引理

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

开证

$${f}(x) = x ^ a \enspace (a\in\mathbb{R})$$ $${f}'(x) = \lim\limits_{\Delta{x} \to 0}\frac{{f}(x + \Delta{x}) - {f}(x)}{\Delta{x}}\\ = \lim\limits_{\Delta{x} \to 0}\frac{(x + \Delta{x})^a - x^a}{\Delta{x}}\\ = x^{a-1}\lim\limits_{\Delta{x} \to 0}\frac{(1 + \frac{\Delta{x}}{x})^a - 1}{\frac{\Delta{x}}{x}}\\ = x^{a-1}\lim\limits_{\Delta{x} \to 0}\frac{(1 + \frac{\Delta{x}}{x})^a - 1}{\log_a (1 + \frac{\Delta{x}}{x})}\cdot\frac{\log_a (1 + \frac{\Delta{x}}{x})}{\frac{\Delta{x}}{x}}\\ $$$$令 t = (1 + \frac{\Delta{x}}{x})^a - 1$$ $$\therefore 1 + \frac{\Delta{x}}{x} = (t + 1)^{\frac{1}{a}} $$$$当 \Delta{x} \to 0 时 t \to 0$$ $$\therefore 原式 = x^{a-1}\lim\limits_{t \to 0}\frac{t}{\log_a (t + 1)^{\frac{1}{a}}}\cdot\lim\limits_{\Delta{x} \to 0}\frac{x}{\Delta{x}}\log_a (1 + \frac{\Delta{x}}{x})\\ = x^{a-1}\lim\limits_{t \to 0}\frac{a}{\log_a (t + 1)^{\frac{1}{t}}}\cdot\lim\limits_{\Delta{x} \to 0}\log_a (1 + \frac{\Delta{x}}{x})^{\frac{x}{\Delta{x}}}\\ = x^{a-1} \cdot a\ln a \cdot \frac{1}{\ln a}\\ = ax^{a-1}\\ $$