{t}{2}}\sin{t}=\frac{1}{2}r^2\frac{1-\cos t}{1+\cos t}\sin t. \\
g(t)=&S_{ABC}-f(t)=\frac{1}{2}r^2(\frac{1-\cos t}{1+\cos t}\sin t-t+\sin t)= \\
=&\frac{1}{2}r^2\frac{2\sin t-t(1+\cos t)}{1+\cos t}\\
\lim_\limits{t\to 0} \frac{f(t)}{g(t)}=&\lim_\limits{t\to 0} \frac{(t-\sin{t})(1+\cos t)}{2\sin t-t(1+\cos t)}=2.\end{align}$$
A: I also got $f(t) = \frac{r^2}{2}(t - sint)$.
However, for $g(t)$ I obtained $$g(t) = r^2\left[sin^2\left(\frac{t}{2}\right)tan\left(\frac{t}{2}\right) - \frac{1}{2}(t - sint)\right] = \frac{r^2}{2}\left[\frac{2sint}{cost + 1} - t\right].$$
Let me explain how I reached this conclusion for $g(t)$. First, we have to find the area of triangle $ABC$. To do this we need to find angle $A$ or angle $B$ so that we can obtain the height of $ABC$. In this case, the height of $ABC$ is the line drawn from $C$ that is perpendicular to $AB$. By Thales Theorem we can determine that angle $A$ is equal to $\frac{t}{2}$, since we had to bisect angle $OAB$ (aka $t$) to get the equation for $f(t)$.
Now we need to recognize that the height of $ABC$ gives us two right triangles. We use that which contains point $A$, point $C$, and the point mid way between $A$ and $B$. Now we can see that $$tan\left(\frac{t}{2}\right) = \frac{H}{\frac{AB}{2}}$$ where $H$ is the height of $ABC$. This simplifies to $$H = rsin\left(\frac{t}{2}\right)tan\left(\frac{t}{2}\right)$$.
By definition of triangle area we get $$A_{ABC} = r^2sin^2\left(\frac{t}{2}\right)tan\left(\frac{t}{2}\right)$$
I can edit in more steps later, but at the moment my browser is bugging out so I will submit this for now.
Thus, $$\frac{f(t)}{g(t)} = \frac{t - sint}{\frac{2sint}{cost + 1} - t}.$$
Notice that taking the limit of this gives $\frac{0}{0}$, so L'Hospital's rule needs to be used.
First some simplification: $$\frac{f(t)}{g(t)} = \frac{(t - sint)(cost + 1)}{2sint - t(cost + 1)} = \frac{tcost + t - sintcost - sint}{2sint - tcost - t}$$.
Now with L'Hospital's rule we get $$\lim_{t\to 0} \frac{f(t)}{g(t)} = \lim_{t\to 0} \frac{f'(t)}{g'(t)} = \lim_{t\to 0} \frac{cost - tsint + 1 - cos^2t + sin^2t - cost}{2cost - cost + tsint - 1} = \frac{0}{0}$$.
After several more instances of L'Hospital's rule we find that the limit is 2.
Credit to farruhota, who realized it before I did.
