다항함수 극한 조건

$\displaystyle \lim\limits _{x \rightarrow \infty} {\frac{f ( x ) - x ^{2}}{ax + 1}} = 2$에서 $f ( x ) = x ^{2} + 2 ax + b$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{x - 1}{f ( x )}} = \frac{1}{4}$에서 $f ( 1 ) = 0$이므로 $1 + 2 a + b = 0$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{x - 1}{f ( x )}} = \lim\limits _{x \rightarrow 1} {\frac{x - 1}{x ^{2} + 2 ax + b}}$$\displaystyle = \lim\limits _{x \rightarrow 1} {\frac{x - 1}{x ^{2} + 2 ax - 2 a - 1}} = \lim\limits _{x \rightarrow 1} {\frac{x - 1}{( x - 1 ) ( x + 2 a + 1 )}}$ $\displaystyle = \frac{1}{2 a + 2} = \frac{1}{4}$ 따라서 $a = 1 , b = - 3$ $\therefore f ( 3 ) = 12$