연속과 합성함수 보기

$\displaystyle f ( x ) = {\begin{cases} \frac{1}{2} x ^{2} & ( x \geq 0 ) \\ - \frac{1}{2} x ^{2} & ( x < 0 ) \end{cases}}$이므로 ㄱ. $\displaystyle \lim\limits _{x \rightarrow + 0} {\frac{f ( x )}{x}} = \lim\limits _{x \rightarrow + 0} {\frac{\frac{1}{2} x ^{2}}{x}} = \lim\limits _{x \rightarrow + 0} {\frac{1}{2} x} = 0$ $\displaystyle \lim\limits _{x \rightarrow - 0} {\frac{f ( x )}{x}} = \lim\limits _{x \rightarrow - 0} {\frac{- \frac{1}{2} x ^{2}}{x}} = \lim\limits _{x \rightarrow - 0} {\left( - \frac{1}{2} x \right)} = 0$ $\therefore$참 ㄴ. $\displaystyle \lim\limits _{x \rightarrow \sqrt{2} + 0} {g ( f ( x ) ) = \lim\limits _{t \rightarrow 1 + 0} {g ( t ) = 2}}$이고 $\displaystyle \lim\limits _{x \rightarrow \sqrt{2} - 0} {g ( f ( x ) ) = \lim\limits _{t \rightarrow 1 - 0} {g ( t ) = 1}}$이므로 $\therefore$거짓 ㄷ. $\displaystyle \lim\limits _{x \rightarrow 1 + 0} {g ( g ( x ) ) = \lim\limits _{t \rightarrow 2 - 0} {g ( t ) = 1}}$ $\displaystyle \lim\limits _{x \rightarrow 1 - 0} {g ( g ( x ) ) = 1}$ $\displaystyle \lim\limits _{x \rightarrow 1} {g ( g ( x ) ) = 1}$이고 $g ( g ( 1 ) ) = 1$이므로 $x = 1$에서 연속이다. $\therefore$참