도형 넓이 극한

[출제의도] 도형의 성질과 삼각함수의 극한의 성질을 이용하여 극한값을 구한다.

$\displaystyle \mathrm{\overline{PR}} = {\overline{OP}} - {\overline{OR}}$=$1 - \cos \theta$ $\displaystyle \mathrm{\overline{PQ}} = \sin \theta$ $\displaystyle \mathrm{\angle} QPR = \frac{\pi}{2} - \theta$이므로 $\displaystyle S \left( \theta \right) = \frac{1}{2} \times \left( 1 - \cos \theta \right) \times \sin \theta \times \sin \left( \frac{\pi}{2} - \theta \right)$ $\displaystyle = \frac{\left( 1 - \cos \theta \right) \sin \theta \cos \theta}{2}$ ∴ $\displaystyle \lim\limits _{\theta \rightarrow + 0} {\frac{S \left( \theta \right)}{\theta ^{3}}}$$\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\left( 1 - \cos \theta \right) \sin \theta \cos \theta}{2 \theta ^{3}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\sin ^{3} \theta \cos \theta}{2 \theta ^{3} \left( 1 + \cos \theta \right)}}$$\displaystyle = \frac{1}{4}$ [다른 풀이] (삼각형 $\mathrm{PRQ}$의 넓이) $=$(삼각형 $\mathrm{POQ}$의 넓이)$-$(삼각형 $\mathrm{ROQ}$의 넓이) 이므로 $\displaystyle S \left( \theta \right) = \frac{1}{2} \sin \theta \cos \theta - \frac{1}{2} \cos ^{2} \theta \sin \theta$ $\displaystyle = \frac{1}{2} \sin \theta \cos \theta \left( 1 - \cos \theta \right)$ ∴ $\displaystyle \lim\limits _{\theta \rightarrow + 0} {\frac{S \left( \theta \right)}{\theta ^{3}}}$$\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\left( 1 - \cos \theta \right) \sin \theta \cos \theta}{2 \theta ^{3}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow + 0} {\frac{\sin ^{3} \theta \cos \theta}{2 \theta ^{3} \left( 1 + \cos \theta \right)}}$$\displaystyle = \frac{1}{4}$