곱의 미분과 극한

[출제의도] 곱의 미분법 이해하기 $g \left( x \right) = \left( x ^{2} - 2 x + 2 \right) f \left( x \right)$에서 $g \left( 2 \right) = 2 f \left( 2 \right)$ $g' \left( x \right) = \left( 2 x - 2 \right) f \left( x \right) + \left( x ^{2} - 2 x + 2 \right) f' \left( x \right)$에서 $g' \left( 2 \right) = 2 f \left( 2 \right) + 2 f' \left( 2 \right)$ $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - 1}{2 f \left( x \right) - 1}} = \mathit{-} 2$에서 $\displaystyle f \left( 2 \right) \neq \frac{1}{2}$이라 하면 $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - 1}{2 f \left( x \right) - 1}} = \frac{g \left( 2 \right) - 1}{2 f \left( 2 \right) - 1} = \frac{2 f \left( 2 \right) - 1}{2 f \left( 2 \right) - 1} = 1 \neq - 2$ 이므로 $\displaystyle f \left( 2 \right) = \frac{1}{2}$ 그러므로 $g \left( 2 \right) = 2 f \left( 2 \right) = 1$이고 $g' \left( 2 \right) = 2 f' \left( 2 \right) + 1$ $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - 1}{2 f \left( x \right) - 1}}$$\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - g \left( 2 \right)}{2 \left\{ f \left( x \right) - f \left( 2 \right) \right\}}}$$\displaystyle = \lim\limits _{x \rightarrow 2} {\frac{\frac{g \left( x \right) - g \left( 2 \right)}{x - 2}}{2 \times \frac{f \left( x \right) - f \left( 2 \right)}{x - 2}}}$ 이때 $f' \left( 2 \right) = 0$이라 하면 $g' \left( 2 \right) = 1$이므로 $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - 1}{2 f \left( x \right) - 1}}$의 값이 존재하지 않는다. 그러므로 $f' \left( 2 \right) \neq 0$ $\displaystyle \lim\limits _{x \rightarrow 2} {\frac{g \left( x \right) - 1}{2 f \left( x \right) - 1}}$$\displaystyle = \frac{g' \left( 2 \right)}{2 f' \left( 2 \right)}$$\displaystyle = \frac{g' \left( 2 \right)}{g' \left( 2 \right) - 1}$$= \mathit{-} 2$ 따라서 $\displaystyle g' \left( 2 \right) = \frac{2}{3}$