사다리꼴 넓이 극한

점 $\mathrm{A} , F$에서 선분 $\mathrm{B} C$에 내린 수선의 발을 $\mathrm{E} , F$라 하면

$\displaystyle {\overline{AE}} = x \tan 2 \theta$ $\displaystyle {\overline{DF}} = ( \sin \theta - x ) \tan 3 \theta$ $\displaystyle {\overline{AE}} = {\overline{DF}}$에서 $\displaystyle x = \frac{\sin \theta \tan 3 \theta}{\tan 2 \theta + \tan 3 \theta}$ $\displaystyle S ( \theta ) = \frac{1}{2} \times 3 \sin \theta \times x \tan 2 \theta$ $\displaystyle = \frac{1}{2} \times 3 \sin \theta \times \frac{\sin \theta \tan 3 \theta}{\tan 2 \theta + \tan 3 \theta} \times \tan 2 \theta$ $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {\frac{S ( \theta )}{\theta ^{3}}} = \lim\limits _{\theta \rightarrow 0 +} {\frac{1}{2} \cdot \frac{3 \sin \theta \tan \theta \sin \theta \tan 3 \theta}{\theta ^{3} ( \tan 2 \theta + \tan 3 \theta )}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\frac{1}{2} \frac{3 \sin \theta \tan 2 \theta \sin \theta \tan 3 \theta}{\theta ^{4}}} \times \frac{\theta}{\tan 2 \theta + \tan 3 \theta}$ $\displaystyle = \frac{9}{5}$ $\therefore p + q = 14$