夹逼定理简介

本文介绍了极限的基本概念及重要极限公式,如当x趋近于0时,正弦x/x的极限为1,(1+x)^(1/x)的极限为e。此外,还详细讲解了夹逼定理的应用,通过两个实例展示了如何利用夹逼定理求解特定类型的极限问题。

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极限知识

  • lim⁡x→0sin⁡xx=1,lim⁡x→0(1+x)1x=e\lim\limits_{x\rightarrow 0} \frac{\sin x}{x}=1,\lim\limits_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=ex0limxsinx=1,x0lim(1+x)x1=e
  • x→0x\rightarrow 0x0
    (1) x∼sin⁡x∼tan⁡x∼arcsin⁡x∼arctan⁡x∼ln(1+x)∼ex−1x\sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x \sim ln(1+x) \sim e^x-1xsinxtanxarcsinxarctanxln(1+x)ex1
    (2)1−cos⁡x∼12x21−cosax∼a2x21-\cos x \sim \frac12x^2 \quad 1-cos^a x \sim \frac a2 x^21cosx21x21cosax2ax2
    (3)1+x−1∼12x(1+x)a−1∼ax\sqrt{1+x}-1\sim \frac12x \quad (1+x)^a-1 \sim ax1+x121x(1+x)a1ax

夹逼定理

若在x0x_0x0的邻域内,恒有g(x)≤f(x)≤h(x)g(x)\leq f(x)\leq h(x)g(x)f(x)h(x),且lim⁡x→x0g(x)=lim⁡x→x0h(x)=A\lim\limits_{x\rightarrow x_0} g(x)=\lim\limits_{x\rightarrow x_0}h(x)=Axx0limg(x)=xx0limh(x)=A,则lim⁡x→x0f(x)=A\lim\limits_{x\rightarrow x_0}f(x)=Axx0limf(x)=A

例1:

lim⁡x→∞(1n2+1+2n2+2+⋯+nn2+n)\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})xlim(n2+11+n2+22++n2+nn)

解:∵lim⁡x→∞(1n2+n+2n2+n+⋯+nn2+n)\because \lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+n}+\dfrac{2}{n^2+n}+\dots+\dfrac{n}{n^2+n})xlim(n2+n1+n2+n2++n2+nn)

≤lim⁡x→∞(1n2+1+2n2+2+⋯+nn2+n)\qquad \leq \lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})xlim(n2+11+n2+22++n2+nn)

≤lim⁡x→∞(1n2+1+2n2+1+⋯+nn2+1)\qquad \leq\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+1}+\dots+\dfrac{n}{n^2+1})xlim(n2+11+n2+12++n2+1n)

左侧:lim⁡x→∞(1n2+n+2n2+n+⋯+nn2+n)=lim⁡x→∞n2+n2n2+n=12\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+n}+\dfrac{2}{n^2+n}+\dots+\dfrac{n}{n^2+n})=\lim\limits_{x\rightarrow \infty}\dfrac{\frac{n^2+n}{2}}{n^2+n}=\dfrac12xlim(n2+n1+n2+n2++n2+nn)=xlimn2+n2n2+n=21

右侧:lim⁡x→∞(1n2+1+2n2+1+⋯+nn2+1)=lim⁡x→∞n2+n2n2+1=12\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+1}+\dots+\dfrac{n}{n^2+1})=\lim\limits_{x\rightarrow \infty}\dfrac{\frac{n^2+n}{2}}{n^2+1}=\dfrac12xlim(n2+11+n2+12++n2+1n)=xlimn2+12n2+n=21

∴lim⁡x→∞(1n2+1+2n2+2+⋯+nn2+n)=12\therefore\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})=\dfrac12xlim(n2+11+n2+22++n2+nn)=21

例2:

lim⁡x→∞(1n2+1+1n2+2+⋯+1n2+n)\lim\limits_{x\rightarrow \infty}(\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})xlim(n2+11+n2+21++n2+n1)

解:∵nn2+n≤(1n2+1+1n2+2+⋯+1n2+n)≤nn2+1\because\dfrac{n}{\sqrt{n^2+n}}\leq (\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})\leq \dfrac{n}{\sqrt{n^2+1}}n2+nn(n2+11+n2+21++n2+n1)n2+1n

左侧:lim⁡x→∞nn2+n=lim⁡x→∞11+1n=1\lim\limits_{x\rightarrow \infty}\dfrac{n}{\sqrt{n^2+n}}=\lim\limits_{x\rightarrow \infty}\dfrac{1}{\sqrt{1+\dfrac{1}{n}}}=1xlimn2+nn=xlim1+n11=1

右侧:lim⁡x→∞nn2+1=lim⁡x→∞11+1n2=1\lim\limits_{x\rightarrow \infty}\dfrac{n}{\sqrt{n^2+1}}=\lim\limits_{x\rightarrow \infty}\dfrac{1}{\sqrt{1+\dfrac{1}{n^2}}}=1xlimn2+1n=xlim1+n211=1

∴lim⁡x→∞(1n2+1+1n2+2+⋯+1n2+n)=1\therefore \lim\limits_{x\rightarrow \infty}(\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})=1xlim(n2+11+n2+21++n2+n1)=1

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