구간별 sin 함수 보기

ㄱ. $\displaystyle \lim\limits _{x \rightarrow 0 -} {} f ( x )$$\displaystyle = \lim\limits _{x \rightarrow 0 -} {} ( 1 + \sin x )$$= 1$ $\displaystyle \lim\limits _{x \rightarrow 0 -} {} f ( - x )$$\displaystyle = \lim\limits _{t \rightarrow 0 +} {f ( t )}$$\displaystyle = \lim\limits _{t \rightarrow 0 +} {} ( - 1 + \sin t )$$= - 1$ ∴ $\displaystyle \lim\limits _{x \rightarrow 0 -} {} f ( x ) f ( - x )$$= 1 \cdot ( - 1 ) = - 1$ $\displaystyle \lim\limits _{x \rightarrow 0 +} {} f ( x )$$\displaystyle = \lim\limits _{x \rightarrow 0 +} {} ( - 1 + \sin x )$$= - 1$ $\displaystyle \lim\limits _{x \rightarrow 0 +} {} f ( - x )$$\displaystyle = \lim\limits _{t \rightarrow 0 -} {f ( t )}$$\displaystyle = \lim\limits _{t \rightarrow 0 -} {} ( 1 + \sin t )$$= 1$ ∴ $\displaystyle \lim\limits _{x\rightarrow 0} f ( x ) f ( - x ) = - 1$ (참) ㄴ. $\displaystyle x \rightarrow \frac{\pi}{2}$일 때, $f ( x ) \rightarrow 0 -$이므로 $\displaystyle \lim\limits _{x \rightarrow \frac{\pi}{2}} f ( f ( x ) )$$\displaystyle = \lim\limits _{t \rightarrow 0 -} {} f ( t )$$\displaystyle = \lim\limits _{t \rightarrow 0 -} {} ( 1 + \sin t )$$= 1$ $\displaystyle f \left( f \left( \frac{\pi}{2} \right) \right)$$= f ( 0 )$$= 1$ $\displaystyle \lim\limits _{x \rightarrow \frac{\pi}{2}} f ( f ( x ) )$$\displaystyle = f \left( f \left( \frac{\pi}{2} \right) \right)$이므로 함수 $f ( f ( x ) )$는 $\displaystyle x = \frac{\pi}{2}$에서 연속이다. (참) ㄷ. $g ( x ) = \left\{ f ( x ) \right\} ^{2}$라 하면 $g ( 0 ) = 1$이므로 $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle = \lim\limits _{x \rightarrow 0 -} {} \frac{( 1 + \sin x ) ^{2} - 1}{x}$ $\displaystyle = \lim\limits _{x \rightarrow 0 -} {} \frac{2 \sin x + \sin ^{2} x}{x}$ $= 2$ $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$∵ $\displaystyle \lim\limits _{x\rightarrow 0} \frac{\sin x}{x} = 1$$\displaystyle \begin{matrix} \\ \end{matrix} )$ $\displaystyle \lim\limits _{x \rightarrow 0 +} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle = \lim\limits _{x \rightarrow 0 +} {} \frac{( - 1 + \sin x ) ^{2} - 1}{x}$ $\displaystyle = \lim\limits _{x \rightarrow 0 +} {} \frac{- 2 \sin x + \sin ^{2} x}{x}$ $= - 2$ $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$∵ $\displaystyle \lim\limits _{x\rightarrow 0} \frac{\sin x}{x} = 1$$\displaystyle \begin{matrix} \\ \end{matrix} )$ $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle \ne \lim\limits _{x \rightarrow 0 +} {} \frac{g ( x ) - g ( 0 )}{x}$이므로 $\displaystyle \lim\limits _{x\rightarrow 0} \frac{g ( x ) - g ( 0 )}{x}$는 존재하지 않는다. 따라서 함수 $g ( x )$는 $x = 0$에서 미분불가능하다. (거짓) 따라서 옳은 것은 ㄱ, ㄴ이다. [다른 풀이] ㄷ. 미분가능한 두 함수 $\left( 1 + \sin x \right) ^{2}$, $\left( - 1 + \sin x \right) ^{2}$의 도함수는 각각 $2 ( 1 + \sin x ) \cos x$, $2 ( - 1 + \sin x ) \cos x$이고 이 두 함수는 실수 전체의 집합에서 연속이다. $g ( x ) = \left\{ f ( x ) \right\} ^{2}$라 하면 $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle = \lim\limits _{x \rightarrow 0 -} {} 2 ( 1 + \sin x ) \cos x = 2$ $\displaystyle \lim\limits _{x \rightarrow 0 +} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle = \lim\limits _{x \rightarrow 0 +} {2 ( - 1 + \sin x ) \cos x}$$= - 2$ $\displaystyle \lim\limits _{x \rightarrow 0 -} {} \frac{g ( x ) - g ( 0 )}{x}$$\displaystyle \ne \lim\limits _{x \rightarrow 0 +} {} \frac{g ( x ) - g ( 0 )}{x}$이므로 $\displaystyle \lim\limits _{x\rightarrow 0} \frac{g ( x ) - g ( 0 )}{x}$는 존재하지 않는다. 따라서 함수 $g ( x )$는 $x = 0$에서 미분불가능하다. (거짓)