ln 극한 이차함수

$\displaystyle \lim\limits _{x\rightarrow 0} \frac{\ln f ( x )}{x}$가 존재하므로 $\displaystyle \lim\limits _{x\rightarrow 0} \ln f ( x ) = 0$이어야 한다. ∴ $f ( 0 ) = 1$ $\displaystyle \lim\limits _{x\rightarrow 0} \frac{\ln f ( x )}{x}$$\displaystyle = \lim\limits _{x\rightarrow 0} \left( \frac{\ln f ( x )}{f ( x ) - 1} \times \frac{f ( x ) - 1}{x} \right)$ $\displaystyle = \lim\limits _{t\rightarrow 0} \frac{\ln \left( 1 + t \right)}{t} \times \lim\limits _{x\rightarrow 0} \frac{f ( x ) - f ( 0 )}{x}$ (∵ $f ( 0 ) = 1$) $= 1 \times f' ( 0 )$$= f' ( 0 )$ ∴ $\displaystyle f' ( 1 ) = f' ( 0 ) + \frac{1}{2}$ $f ( 0 ) = 1$이므로 $f ( x ) = ax ^{2} + bx + 1$($a \ne 0$)이라 하면 $f ( 1 ) = 2$에서 $a + b = 1$ ⋯⋯ ㉠ $f' ( x ) = 2 ax + b$이므로 $\displaystyle f' ( 1 ) = f' ( 0 ) + \frac{1}{2}$에서 $\displaystyle 2 a = \frac{1}{2}$ ⋯⋯ ㉡ ㉠, ㉡에서 $\displaystyle a = \frac{1}{4}$, $\displaystyle b = \frac{3}{4}$ ∴ $\displaystyle f ( x ) = \frac{1}{4} x ^{2} + \frac{3}{4} x + 1$ ∴ $f ( 8 )$$= 16 + 6 + 1$$= 23$