# 夹挤原理与两个重要极限（一）

$\lim_{x\to0}\frac{\sin x}{x}=1$

$A_{\Delta OAC}<A_{\frown\atop OAB}<A_{\Delta OAB}$

$A_{\Delta OAC}=\frac{1}{2}\sin x, A_{\frown \atop OAB}=\frac{1}{2}x, A_{\Delta OAB}=\frac{1}{2}\tan x$因为三角形 $$\Delta OAD$$ 的高为 $$\sin x$$，三角形 $$OAB$$的高为$$\tan x$$。所以我们有

$\sin x<x<\tan x\Longleftrightarrow \frac{1}{\sin x}>\frac{1}{x}>\frac{\cos x}{\sin x}$

$1>\frac{\sin x}{x}>\cos x$因为当 $$x\to0$$ 时，$$\cos x\to 1$$，由夹挤原理，我们证明了当 $$0<x<\frac{\pi}{2}$$ 时，$$\lim_{x\to0^+}\frac{\sin x}{x}=1$$。

\begin{align*}\lim_{x\to0}\frac{\sin 3x}{\tan 5x}&= \lim_{x\to0}\frac{\sin 3x}{3x}\cdot\frac{5x}{\tan 5x}\cdot\frac{3x}{5x}\\ &=\lim_{x\to0}\frac{\sin 3x}{3x}\cdot\frac{5x}{\sin 5x}\cdot\cos 5x\cdot\frac{3x}{5x}\\&=\frac{3}{5}\end{align*}

$\lim_{n\to\infty}2^n\sin\frac{x}{2^n}=\lim_{n\to\infty}\frac{\sin\frac{x}{2^n}}{\frac{x}{2^n}}\cdot x=x$

$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}<\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+1}}=\frac{n}{\sqrt{n^2+1}}$

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}>\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\cdots+\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}$

$\frac{n}{\sqrt{n^2+n}}<\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}<\frac{n}{\sqrt{n^2+1}}$

$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1$