구간별 함수 곱의 극한과 연속

ㄱ. (참) $\displaystyle \lim\limits _{x \rightarrow 1 + 0} {f ( x ) g ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 1 + 0} {f ( x )} \times \lim\limits _{x \rightarrow 1 + 0} {g ( x )}$ $= ( - 1 ) \times 1 = - 1$ $\displaystyle \lim\limits _{x \rightarrow 1 - 0} {f ( x ) g ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 1 - 0} {f ( x )} \times \lim\limits _{x \rightarrow 1 - 0} {g ( x )}$ $= 1 \times ( - 1 ) = - 1$ ㄴ. (거짓) $\displaystyle \lim\limits _{x \rightarrow + 0} {g ( x + 1 )} = \lim\limits _{t \rightarrow 1 + 0} {g ( t )} = 1$ $\displaystyle \lim\limits _{x \rightarrow - 0} {g ( x + 1 )} = \lim\limits _{t \rightarrow 1 - 0} {g ( t )} = - 1$ 즉, $\displaystyle \lim\limits _{x \rightarrow + 0} {g ( x + 1 )} \ne \lim\limits _{x \rightarrow - 0} {g ( x + 1 )}$이므로 함수 $g ( x + 1 )$은 $x = 0$에서 불연속이다. ㄷ. (참) $\displaystyle \lim\limits _{x \rightarrow - 1 - 0} {f ( x ) g ( x + 1 )}$$\displaystyle = \lim\limits _{x \rightarrow - 1 - 0} {f ( x )} \times \lim\limits _{x \rightarrow - 1 - 0} {g ( x + 1 )}$ $\displaystyle = \lim\limits _{x \rightarrow - 1 - 0} {f ( x )} \times \lim\limits _{t \rightarrow - 0} {g ( t )}$$= - 1 \times 0 = 0$ $\displaystyle \lim\limits _{x \rightarrow - 1 + 0} {f ( x ) g ( x + 1 )}$$\displaystyle = \lim\limits _{x \rightarrow - 1 + 0} {f ( x )} \times \lim\limits _{x \rightarrow - 1 + 0} {g ( x + 1 )}$ $\displaystyle = \lim\limits _{x \rightarrow - 1 + 0} {f ( x )} \times \lim\limits _{t \rightarrow + 0} {g ( t )}$$= 1 \times 0 = 0$ $\displaystyle \lim\limits _{x \rightarrow - 1 - 0} {f ( x ) g ( x + 1 )} = \lim\limits _{x \rightarrow - 1 + 0} {f ( x ) g ( x + 1 )} = 0$이고 $f ( 0 ) g ( 1 ) = 1 \times 0 = 0$이므로 함수 $f ( x ) g ( x + 1 )$은 $x = - 1$에서 연속이다. 따라서 옳은 것은 ㄱ, ㄷ이다.