高数强化NO4|数列极限的计算|可转化为未定式的极限|夹逼准则|运用定积分定义|单调有界准则

数列极限的计算

$$\begin{array}{rl}& \text{直接计算：转化成函数，奇偶子数列（13、14、15）}\\ & \text{夹逼准则：数值型（16）、抽象型（17、18）}\\ & \text{定积分定义：用公式（19），和夹逼相结合（20）}\\ & \text{单调有界：例21，22．}\end{array}$$

可转化为未定式的极限

$$\begin{array}{rl}& \text{解：考虑}\underset{x\to +\infty }{\lim }{\tan }^x\left(\frac{\pi }{4}+\frac{2}{x}\right)=e^{{\lim }_{x\to +\infty }x\cdot \left(\tan \left(\frac{\pi }{4}+\frac{2}{x}\right)-1\right)}\\ & \stackrel{t=\frac{1}{x}}{=}{e}^{{\lim }_{t\to 0^{+}}\frac{\tan \left(\frac{\pi }{4}+2t\right)-1}{t}}\\ & \stackrel{\text{洛必达}}{=}{e}^{{\lim }_{t\to 0^{+}}\frac{{\sec }^2\left(\frac{\pi }{4}+2t\right)\cdot 2}{1}}=e^4\end{array}$$
$$\begin{array}{rl}& \text{小结：① 函数极限存在，数列极限一定存在}\\ & (n\to \infty \text{是}x\to +\infty \text{上部分的离散点列})\\ & \\ & ②\text{数列极限存在，函数极限未必存在}\\ & \underset{n\to \infty }{\lim }\sin n\pi =\underset{n\to \infty }{\lim }0=0,\text{但}\underset{x\to +\infty }{\lim }\sin \pi x\text{不存在}\\ & \underset{n\to \infty }{\lim }{\left(\frac{n}{n+1}\right)}^{(-1{)}^n}=1,\text{但}\underset{x\to +\infty }{\lim }{\left(\frac{x}{x+1}\right)}^{(-1{)}^x}\text{不存在}\end{array}$$

$$\begin{array}{rl}& \text{解：}\underset{n\to \infty }{\lim }{\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots +\frac{1}{n}-\frac{1}{n+1}\right)}^n\\ & =\underset{n\to \infty }{\lim }{\left(1-\frac{1}{n+1}\right)}^n\\ & =e^{{\lim }_{n\to \infty }n\cdot \left(-\frac{1}{n+1}\right)}=e^{-1}\end{array}$$

$$\begin{array}{rl}& \text{解1：抓大头：}\underset{n\to \infty }{\lim }(a^{-n}+b^{-n}{)}^{\frac{1}{n}}=\underset{n\to \infty }{\lim }(a^{-n}{)}^{\frac{1}{n}}=a^{-1}(a>b>0)\\ & \\ & \text{解2：抓大头：原式}=\underset{n\to \infty }{\lim }(a^{-n}{)}^{\frac{1}{n}}{\left(1+{\left(\frac{b^{-1}}{a^{-1}}\right)}^n\right)}^{\frac{1}{n}}\\ & =a^{-1}\cdot \underset{n\to \infty }{\lim }{\left(1+{\left(\frac{b}{a}\right)}^n\right)}^{\frac{1}{n}}=a^{-1}(a>b>0)\\ & \\ & \text{拓：}\underset{x\to +\infty }{\lim }(e^x+1{)}^{\frac{1}{x}}=\underset{x\to +\infty }{\lim }(e^x{)}^{\frac{1}{x}}=e\\ & =\underset{x\to +\infty }{\lim }(e^x{)}^{\frac{1}{x}}{\left(1+e^{-x}\right)}^{\frac{1}{x}}=e\end{array}$$

