几个等价无穷小及证明

$\sin x\sim x$

这是最重要的等价无穷小，后文其它几个等价无穷小和$\sin x$的导数推导均需要用到$\sin x\sim x$。下面引用知友@半个冯博士的简洁证明。

单位圆

$$S_{\triangle AOB} < S_{扇AOB} < S_{\triangle AOD}$$

代入面积公式：

$$\frac{1}{2} \sin x < \frac{1}{2} x < \frac{1}{2} \tan x$$

整理：

$$1<\frac{x}{\sin x} < \frac{1}{\cos x} \quad\left(0 < x < \frac{\pi}{2}\right)$$

$$\cos x < \frac{\sin x}{x} < 1 \quad\left(0 < x < \frac{\pi}{2}\right)$$

根据夹逼准则，两边取极限：

$$1=\lim_{x \rightarrow 0} \cos x ≤ \lim_{x \rightarrow 0} \frac{\sin x}{x} ≤ 1$$

$$\therefore \lim_{x \rightarrow 0} \frac{\sin x}{x}=1$$

当$x$为负时同理。

$\tan x \sim x$

$$\lim\limits_{x\to 0 } \dfrac{\tan x}{x}=\lim\limits_{x \to 0 } \dfrac{\sin x}{x\cos x}=\lim \limits_{x\to 0 } \dfrac{\sin x}{x} \cdot \lim \limits_{x \to 0 } \dfrac{1}{\cos x}=1$$

$1-\cos x \sim \dfrac{1}{2}x^2$

$$\lim \limits_{x \to 0 } \dfrac{1-\cos x}{x^{2}}=\lim \limits_{x \to 0 } \dfrac{2 \sin ^{2} \frac{x}{2}}{4 \cdot\left(\frac{x}{2}\right)^{2}}=\dfrac{1}{2} \cdot\left(\lim \limits_{x \to 0 } \dfrac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}=\dfrac{1}{2}$$

另解:

$$\lim \limits_{x \to 0 } \dfrac{1-\cos x}{x^{2}}=\lim \limits_{x \to 0 }\left(\dfrac{\sin ^{2} x}{x^{2}} \cdot \dfrac{1}{1+\cos x}\right)=\dfrac{1}{2}$$

$\arcsin x \sim x$

令$t=\arcsin x$，则$x=\sin t$

$\lim \limits_{x\rightarrow0} \dfrac{\arcsin x}{x}=\lim \limits_{t\rightarrow0}\dfrac{t}{\sin t}=\lim \limits_{t\rightarrow0}\dfrac{1}{\dfrac{\sin t}{t}}=1$

$\arctan x \sim x$

令$t=\arctan x$, 则$x=\tan t$，当$x \rightarrow 0$ 时，$t \rightarrow 0$.

$$\lim \limits_{x \to 0 } \dfrac{\arctan x}{x}=\lim \limits_{t \to 0 } \dfrac{t}{\tan t}=\lim \limits_{t \to 0 } \dfrac{t \cos t}{\sin t}=\lim \limits_{t \to 0 } \dfrac{t}{\sin t} \cdot \lim \limits_{t \to 0 } \cos t=1$$

$\sec x-1 \sim \dfrac{x^{2}}{2}$

$$\lim \limits_{x \to 0 } \dfrac{\dfrac{1}{\cos x}-1}{x^{2}}=\lim \limits_{x \to 0 } \dfrac{1-\cos x}{x^{2} \cdot \cos x}=\dfrac{\dfrac{1}{2} x^{2}}{x^{2} \cdot \cos x}=\dfrac{1}{2}$$

$\tan x-\sin x \sim \dfrac{1}{2} x^{3}$

$$\lim \limits_{x \to 0 } \dfrac{\tan x-\sin x}{\dfrac{1}{2} x^{3}}=\lim \limits_{x \to 0 } \dfrac{\tan x}{x} \cdot \dfrac{1-\cos x}{\dfrac{1}{2} x^{2}}=\lim \limits_{x \to 0 } \dfrac{\tan x}{x} \cdot \lim \limits_{x \to 0 } \dfrac{1-\cos x}{\dfrac{1}{2} x^{2}}=1$$