로그 극한

[출제의도] 함수의 극한의 성질을 이해하고 극한값을 구한다. $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{\ln \left\{ 1 + f ( 2 x ) \right\}}{x}} = 10$이고, $\displaystyle \lim\limits _{x \rightarrow 0} {} x = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} {\ln \left\{ 1 + f ( 2 x ) \right\}} = 0$ $\displaystyle \lim\limits _{x \rightarrow 0} {f ( 2 x )} = 0$이므로 $t = f ( 2 x )$로 놓으면 $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{\ln \left\{ 1 + f ( 2 x ) \right\}}{f ( 2 x )}} = \lim\limits _{t \rightarrow 0} {\frac{\ln ( 1 + t )}{t}}$ $\displaystyle = \lim\limits _{t \rightarrow 0} {\ln ( 1 + t ) ^{\frac{1}{t}}} = \ln e = 1$ 따라서 $x = 2 y$로 놓으면 $\displaystyle \lim\limits _{x \rightarrow 0} {\frac{f ( x )}{x}}$$\displaystyle = \lim\limits _{y \rightarrow 0} {\frac{f ( 2 y )}{2 y}}$ $\displaystyle = \lim\limits _{y \rightarrow 0} {\frac{f ( 2 y )}{\ln \left\{ 1 + f ( 2 y ) \right\}} \cdot \frac{\ln \left\{ 1 + f ( 2 y ) \right\}}{y}} \cdot \frac{1}{2}$ $\displaystyle = 1 \times 10 \times \frac{1}{2}$$= 5$