그래프 극한 보기

[출제의도] 함수의 연속성 추론하기 ㄱ. $\displaystyle \lim\limits _{x \rightarrow + 0} {f \left( x \right)} = - 2$, $\displaystyle \lim\limits _{x \rightarrow + 0} {f \left( - x \right) = \lim\limits _{x \rightarrow - 0} {f \left( x \right)}} = 2$ $\therefore$$\displaystyle \lim\limits _{x \rightarrow + 0} {\left\{ f \left( x \right) + f \left( - x \right) \right\}} = - 2 + 2 = 0$ (참) ㄴ. $f \left( x \right) = t$라 하면 $x \rightarrow 1 + 0$일 때 $t \rightarrow - 1 + 0$ $\therefore$$\displaystyle \lim\limits _{x \rightarrow 1 + 0} {f \left( f \left( x \right) \right)} = \lim\limits _{t \rightarrow - 1 + 0} {f \left( t \right)} = 0$ (거짓) ㄷ. $\displaystyle \lim\limits _{x \rightarrow 1 + 0} {\left\{ f \left( x - 1 \right) \right\} ^{2} = \lim\limits _{x \rightarrow + 0} {\left\{ f \left( x \right) \right\} ^{2}}}$ $= \left( - 2 \right) ^{2} = 4$ $\displaystyle \lim\limits _{x \rightarrow 1 - 0} {\left\{ f \left( x - 1 \right) \right\} ^{2} = \lim\limits _{x \rightarrow - 0} {\left\{ f \left( x \right) \right\} ^{2}}}$ $= 2 ^{2} = 4$ $\left\{ f \left( 1 - 1 \right) \right\} ^{2} = \left\{ f \left( 0 \right) \right\} ^{2} = 2 ^{2} = 4$ $\therefore$함수 $\left\{ f \left( x - 1 \right) \right\} ^{2}$은 $x = 1$에서 연속이다. (참) 따라서 옳은 것은 ㄱ, ㄷ