运用夹逼准则

$$\begin{array}{rl}& \text{错解：}=\underset{n\to \infty }{\lim }\frac{1}{n^2+n+1}+\underset{n\to \infty }{\lim }\frac{2}{n^2+n+2}+\cdots +\underset{n\to \infty }{\lim }\frac{n}{n^2+n+n}=0+0+\cdots +0=0\\ & \text{（极限的拆分只能用于有限项，无限项不能用）}\\ & \\ & \text{解1：通分}\\ & \underset{n\to \infty }{\lim }\frac{1\cdot (n^2+n+2)\cdots (n^2+n+n)+\cdots +n\cdot (n^2+n+1)\cdots (n^2+n+n-1)}{(n^2+n+1)(n^2+n+2)\cdots (n^2+n+n)}\\ & \text{简化后：}\\ & \underset{n\to \infty }{\lim }\frac{n^{2(n-1)}\cdot (1+2+\cdots +n)}{n^{2n}}=\underset{n\to \infty }{\lim }\frac{n^{2(n-1)}\frac{n(n+1)}{2}}{n^{2n}}=\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \text{解2：}\underset{n\to \infty }{\lim }\left(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}\right)\\ & \text{用夹逼准则：}\\ & \sum _{i=1}^n\frac{i}{n^2+n+n}\le \sum _{i=1}^n\frac{i}{n^2+n+i}\le \sum _{i=1}^n\frac{i}{n^2+n+1}\\ & \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{i}{n^2+n+n}=\underset{n\to \infty }{\lim }\frac{1}{n^2+2n}\sum _{i=1}^{n}i=\underset{n\to \infty }{\lim }\frac{\frac{n(n+1)}{2}}{n^2+2n}=\frac{1}{2}\\ & \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{i}{n^2+n+1}=\underset{n\to \infty }{\lim }\frac{1}{n^2+n+1}\sum _{i=1}^{n}i=\underset{n\to \infty }{\lim }\frac{\frac{n(n+1)}{2}}{n^2+n+1}=\frac{1}{2}\\ & \\ & \text{由夹逼准则可知，原式}=\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \text{eg：}\underset{n\to \infty }{\lim }\frac{1}{n^2+1^2}+\frac{2}{n^2+2^2}+\cdots +\frac{n}{n^2+n^2}=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\frac{\frac{i}{n}}{1+{\left(\frac{i}{n}\right)}^2}={\int }_0^1\frac{x}{1+x^2}dx\\ & =\frac{1}{2}\ln (1+x^2){|}_0^1=\frac{\ln 2}{2}\\ & \\ & \text{思路：}\\ & \sum _{i=1}^n\frac{i}{n^2+n^2}\le \sum _{i=1}^n\frac{i}{n^2+i^2}\le \sum _{i=1}^n\frac{i}{n^2+1^2}\\ & \\ & \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{i}{n^2+1^2}=\underset{n\to \infty }{\lim }\frac{\frac{n(n+1)}{2}}{n^2+1}=\frac{1}{2}\\ & \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{i}{n^2+n^2}=\underset{n\to \infty }{\lim }\frac{\frac{n(n+1)}{2}}{n^2+n^2}=\frac{1}{4}\\ & \text{不能用夹逼准则（左右极限不相等）}\end{array}$$
$$\begin{array}{rl}& \text{结论：}\\ & \text{分母最大值与最小值的极限为1，用夹逼准则；}\\ & \text{极限不是1，用定积分定义。}\end{array}$$

