극한이 존재하는 함수

$\displaystyle \lim\limits _{x \rightarrow 0} xf ( x ) = 1$에서 $xf ( x ) = h ( x )$라 하면 $\displaystyle \lim\limits _{x \rightarrow 0} h ( x ) = 1$, $\displaystyle f ( x ) = \frac{1}{x} h ( x ) ( x \ne 0 )$ $\cdots \cdots$㉠ ㄱ.$\displaystyle \lim\limits _{x \rightarrow 0} f ( x ) \cdot g ( x ) = \lim\limits _{x \rightarrow 0} \frac{1}{x} \cdot h ( x ) \cdot \sin x$ $\displaystyle = \lim\limits _{x \rightarrow 0} \frac{\sin x}{x} \cdot h ( x )$ $= 1 \times 1 = 1$ ㄴ. $\displaystyle \lim\limits _{x \rightarrow 0} f ( x ) \cdot g ( x ) = \lim\limits _{x \rightarrow 0} \frac{1}{x} \cdot h ( x ) \cdot \cos x$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{\cos x}{x} \cdot h ( x )$ $\cdots \cdots$㉡ 그런데, $\displaystyle \lim\limits _{x \rightarrow 0} \frac{\cos x}{x}$의 값이 존재하지 않으므로 ㉡의 값은 존재하지 않는다. ㄷ. $\displaystyle \lim\limits _{x \rightarrow 0} f ( x ) \cdot g ( x ) = \lim\limits _{x \rightarrow 0} \frac{1}{x} \cdot h ( x ) \cdot \ln ( x + 1 )$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{\ln ( x + 1 )}{x} \cdot h ( x )$ $= 1 \times 1$$= 1$ 따라서, $\displaystyle \lim\limits _{x \rightarrow 0} f ( x ) \cdot g ( x )$가 존재하는 것은 ㄱ, ㄷ이다.