극한으로 다항함수 결정

$f \left( x \right)$가 다항함수이고 $\displaystyle \lim\limits _{x \rightarrow \infty} \frac{f \left( x \right) - 2 x ^{3}}{x ^{2}} = 3$이므로 $f \left( x \right) = 2 x ^{3} + 3 x ^{2} + ax + b \left( a , b \right.$는 상수$)$ $\displaystyle \lim\limits _{x \rightarrow 0} \frac{f \left( x \right)}{x} = 3$이고 $\displaystyle \lim\limits _{x \rightarrow 0} x = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} f \left( x \right) = 0$ 그러므로 $b = 0$ $\displaystyle \lim\limits _{x \rightarrow 0} \frac{f \left( x \right)}{x}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{2 x ^{3} + 3 x ^{2} + a x}{x}$ $\displaystyle = \lim\limits _{x \rightarrow 0} {} \left( 2 x ^{2} + 3 x + a \right)$ $= a = 3$ 그러므로 $f \left( x \right) = 2 x ^{3} + 3 x ^{2} + 3 x$ 따라서 $f \left( 2 \right) = 34$