$$\begin{array}{rl}& \text{积分比较大小：}\\ & ①\text{一元定积分：上下限大小关系 + 被积函数大小；}\\ & ②\text{二重积分：被积函数大小。}\end{array}$$
$$\begin{array}{rl}& \text{解：(1) 由}\ln (1+t)<t(\text{当}t>0),\text{得}(\ln (1+t){)}^n<t^n,\\ & \text{故}{\int }_0^1|\ln t|[\ln (1+t){]}^{n}dt<{\int }_0^1{t}^n|\ln t|dt.\\ & \\ & (2){\int }_0^1{t}^n|\ln t|dt=-{\int }_0^1{t}^n\ln tdt=-{\int }_0^1{\left(\frac{1}{n+1}{t}^{n+1}\right)}^{\prime }\ln tdt\\ & =-{\left(\frac{1}{n+1}{t}^{n+1}\cdot \ln t\right)|}_0^1+{\int }_0^1\frac{1}{n+1}{t}^{n+1}\cdot \frac{1}{t}dt\\ & =\frac{1}{n+1}{\int }_0^1{t}^{n}dt=\frac{1}{(n+1{)}^2}.\\ & \\ & \text{又}\underset{n\to \infty }{\lim }\frac{1}{(n+1{)}^2}=0,\text{且}0\le u_n\le \frac{1}{(n+1{)}^2},\\ & \text{由夹逼准则可知：}\underset{n\to \infty }{\lim }{u}_n=0.\end{array}$$
$$\begin{array}{rl}& \text{小结：当出现不等式时，且是计算一个极限，用夹逼准则。}\\ & ②\text{不等式：}\\ & e^x-1>x>\ln (1+x)\\ & \tan x>x>\arctan x\\ & \arcsin x>x>\sin x\end{array}$$
$$\begin{array}{rl}& \text{改题：}\\ & \text{记}{u}_n={\int }_0^1|\ln t|[\sin t{]}^{n}dt(n=1,2,\cdots )，\text{求极限}\underset{n\to \infty }{\lim }{u}_n.\end{array}$$
$$0\le {\int }_0^1|\ln t|[\sin t{]}^{n}dt<{\int }_0^1|\ln t|t^{n}dt=\frac{1}{(n+1{)}^2}$$

$$\begin{array}{rl}& \text{解：(1) 因}|\cos t|\ge 0,\text{故}{\int }_0^{n\pi }|\cos t|dt\le {\int }_0^x|\cos t|dt<{\int }_0^{(n+1)\pi }|\cos t|dt\\ & \\ & {\int }_0^{n\pi }|\cos t|dt=n{\int }_0^{\pi }|\cos t|dt=n\cdot 2{\int }_0^{\frac{\pi }{2}}|\cos t|dt=2n{\int }_0^{\frac{\pi }{2}}\cos tdt=2n\\ & \\ & {\int }_0^{(n+1)\pi }|\cos t|dt=(n+1){\int }_0^{\pi }|\cos t|dt=(n+1)\cdot 2{\int }_0^{\frac{\pi }{2}}\cos tdt=2(n+1)\\ & \\ & \text{故}2n\le S(x)<2(n+1)\end{array}$$
$$\begin{array}{rl}& (2)\frac{2n}{(n+1)\pi }\le \frac{S(x)}{x}<\frac{2(n+1)}{n\pi }\\ & \\ & \underset{n\to \infty }{\lim }\frac{2n}{(n+1)\pi }=\underset{n\to \infty }{\lim }\frac{2(n+1)}{n\pi }=\frac{2}{\pi }\\ & \\ & \text{故由夹逼准则可知：}\underset{x\to \infty }{\lim }\frac{S(x)}{x}=\frac{2}{\pi }\end{array}$$
$$\begin{array}{rl}& \text{改题：}\\ & \text{设函数}S(x)={\int }_0^x|\sin t|dt，\text{求}\underset{x\to +\infty }{\lim }\frac{S(x)}{x}.\end{array}$$
$$\begin{array}{rl}& \text{错解：}\underset{x\to +\infty }{\lim }\frac{{\int }_0^x|\sin t|dt}{x}\stackrel{\text{洛必达}}{=}\underset{x\to +\infty }{\lim }\frac{|\sin x|}{1}\text{不存在}\\ & \text{故}\underset{x\to +\infty }{\lim }\frac{S(x)}{x}\text{不存在}.\end{array}$$
$$\begin{array}{rl}& \text{错因：用洛必达时，若}\frac{f^{\prime }(x)}{g^{\prime }(x)}\text{的极限存在或为}\infty ,\text{则}\lim \frac{f(x)}{g(x)}=\lim \frac{f^{\prime }(x)}{g^{\prime }(x)};\\ & \text{但若}\frac{f^{\prime }}{g^{\prime }}\text{的极限不存在且不为}\infty ,\text{则无法推出原来的极限}\frac{f}{g}\text{不存在}.\end{array}$$
$$\begin{array}{rl}& \text{正解：}n\pi \le x<(n+1)\pi ,2n={\int }_0^{n\pi }|\sin t|dt\le {\int }_0^x|\sin t|dt<{\int }_0^{(n+1)\pi }|\sin t|dt=2(n+1)\\ & \\ & \frac{2n}{(n+1)\pi }<\frac{S(x)}{x}<\frac{2(n+1)}{n\pi },\underset{n\to \infty }{\lim }\frac{2n}{(n+1)\pi }=\underset{n\to \infty }{\lim }\frac{2(n+1)}{n\pi }=\frac{2}{\pi }\\ & \\ & \text{故由夹逼准则可知：}\underset{x\to +\infty }{\lim }\frac{S(x)}{x}=\frac{2}{\pi }\end{array}$$

