引理

$$\lim\limits_{x \to \infty}(1 + \frac{1}{x}) ^ x = e\\ \lim\limits_{x \to 0}\sin{x} = x\\ $$

开证

$${f}(x) = a ^ x$$ $${f}'(x) = \lim\limits_{\Delta{x} \to 0}\frac{{f}(x + \Delta{x}) - {f}(x)}{\Delta{x}}\\ = \lim\limits_{\Delta{x} \to 0}\frac{a ^ {x + \Delta{x}} - a ^ x}{\Delta{x}}\\ = a ^ x\lim\limits_{\Delta{x} \to 0}\frac{a ^ {\Delta{x}} - 1}{\Delta{x}} $$$$令 t = a ^ {\Delta{x}} - 1$$ $$\therefore \Delta{x} = \log_a{1 + t}$$ $$\therefore 1 + \frac{\Delta{x}}{x} = (t + 1)^{\frac{1}{a}} $$$$\therefore 原式 = a ^ x\lim\limits_{t \to 0}\frac{t}{\log_a{(1 + t)}}\\ = a ^ x\lim\limits_{t \to 0}\frac{1}{\frac{1}{t}\log_a{(1 + t)}}\\ = a ^ x\lim\limits_{t \to 0}\frac{1}{\log_a{(1 + t)} ^ \frac{1}{t}}\\ = a ^ x \frac{1}{\log_a{e}}\\ = a ^ x \ln a $$