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)=logax
$${f}'(x) = \lim\limits_{\Delta{x} \to 0}\frac{{f}(x + \Delta{x}) - {f}(x)}{\Delta{x}}\\
= \lim\limits_{\Delta{x} \to 0}\frac{\log_a{(x + \Delta{x})} - \log_a{x}}{\Delta{x}}\\
= \lim\limits_{\Delta{x} \to 0}\frac{\log_a{(1 +\frac{\Delta{x}}{x}})}{\Delta{x}}\\
= \lim\limits_{\Delta{x} \to 0}\frac{1}{\Delta{x}}\log_a{(1 +\frac{\Delta{x}}{x}})\\
= \lim\limits_{\Delta{x} \to 0}\frac{\log_a{(1 +\frac{\Delta{x}}{x}})^\frac{x}{{\Delta{x}}}}{x}\\
= \lim\limits_{\Delta{x} \to 0}\frac{\log_a{e}}{x}\\
= \lim\limits_{\Delta{x} \to 0}\frac{1}{x\ln a}
$$