분수꼴 함수의 극한

$\displaystyle \lim\limits _{x \rightarrow 3} {} g \left( x \right) = g \left( 3 \right) - 1$이므로 함수 $g \left( x \right)$는 $x = 3$에서 불연속이다. 함수 $f \left( x \right)$는 최고차항의 계수가 $1$인 삼차함수이다. $f \left( k \right) \ne 0$인 임의의 실수 $k$에 대하여 $\displaystyle \lim\limits _{x \rightarrow k} {} g \left( x \right) = \lim\limits _{x \rightarrow k} {} \frac{f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\}}{f \left( x \right)} = \frac{f \left( k + 3 \right) \left\{ f \left( k \right) + 1 \right\}}{f \left( k \right)}$ $\displaystyle \lim\limits _{x \rightarrow k} {} g \left( x \right) = g \left( k \right)$이므로 함수 $g \left( x \right)$는 $x = k$에서 연속이다. $\therefore$$f \left( 3 \right) = 0$ $g \left( 3 \right) = 3$이므로 $\displaystyle \lim\limits _{x \rightarrow 3} {} g \left( x \right) = 2$이다. $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\}}{f \left( x \right)} = 2$ $x \rightarrow 3$일 때 (분모)$\rightarrow 0$이고 극한값이 존재하므로 (분자)$\rightarrow 0$이다. $\displaystyle \lim\limits _{x \rightarrow 3} {} \left[ f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\} \right] = 0$ $f \left( 6 \right) = 0$ 또는 $f \left( 3 \right) = - 1$ $\therefore$$f \left( 6 \right) = 0$ $f \left( 3 \right) = 0$, $f \left( 6 \right) = 0$이므로 $f \left( x \right) = \left( x - 3 \right) \left( x - 6 \right) \left( x + a \right)$($a$는 상수) 라 하자. $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\}}{f \left( x \right)} = 2$에서 $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\}}{f \left( x \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 3} {} \frac{x \left( x - 3 \right) \left( x + a + 3 \right) \left\{ \left( x - 3 \right) \left( x - 6 \right) \left( x + a \right) + 1 \right\}}{\left( x - 3 \right) \left( x - 6 \right) \left( x + a \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 3} {} \frac{x \left( x + a + 3 \right) \left\{ \left( x - 3 \right) \left( x - 6 \right) \left( x + a \right) + 1 \right\}}{\left( x - 6 \right) \left( x + a \right)}$ (ⅰ) $a = - 3$일 때 $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{x \left( x + a + 3 \right) \left\{ \left( x - 3 \right) \left( x - 6 \right) \left( x + a \right) + 1 \right\}}{\left( x - 6 \right) \left( x + a \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 3} {} \frac{x ^{2} \left\{ \left( x - 3 \right) ^{2} \left( x - 6 \right) + 1 \right\}}{\left( x - 6 \right) \left( x - 3 \right)}$ $x \rightarrow 3$일 때 (분모)$\rightarrow 0$이고, (분자)$\rightarrow 9$이므로 $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{f \left( x + 3 \right) \left\{ f \left( x \right) + 1 \right\}}{f \left( x \right)}$의 값이 존재하지 않는다. (ⅱ) $a \ne - 3$일 때 $\displaystyle \lim\limits _{x \rightarrow 3} {} \frac{x \left( x + a + 3 \right) \left\{ \left( x - 3 \right) \left( x - 6 \right) \left( x + a \right) + 1 \right\}}{\left( x - 6 \right) \left( x + a \right)}$ $\displaystyle = \frac{3 \left( a + 6 \right)}{- 3 \left( a + 3 \right)}$ $\displaystyle = - \frac{a + 6}{a + 3}$ $\displaystyle - \frac{a + 6}{a + 3} = 2$이므로 $a = - 4$ (ⅰ), (ⅱ)에 의하여 함수 $f \left( x \right)$는 $f \left( x \right) = \left( x - 3 \right) \left( x - 6 \right) \left( x - 4 \right)$ $f \left( 5 \right) \ne 0$이므로 $\displaystyle g \left( 5 \right) = \frac{f \left( 8 \right) \left\{ f \left( 5 \right) + 1 \right\}}{f \left( 5 \right)} = 20$