<s>[INST] 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. 
 [/INST] Alternatively:
$$\begin{align}
f(t)=&\frac{1}{2}r^2t-\frac{1}{2}r^2\sin{t}=\frac{1}{2}r^2(t-\sin t). \\
AC=&r\tan{\frac{t}{2}}. \\
S_{ABC}=&\frac{1}{2}AC^2\sin{(\pi-t)}=\frac{1}{2}r^2\tan^2{\frac{t}{2}}\sin{t}=\frac{1}{2}r^2\frac{1-\cos t}{1+\cos t}\sin t. \\
g(t)=&S_{ABC}-f(t)=\frac{1}{2}r^2(\frac{1-\cos t}{1+\cos t}\sin t-t+\sin t)= \\
=&\frac{1}{2}r^2\frac{2\sin t-t(1+\cos t)}{1+\cos t}\\
\lim_\limits{t\to 0} \frac{f(t)}{g(t)}=&\lim_\limits{t\to 0} \frac{(t-\sin{t})(1+\cos t)}{2\sin t-t(1+\cos t)}=2.\end{align}$$
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