반원 넓이비 극한

$\displaystyle S ( x ) = \frac{1}{2} x \left( \mathrm{\overline{AB}} + \overline{CD} \right)$, $T ( x )$$\displaystyle = x \cdot \overline{\mathrm{CD}}$이므로 $\displaystyle \frac{T ( x )}{S ( x )} = \frac{\mathrm{2 \cdot \overline{CD}}}{\overline{AB} + \overline{CD}}$$\displaystyle = \frac{\mathrm{2 \cdot \overline{QD}}}{\overline{PB} + \overline{QD}}$이다. $\displaystyle \mathrm{\overline{OP}} = \mathit{1} - 2 x$이므로 직각삼각형 $\mathrm{OPB}$에서 $\displaystyle \overline{\mathrm{PB}}$$\displaystyle = \sqrt{1 - ( 1 - 2 x ) ^{2}} = \sqrt{4 x - 4 x ^{2}}$ $\displaystyle \mathrm{\overline{OQ}} = \mathit{1} - x$이므로 직각삼각형 $\mathrm{OQD}$에서 $\displaystyle \overline{\mathrm{QD}}$$\displaystyle = \sqrt{1 - ( 1 - x ) ^{2}} = \sqrt{2 x - x ^{2}}$ ∴ $\displaystyle \frac{\lim\limits _{x\rightarrow 0{T ( x )}}}{S ( x )}$$\displaystyle = \lim\limits _{x\rightarrow 0} \frac{2 \sqrt{2 x - x ^{2}}}{\sqrt{4 x - 4 x ^{2}} + \sqrt{2 x - x ^{2}}}$ $\displaystyle = \lim\limits _{x\rightarrow 0} \frac{2 \sqrt{2 - x}}{\sqrt{4 - 4 x} + \sqrt{2 - x}}$ $\displaystyle = \frac{2 \sqrt{2}}{2 + \sqrt{2}}$$\displaystyle = 2 \sqrt{2} - 2$