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，\(t \rightarrow 0\).

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

\(\sec x-1 \sim \dfrac{x^{2}}{2}\)

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

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

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