e 정의 극한 보기

ㄱ. $\displaystyle \lim\limits _{x \rightarrow \infty} {f ( x )}$$\displaystyle = \lim\limits _{x \rightarrow \infty} \left( \frac{x}{x - 1} \right) ^{x}$$\displaystyle = \lim\limits _{x \rightarrow \infty} \left( 1 + \frac{1}{x - 1} \right) ^{x}$ $\displaystyle = \lim\limits _{x \rightarrow \infty} \left\{ \left( 1 + \frac{1}{x - 1} \right) ^{x-1} \right\} ^{\frac{x}{x - 1}}$ 그런데, $\displaystyle \lim\limits _{x \rightarrow \infty} \left( 1 + \frac{1}{x - 1} \right) ^{x-1} = e$, $\displaystyle \lim\limits _{x \rightarrow \infty} {\frac{x}{x - 1}} = 1$이므로 $\displaystyle \lim\limits _{x \rightarrow \infty} \left\{ \left( 1 + \frac{1}{x - 1} \right) ^{x-1} \right\} ^{\frac{x}{x - 1}} = e ^{1} = e$ ∴ 참 ㄴ. $\displaystyle \lim\limits _{x \rightarrow \infty} {f ( x )} = e$ 이므로 $x + 1 = t$라 하면 $\displaystyle \lim\limits _{x \rightarrow \infty} {f ( x + 1 )} = \lim\limits _{t \rightarrow \infty} {f ( t )} = e$ 수렴하는 함수의 곱은 수렴하므로 $\displaystyle \lim\limits _{x \rightarrow \infty} {} f ( x ) \cdot f ( x + 1 )$$\displaystyle = \lim\limits _{x \rightarrow \infty} {f ( x )} \times \lim\limits _{x \rightarrow \infty} {f ( x + 1 )}$ $= e \times e = e ^{2}$ ∴ 참 ㄷ. $\displaystyle \lim\limits _{x \rightarrow \infty} {f ( kx )}$$\displaystyle = \lim\limits _{x \rightarrow \infty} \left( \frac{kx}{kx - 1} \right) ^{kx}$$\displaystyle = \lim\limits _{x \rightarrow \infty} \left( 1 + \frac{1}{kx - 1} \right) ^{kx}$ $\displaystyle = \lim\limits _{x \rightarrow \infty} \left\{ \left( 1 + \frac{1}{kx - 1} \right) ^{kx-1} \right\} ^{\frac{kx}{kx - 1}}$ 그런데, $x \rightarrow \infty$이면 $kx \rightarrow \infty$ 이므로 $\displaystyle \lim\limits _{x \rightarrow \infty} \left( 1 + \frac{1}{kx - 1} \right) ^{kx-1} = e$ $\displaystyle \lim\limits _{x \rightarrow \infty} {} \frac{kx}{kx - 1} = 1$ 이므로 $\displaystyle \therefore \lim\limits _{x \rightarrow \infty} {} \left\{ \left( 1 + \frac{1}{kx - 1} \right) ^{kx-1} \right\} ^{\frac{kx}{kx - 1}} = e ^{1} = e$ ∴ 거짓 따라서, 옳은 것은 ㄱ, ㄴ이다.