곱함수의 연속

(a) $x = 0$에서 연속이어야 하므로 $h ( 0 ) = f ( 0 ) g ( 0 ) = b \times ( - 1 ) = - b$ $\displaystyle \lim\limits _{x \rightarrow + 0} {h ( x ) = \lim\limits _{x \rightarrow + 0} {f ( x ) g ( x )}}$ $\displaystyle = \lim\limits _{x \rightarrow + 0} {f ( x ) \lim\limits _{x \rightarrow + 0} {g ( x ) = b \times 1 = b}}$ $\displaystyle \lim\limits _{x \rightarrow - 0} {h ( x ) = \lim\limits _{x \rightarrow - 0} {f ( x ) g ( x )}}$ $\displaystyle = \lim\limits _{x \rightarrow - 0} {f ( x ) \lim\limits _{x \rightarrow - 0} {g ( x ) = b \times ( - 1 ) = - b}}$ 따라서, $b = - b$ 이므로 $b = 0$ (b) $x = 2$에서 연속이어야 하므로 $h ( 2 ) = f ( 2 ) g ( 2 ) = ( 4 + 2 a ) \times 1 = 4 + 2 a$ $\displaystyle \lim\limits _{x \rightarrow 2 + 0} {h ( x ) = \lim\limits _{x \rightarrow 2 + 0} {f ( x ) g ( x )}}$ $\displaystyle = \lim\limits _{x \rightarrow 2 + 0} {f ( x ) \lim\limits _{x \rightarrow 2 + 0} {g ( x ) = ( 4 + 2 a ) \times 1 = 4 + 2 a}}$ $\displaystyle \lim\limits _{x \rightarrow 2 - 0} {h ( x ) = \lim\limits _{x \rightarrow 2 - 0} {f ( x ) g ( x )}}$ $\displaystyle = \lim\limits _{x \rightarrow 2 - 0} {f ( x ) \lim\limits _{x \rightarrow 2 - 0} {g ( x ) = ( 4 + 2 a ) \times ( - 1 ) = - 4 - 2 a}}$ 따라서, $4 + 2 a = - 4 - 2 a$ 이므로 $a = - 2$ (a), (b)에 의하여 $f ( x ) = x ^{2} - 2 x$ 이므로 $f ( 5 ) = 15$