극한 존재와 사차함수

$f \left( x \right) = 0$에서 $x = 1$ 또는 $x = 2$이고, 모든 실수 $a$에 대하여 $\displaystyle \lim\limits _{x \rightarrow a} \frac{g \left( x \right) \times \left| f \left( x \right) \right|}{f \left( x \right)}$의 값이 존재한다. (ⅰ) $a = 1$일 때 $\displaystyle \lim\limits _{x \rightarrow 1 -} \frac{g \left( x \right) \times \left| f \left( x \right) \right|}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1 -} \frac{g \left( x \right) f \left( x \right)}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1 -} g \left( x \right)$ $\displaystyle \lim\limits _{x \rightarrow 1 +} \frac{g \left( x \right) \times \left| f \left( x \right) \right|}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1 +} \frac{g \left( x \right) \left\{ - f \left( x \right) \right\}}{f \left( x \right)}$$\displaystyle = - \lim\limits _{x \rightarrow 1 +} g \left( x \right)$ 에서 $\displaystyle \lim\limits _{x \rightarrow 1 -} g \left( x \right)$$\displaystyle = - \lim\limits _{x \rightarrow 1 +} g \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 1} g \left( x \right)$ $\therefore$ $\displaystyle \lim\limits _{x \rightarrow 1} g \left( x \right)$$= 0$ (ⅱ) $a = 2$일 때 $\displaystyle \lim\limits _{x \rightarrow 2 -} \frac{g \left( x \right) \times \left| f \left( x \right) \right|}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 -} \frac{g \left( x \right) \left\{ - f \left( x \right) \right\}}{f \left( x \right)}$$\displaystyle = - \lim\limits _{x \rightarrow 2 -} g \left( x \right)$ $\displaystyle \lim\limits _{x \rightarrow 2 +} \frac{g \left( x \right) \times \left| f \left( x \right) \right|}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{g \left( x \right) f \left( x \right)}{f \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} g \left( x \right)$ 에서 $\displaystyle - \lim\limits _{x \rightarrow 2 -} g \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 2 +} g \left( x \right)$$\displaystyle = \lim\limits _{x \rightarrow 2} g \left( x \right)$ $\therefore$ $\displaystyle \lim\limits _{x \rightarrow 2} g \left( x \right)$$= 0$ $g \left( x \right)$는 최고차항의 계수가 $1$인 사차함수이므로 (ⅰ), (ⅱ)에 의해 $g \left( 1 \right) = g \left( 2 \right) = 0$ 따라서 $g \left( x \right)$$= \left( x - 1 \right) \left( x - 2 \right) h \left( x \right)$$= f \left( x \right) h \left( x \right)$ 라 놓을 수 있다. 모든 실수 $a$에 대하여 $\displaystyle \lim\limits _{x \rightarrow a} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$의 값이 존재해야하고 $\displaystyle \lim\limits _{x \rightarrow a} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow a} \frac{\left| f \left( x \right) h \left( x \right) - f \left( x \right) \right|}{f \left( x \right) h \left( x \right)}$ $\displaystyle = \lim\limits _{x \rightarrow a} \frac{\left| f \left( x \right) \right| \left| h \left( x \right) - 1 \right|}{f \left( x \right) h \left( x \right)}$ 이므로 (ⅰ) $a = 1$일 때 $\displaystyle \lim\limits _{x \rightarrow 1 -} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1 -} \frac{f \left( x \right) \left| h \left( x \right) - 1 \right|}{f \left( x \right) h \left( x \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 1 -} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ $\displaystyle \lim\limits _{x \rightarrow 1 +} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 1 +} \frac{- f \left( x \right) \left| h \left( x \right) - 1 \right|}{f \left( x \right) h \left( x \right)}$ $\displaystyle = - \lim\limits _{x \rightarrow 1 +} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ 에서 $\displaystyle \lim\limits _{x \rightarrow 1 -} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$$\displaystyle = - \lim\limits _{x \rightarrow 1 +} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ $\cdots$$\cdots$ ㉠ $h \left( 1 \right) = 0$이면 ㉠의 값이 존재하지 않으므로 $h \left( 1 \right) \ne 0$ $\therefore$ $h \left( 1 \right) = 1$ (ⅱ) $a = 2$일 때 $\displaystyle \lim\limits _{x \rightarrow 2 -} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 -} \frac{- f \left( x \right) \left| h \left( x \right) - 1 \right|}{f \left( x \right) h \left( x \right)}$ $\displaystyle = - \lim\limits _{x \rightarrow 2 -} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ $\displaystyle \lim\limits _{x \rightarrow 2 +} \frac{\left| g \left( x \right) - f \left( x \right) \right|}{g \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{f \left( x \right) \left| h \left( x \right) - 1 \right|}{f \left( x \right) h \left( x \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 1 +} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ 에서 $\displaystyle - \lim\limits _{x \rightarrow 2 -} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{\left| h \left( x \right) - 1 \right|}{h \left( x \right)}$ $\cdots$$\cdots$ ㉡ $h \left( 2 \right) = 0$이면 ㉡의 값이 존재하지 않으므로 $h \left( 2 \right) \ne 0$ $\therefore$ $h \left( 2 \right) = 1$ (ⅰ), (ⅱ)에서 $h \left( 1 \right) = 1$, $h \left( 2 \right) = 1$이고 $h \left( x \right)$는 최고차항의 계수가 $1$인 이차식이므로 $h \left( x \right) = \left( x - 1 \right) \left( x - 2 \right) + 1$ $g \left( x \right)$$= f \left( x \right) h \left( x \right)$에 대입하면 $g \left( x \right)$$= \left( x - 1 \right) \left( x - 2 \right) \left\{ \left( x - 1 \right) \left( x - 2 \right) + 1 \right\}$ $= \left( x - 1 \right) ^{2} \left( x - 2 \right) ^{2} + \left( x - 1 \right) \left( x - 2 \right)$ $\therefore$ $g \left( - 1 \right) = 36 + 6 = 42$