삼각함수 극한 상수

$\displaystyle \lim\limits _{x \rightarrow 0} \frac{x ^{2}}{a \cos ^{2} x + b} = \frac{1}{2}$에서 $\displaystyle \lim\limits _{x \rightarrow 0} x ^{2} = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} \left( a \cos ^{2} x + b \right) = 0$이어야 한다. 즉, $a + b = 0$ $\displaystyle \therefore \lim\limits _{x \rightarrow 0} \frac{x ^{2}}{a \cos ^{2} x + b} = \lim\limits _{x \rightarrow 0} \frac{x ^{2}}{a \cos ^{2} x - a}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{x ^{2}}{- a \sin ^{2} x}$ $\displaystyle = - \frac{1}{a} \left( \because \lim\limits _{x \rightarrow 0} {} \frac{x}{\sin x} = 1 \right)$ $\displaystyle - \frac{1}{a} = \frac{1}{2}$에서 $a = - 2$ $\therefore b = 2 ( \because a + b = 0 )$ $\therefore ab = - 4$