지수와 삼각 극한 보기

$\displaystyle \lim\limits _{x \rightarrow 0} \frac{f ( x )}{x} = \lim\limits _{x \rightarrow 0} \frac{e ^{-x} \sin x + g ( x )}{x} = 1$에서 $\displaystyle \lim\limits _{x \rightarrow 0} \frac{e ^{-x} \sin x}{x} = 1$, $\displaystyle \lim\limits _{x \rightarrow 0} \frac{g ( x )}{x} = 0$ …… ㉠ $\displaystyle \lim\limits _{x \rightarrow \infty} \frac{f ( x )}{x ^{2}} = \lim\limits _{x \rightarrow \infty} \frac{e ^{-x} \sin x + g ( x )}{x ^{2}} = 1$에서 $\displaystyle \lim\limits _{x \rightarrow \infty} \frac{e ^{-x} \sin x}{x ^{2}} = 0$,$\displaystyle \lim\limits _{x \rightarrow \infty} \frac{g ( x )}{x ^{2}} = 1$ …… ㉡ $g ( x )$는 다항함수이므로 ㉠, ㉡에 의해 $g ( x ) = x ^{2}$ ㄱ. $g ( x ) = x ^{2}$이므로 $g ( 0 ) = 0$ $\therefore$ 참 ㄴ. $\displaystyle \lim\limits _{x \rightarrow \infty} \frac{g ( x )}{x ^{2}}$$\displaystyle = \lim\limits _{x \rightarrow \infty} \frac{x ^{2}}{x ^{2}} = 1$ $\therefore$ 참 ㄷ. $\displaystyle \lim\limits _{x \rightarrow \infty} \frac{f ( x )}{g ( x )} = \lim\limits _{x \rightarrow 0} \frac{e ^{-x} \sin x + x ^{2}}{x ^{2}} =$(발산) $\therefore$ 거짓 따라서, 옳은 것은 ㄱ, ㄴ이다.