运用定积分的定义

$$\begin{array}{rl}& \text{定积分定义}\{\begin{array}{l}\text{①计算极限数值型：}\\ ①{\int }_0^{1}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^n\frac{1}{n}f\left(\frac{i}{n}\right)\\ ②{\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^n\frac{b-a}{n}f\left(a+\frac{b-a}{n}i\right)\\ \text{②抽象型：}\\ {\int }_0^{1}f(x)dx={\lim }_{n\to \infty }\frac{1}{kn}{\sum }_{i=1}^{kn}f({\xi }_i),{\xi }_i\in \left[\frac{i-1}{kn},\frac{i}{kn}\right]\end{array}\end{array}$$

$$\begin{array}{rl}& \text{解：}\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\sqrt{1+\cos \frac{i\pi }{n}}={\int }_0^1\sqrt{1+\cos \pi x}dx={\int }_0^1\sqrt{2{\cos }^2\frac{\pi x}{2}}dx\\ & =\sqrt{2}{\int }_0^1\cos \frac{\pi }{2}xdx=\sqrt{2}\cdot \frac{2}{\pi }=\frac{2\sqrt{2}}{\pi }\end{array}$$
$$\begin{array}{rl}& \text{拓：}\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\frac{1}{1+\frac{i}{n}}={\int }_0^1\frac{1}{1+x}dx=\ln (1+x){|}_0^1=\ln 2\\ & \\ & {\int }_0^{2}f(x)dx=\underset{n\to \infty }{\lim }\frac{2}{n}\sum _{i=1}^{n}f\left(\frac{2}{n}i\right)\\ & \\ & \text{也可表示为：}\underset{n\to \infty }{\lim }\frac{1}{2}\cdot \frac{2}{n}\sum _{i=1}^n\frac{1}{1+\frac{1}{2}\cdot \frac{2i}{n}}=\frac{1}{2}{\int }_0^2\frac{1}{1+\frac{1}{2}x}dx\\ & =\ln \left(1+\frac{1}{2}x\right){|}_0^2=\ln 2\end{array}$$
$$\begin{array}{rl}& \text{定积分定义和夹逼准则的区分：}\\ & ①\text{有分母时，最大分母与最小分母的极限为1，用夹逼准则；}\\ & ②\text{无分母时，令}i=nx\text{代入和式，若}n\text{消掉了，则可用定积分定义，所剩函数即为}f(x).\\ & \\ & \text{例：}\sqrt{1+\cos \frac{i\pi }{n}}\stackrel{i=nx}{\to }\sqrt{1+\cos \frac{nx\cdot \pi }{n}}=\sqrt{1+\cos \pi x}\end{array}$$

