# 关于自然常数#

$e=1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}+\cdots=\sum_{n=0}^{\infty}\frac{1}{n!}\tag{1}$

## 极限证明#

$$e_n=\left(1+\frac{1}{n}\right)^n$$ ，根据二项式定理，可得：

$e_n=1+\frac{n}{1!}\frac{1}{n}+\frac{n(n-1)}{2!}\frac{1}{n^2}+\frac{n(n-1)(n-2)}{3!}\frac{1}{n^3}+\cdots+\frac{1}{n^n}$

$$a_{n,k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\frac{1}{n^k}$$

$$e_n=1+\sum_{k=1}^na_{n,k}$$

$a_{n,k}\lt{a_{n+1,k}}\lt\frac{1}{k!}$

$e_n\lt{e_{n+1}}\lt 1+\sum_{k=1}^{\infty}\frac{1}{k!}=e$

$$\{e_n\}$$ 是单调递增数列，且 $$\lim_{n\to\infty}e_n\le e$$

$\lim_{n\to\infty}e_n\ge\lim_{n\to\infty}\left(1+\sum_{k=1}^ma_{n,k}\right)=1+\sum_{k=1}^m\frac{1}{k!}$

$$\lim_{n\to\infty}e_n=e$$ ，即 $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ 成立

## 指数函数 $$e^x$$#

$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n \tag{2}$

$e=\lim_{n\to+\infty}\left(1+\frac{1}{t}\right)^t \tag{3}$

$\left(1+\frac{1}{n+1}\right)^n\lt\left(1+\frac{1}{t}\right)^t\lt\left(1+\frac{1}{n}\right)^{n+1}$

$$t\to+\infty$$ 时，$$n\to+\infty$$ ，并且 $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+1}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)=e$$

$e=\lim_{n\to+\infty}\left(1-\frac{1}{t}\right)^{-t} \tag{4}$

$$\frac{1}{1-\frac{1}{t}}=1+\frac{1}{s}$$ ，则 $$s=t-1$$ 。当 $$t\to+\infty$$ 时，$$s\to+\infty$$ ，得：

$\lim_{n\to+\infty}\left(1-\frac{1}{t}\right)^{-t}=\lim_{s\to+\infty}\left(1+\frac{1}{s}\right)^{s+1}=e$

$$x\gt0$$ 时，令 $$s=tx$$

$e^x=\lim_{t\to+\infty}\left(1+\frac{1}{t}\right)^{tx}=\lim_{s\to+\infty}\left(1+\frac{x}{s}\right)^s=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$

$$x\lt0$$ 时，令 $$x=-y, s=ty$$ ，则 $$\frac{1}{t}=\frac{y}{s}$$

$e^x=\lim_{t\to+\infty}\left(1-\frac{1}{t}\right)^{-tx}=\lim_{s\to+\infty}\left(1-\frac{y}{s}\right)^s=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$

$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} \tag{5}$

## 对数函数的导数#

$\frac{1}{h}(\log_a(x+h)-\log_ax)=\frac{1}{h}\log_a\left(\frac{x+h}{x}\right)=\log_a\left(\frac{x+h}{x}\right)^{1/h}$

$$s=\frac{h}{x}$$ ，则上式变化为：

$\frac{1}{h}(\log_a(x+h)-\log_ax)=\log_a(1+s)^{\frac{1}{sx}}=\frac{1}{x}\log_a(1+s)^{\frac{1}{s}}$

$\lim_{h\to0}\frac{1}{h}(\log_a(x+h)-\log_ax)=\frac{1}{x}\lim_{s\to0}\log_a(1+s)^{\frac{1}{s}}=\frac{1}{x}\log_ae$

$\frac{d}{dx}\log_ax=(\log_ae)\frac{1}{x} \tag{6}$

$\frac{d}{dx}\ln x=\frac{1}{x} \tag{7}$

## 参考文献#

1. 微积分入门(I)一元微积分. [日]小平邦彦. 北京：人民邮电出版社，2008.4.第1版
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