곱의 미분법

[출제의도] 곱의 미분법을 활용하여 문제 해결하기 $\displaystyle \lim\limits _{x \rightarrow 2} {} \frac{f \left( x \right) - 4}{x ^{2} - 4}$$= 2$에서 (분모)$\rightarrow$$0$이고 극한값이 존재하므로 (분자)$\rightarrow$$0$ $\displaystyle \lim\limits _{x \rightarrow 2} {} \left\{ f \left( x \right) - 4 \right\}$$= 0$이므로 $f \left( 2 \right)$$= 4$ $\displaystyle \lim\limits _{x \rightarrow 2} {} \frac{f \left( x \right) - 4}{x ^{2} - 4}$$\displaystyle = \lim\limits _{x \rightarrow 2} {} \left\{ \frac{1}{x + 2} \times \frac{f \left( x \right) - f \left( 2 \right)}{x - 2} \right\}$ $\displaystyle = \frac{1}{4} f' \left( 2 \right)$$= 2$ $f' \left( 2 \right)$$= 8$ $\displaystyle \lim\limits _{x \rightarrow 2} {} \frac{g \left( x \right) + 1}{x - 2}$$= 8$에서 (분모)$\rightarrow$$0$이고 극한값이 존재하므로 (분자)$\rightarrow$$0$ $\displaystyle \lim\limits _{x \rightarrow 2} {} \left\{ g \left( x \right) + 1 \right\}$$= 0$이므로 $g \left( 2 \right)$$= \mathit{-} 1$ $\displaystyle \lim\limits _{x \rightarrow 2} {} \frac{g \left( x \right) + 1}{x - 2}$$\displaystyle = \lim\limits _{x \rightarrow 2} {} \frac{g \left( x \right) - g \left( 2 \right)}{x - 2}$$= g' \left( 2 \right)$$= 8$ $h' \left( x \right)$$= f' \left( x \right) g \left( x \right) + f \left( x \right) g' \left( x \right)$ 따라서 $h' \left( 2 \right)$$= f' \left( 2 \right) g \left( 2 \right) + f \left( 2 \right) g' \left( 2 \right)$$= 24$