지수함수 극한

ㄱ. $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{2 x}}$$=$$\displaystyle \frac{1}{2} \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{x}} = \frac{1}{2}$ ㄴ. $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{e ^{2 x} - 1}}$$=$$\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{( e ^{x} - 1 ) ( e ^{x} + 1 )}}$ $\displaystyle = \lim\limits _{x \rightarrow 0} {\frac{1}{e ^{x} + 1}} = \frac{1}{2}$ ㄷ. $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{1 - \cos x}}$ $=$$\displaystyle \lim\limits _{x \rightarrow 0} {\frac{( e ^{x} - 1 ) ( 1 + \cos x )}{1 - \cos ^{2} x}}$ $=$$\displaystyle \lim\limits _{x \rightarrow 0} {\left( \frac{e ^{x} - 1}{x} \cdot \frac{x ^{2}}{\sin ^{2} x} \cdot \frac{1 + \cos x}{x} \right)}$ 이 때, $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{x}} = \lim\limits _{x \rightarrow 0} {\frac{x}{\sin x}} = 1$이지만 $x \rightarrow 0$일 때, $\displaystyle \frac{1 + \cos x}{x}$의 극한값은 존재하지 않는다. 따라서 $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{e ^{x} - 1}{f ( x )}}$이 존재하는 것은 ㄱ, ㄴ이다.