# Quod Erat Demonstrandum

## 2010/05/11

### 無聊中四數學課紀錄：(1 + 1/n)^n

Filed under: NSS,Teaching — johnmayhk @ 10:57 下午
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$e = \displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n$

(a) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{2n})^{2n} = ?$

(b) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^{2n} = ?$

(c) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{2n})^{n} = ?$

(d) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^{2} = ?$

(e) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{2})^{n} = ?$

(f) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n + 1})^{n - 1} = ?$

(g) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{3}{2n})^{n} = ?$

(h) $\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n - 1})^{3 - 4n} = ?$

[method 1]

$\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n + 1})^{n - 1}$
$= \displaystyle\lim_{n \rightarrow \infty}\frac{(1 + 1/(n + 1))^{n + 1}}{(1 + 1/(n + 1))^2}$ … … … (1)
$= \frac{\displaystyle\lim_{n \rightarrow \infty}(1 + 1/(n + 1))^{n + 1}}{\displaystyle\lim_{n \rightarrow \infty}(1 + 1/(n + 1))^2}$ … … … (2)
$= \frac{e}{1} = e$

[method 2]

$\displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n + 1})^{n - 1}$
$= \displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n + 1})^{(n + 1)(n - 1)/(n + 1)}$
$= \displaystyle\lim_{n \rightarrow \infty}[(1 + \frac{1}{n + 1})^{n + 1}]^{(n - 1)/(n + 1)}$ … … … (3)
$= \displaystyle\lim_{n \rightarrow \infty}[(1 + \frac{1}{n + 1})^{n + 1}]^{\displaystyle\lim_{n \rightarrow \infty}(n - 1)/(n + 1)}$ … … … (4)
$= e^1 = e$

$\displaystyle\lim_{n \rightarrow \infty}(-1)^{\frac{1}{n}} = \displaystyle\lim_{n \rightarrow \infty}(-1)^{\displaystyle\lim_{n \rightarrow \infty}\frac{1}{n}} = (-1)^0 = 1$

$e = \displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n$

1. n 不一定是正整數，n 是實數也可。
2. n 不只趨向正無窮，趨向負無窮也可，即 $\displaystyle\lim_{n \rightarrow -\infty}(1 + \frac{1}{n})^n$ 也是 e。

$\displaystyle\lim_{x \rightarrow 0}\frac{e^x - 1}{x} = 1$

$\displaystyle\lim_{y \rightarrow 0}(1 + y)^{\frac{1}{y}}$

$z = \frac{1}{y}$

$\displaystyle\lim_{y \rightarrow 0}(1 + y)^{\frac{1}{y}}$
$= \displaystyle\lim_{z \rightarrow \infty}(1 + \frac{1}{z})^z$ … … … … (5)
$= e$ … … … … (6)

$\displaystyle\lim_{y \rightarrow 0}(1 + y)^{\frac{1}{y}}$

$e = \displaystyle\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n$

（答曰：$\displaystyle\lim_{y \rightarrow 0+} z \rightarrow +\infty$$\displaystyle\lim_{y \rightarrow 0-} z \rightarrow -\infty$，一言敝之曰：does not exist）

$\displaystyle\lim_{x \rightarrow 0}\frac{e^x - 1}{x}$
$= \displaystyle\lim_{x \rightarrow 0}(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ... - 1)/x$
$= \displaystyle\lim_{x \rightarrow 0}(x + \frac{x^2}{2} + \frac{x^3}{6} + ...)/x$
$= \displaystyle\lim_{x \rightarrow 0}(1 + \frac{x}{2} + \frac{x^2}{6} + ...)$
$= 1$

（因除第一項，以後各項皆趨於零）

$\displaystyle\lim_{n \rightarrow \infty}\frac{1 + 2 + 3 + \dots + n}{n^2} \neq 0 + 0 + 0 + \dots + 0 = 0$（正確極值是 $\frac{1}{2}$，不是零。）

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$

$e^x = \frac{x^0}{0!} + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$