MATH FACTORY

• $ e = 2.718281828459045\cdots $
• $ \displaystyle \lim_{x \rightarrow 0} (1+cx)^{\frac{1}{x}} = \lim_{x \rightarrow 0} (1+cx)^{\frac{1}{cx} \cdot c} = e^c $
• $ \displaystyle \lim_{x \rightarrow \infty} \left( 1+\frac{c}{x} \right)^{x} = \lim_{x \rightarrow \infty} \left( 1+\frac{c}{x} \right)^{\frac{x}{c} \cdot c} = e^c $

• $ \displaystyle \lim_{x \to 0} \frac{\ln (1+cx)}{x} = \lim_{x \to 0} \frac{\ln (1+cx)}{cx} \cdot c = c $
• $ \displaystyle \lim_{x \rightarrow 0} \frac{ \log_a (1+cx) }{x} = \lim_{x \rightarrow 0} \frac{ \log_a (1+cx) }{cx} \cdot c = \frac{c}{\ln a} $

• $ \displaystyle \lim_{x \to 0} \frac{\ln (1+x)}{x} = \lim_{x \to 0} \ln (1+x)^{\frac{1}{x}} = \lim_{x \to 0} \ln e = 1 $
• $ \displaystyle \lim_{x \to 0} \frac{\log_a (1+x)}{x} = \lim_{x \to 0} \frac{\ln (1+x)}{x \ln a} = \frac{1}{\ln a} $

• $ \displaystyle \lim_{x \rightarrow 0} \frac{ e^{cx} - 1 }{x} = \lim_{x \rightarrow 0} \frac{ e^{cx} - 1 }{cx} \cdot c = c $
• $ \displaystyle \lim_{x \rightarrow 0} \frac{ a^{cx} - 1 }{x} = \lim_{x \rightarrow 0} \frac{ a^{cx} - 1 }{cx} \cdot c = c \ln a $

증명

1. $ t = e^x - 1 $로 놓으면 $ x \to 0 $일 때 $ t \to 0 $이고
   \begin{gather*}
   e^x = 1 + t
   \end{gather*}양변에 자연로그를 취하면
   \begin{gather*}
   x = \ln (1+t)
   \end{gather*}이므로
   \begin{gather*}
   \lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{t \to 0} \frac{t}{\ln (1+t)} = \lim_{t \to 0} \frac{1}{\dfrac{\ln (1+t)}{t}} = 1
   \end{gather*}
2. $ t = a^x - 1 $로 놓으면 $ x \to 0 $일 때 $ t \to 0 $이고
   \begin{gather*}
   a^x = 1 + t
   \end{gather*}양변에 밑을 $ a $로 하는 로그를 취하면
   \begin{gather*}
   x = \log_a (1+t)
   \end{gather*}이므로
   \begin{gather*}
   \lim_{x \to 0} \frac{a^x - 1}{x} = \lim_{t \to 0} \frac{t}{\log_a (1+t)} = \lim_{t \to 0} \frac{1}{\dfrac{\log_a (1+t)}{t}} = \ln a
   \end{gather*}