삼각함수 극한 차수

$\displaystyle \lim\limits _{x \rightarrow 0} \frac{f ( x )}{1 - \cos ( x ^{2} )}$$\displaystyle = \lim\limits _{x \rightarrow 0} {\frac{f ( x ) \left\{ 1 + \cos ( x ^{2} ) \right\}}{\left\{ 1 - \cos ( x ^{2} ) \right\} \left\{ 1 + \cos ( x ^{2} ) \right\}}}$ $\displaystyle = \lim\limits _{x \rightarrow 0} {\frac{2 \cdot f ( x )}{\sin ^{2} ( x ^{2} )}}$$\displaystyle = \lim\limits _{x \rightarrow 0} {\frac{( x ^{2} ) ^{2}}{\sin ^{2} ( x ^{2} )}} \cdot \frac{2 \cdot f ( x )}{( x ^{2} ) ^{2}}$ $\displaystyle = 2 \cdot \lim\limits _{x \rightarrow 0} {\frac{f ( x )}{x ^{4}}} = 2$ 따라서, $\displaystyle \lim\limits _{x \rightarrow 0} {} \frac{f ( x )}{x ^{4}} = 1$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} {} \frac{f ( x )}{x ^{p}} = q$를 반드시 만족하는 상수 $p , q$는 $p = 4 , q = 1$일 때이다.