원 접선 삼각형 극한

[출제의도] 수학 내적 문제 해결능력-삼각함수 직선 $OP$의 기울기가 $\tan \theta$이므로 $\displaystyle \angle POQ = \theta , {\overline{OQ}} = \frac{2}{\cos \theta}$ 점 $Q$의 좌표는 $\displaystyle Q \left( \frac{2}{\cos \theta} , 0 \right)$이므로 점 $R$의 좌표는 $\displaystyle R \left( \frac{2}{\cos \theta} , \frac{2 \tan \theta}{\cos \theta} \right)$ $\displaystyle {\overline{OA}} = 2$이므로 $\displaystyle f ( \theta ) = \frac{1}{2} \left( \frac{2}{\cos \theta} - 2 \right) \times \frac{2 \tan \theta}{\cos \theta}$$\displaystyle = \frac{2 ( 1 - \cos \theta ) \tan \theta}{\cos ^{2} \theta}$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{f ( \theta )}{\theta ^{3}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{2 ( 1 - \cos \theta ) \tan \theta}{\theta ^{3} \times \cos ^{2} \theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left\{ 2 \times \frac{1 - \cos ^{2} \theta}{\theta ^{2} ( 1 + \cos \theta )} \times \frac{\tan \theta}{\theta} \times \frac{1}{\cos ^{2} \theta} \right\}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left( 2 \times \frac{\sin ^{2} \theta}{\theta ^{2}} \times \frac{1}{1 + \cos \theta} \times \frac{\tan \theta}{\theta} \times \frac{1}{\cos ^{2} \theta} \right)$ $\displaystyle = 2 \times 1 \times \frac{1}{2} \times 1 \times 1 = 1$