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{\sin \frac{i}{n}\pi }{n+\frac{1}{i}}=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\frac{n\sin \frac{i}{n}\pi }{n+\frac{1}{i}}\\ & \\ & \frac{n\sin \frac{i}{n}\pi }{n+\frac{1}{i}}=\frac{n\cdot \sin \frac{nx}{n}\pi }{n+\frac{1}{nx}}(\text{令}i=nx)，\text{仍含}n，\text{不能用定积分定义}.\end{array}$$
$$\underset{n\to \infty }{\lim }\frac{n+1}{n+\frac{1}{n}}=1用夹逼准则$$
$$\begin{array}{rl}& \text{解：}\\ & \sum _{i=1}^n\frac{\sin \frac{i}{n}\pi }{n+1}\le \sum _{i=1}^n\frac{\sin \frac{i}{n}\pi }{n+\frac{1}{i}}\le \sum _{i=1}^n\frac{1}{n}\sin \frac{i}{n}\pi \\ & \\ & \underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{1}{n+1}\sin \frac{i}{n}\pi =\underset{n\to \infty }{\lim }\frac{n}{n+1}\cdot \frac{1}{n}\sum _{i=1}^n\sin \frac{i}{n}\pi =\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\sin \frac{i}{n}\pi \\ & ={\int }_0^1\sin \pi xdx=\frac{2}{\pi }\\ & \\ & \text{故由夹逼准则可知：}\\ & \underset{n\to \infty }{\lim }\left(\frac{\sin \frac{\pi }{n}}{n+1}+\frac{\sin \frac{2\pi }{n}}{n+\frac{1}{2}}+\cdots +\frac{\sin \pi }{n+\frac{1}{n}}\right)=\frac{2}{\pi }\end{array}$$

运用单调有界收敛准则

$$\begin{array}{rl}& \text{收敛的数列一定有界，}\\ & \text{但有界的数列不一定收敛。}\\ & \text{单调有界}\Rightarrow \text{收敛。}\end{array}$$
$$\begin{array}{rl}& \text{夹逼准则和单调有界收敛准则的区别：}\\ & ①\text{计算极限值，用夹逼；}\\ & ②\text{证明极限存在，用单调有界。}\end{array}$$

$$\begin{array}{rl}& \text{分析：}\\ & A=\sqrt{A(3-A)}\\ & A^2=3A-A^2\\ & A=0\text{或}A=\frac{3}{2}\end{array}$$
证：xn+1=xn(3−xn)≤xn+3−xn2=32当n>1时，xn≤32xn+1−xn=xn(3−xn)−xn=xn(3−xn−xn)=xn⋅3−2xn3−xn+xn≥0故{xn}单调递增有上界，因此{xn}收敛。设lim⁡n→∞xn=A.A=A(3−A)  ⟹  A2=3A−A2  ⟹  A=0(舍)或A=32 \begin{aligned} &\text{证：}x_{n+1} = \sqrt{x_n(3 - x_n)} \leq \frac{x_n + 3 - x_n}{2} = \frac{3}{2} \\ & \text{当}n > 1\text{时，}x_n \leq \frac{3}{2} \\ &x_{n+1} - x_n = \sqrt{x_n(3 - x_n)} - x_n \\ &= \sqrt{x_n} \left( \sqrt{3 - x_n} - \sqrt{x_n} \right) \\ &= \sqrt{x_n} \cdot \frac{3 - 2x_n}{\sqrt{3 - x_n} + \sqrt{x_n}} \geq 0 \\ & \\ &\text{故}\{x_n\}\text{单调递增有上界，因此}\{x_n\}\text{收敛。设}\lim_{n \to \infty}x_n = A. \\ & \\ &A = \sqrt{A(3 - A)} \implies A^2 = 3A - A^2 \implies A = 0(\text{舍}) \text{或}A = \frac{3}{2} \end{aligned} ​证：xn+1​=xn​(3−xn​)​≤2xn​+3−xn​​=23​当n>1时，xn​≤23​xn+1​−xn​=xn​(3−xn​)​−xn​=xn​​(3−xn​​−xn​​)=xn​​⋅3−xn​​+xn​​3−2xn​​≥0故{xn​}单调递增有上界，因此{xn​}收敛。设n→∞lim​xn​=A.A=A(3−A)​⟹A2=3A−A2⟹A=0(舍)或A=23​​
$$\begin{array}{rl}& \text{小结：去掉有限项后}\\ & \text{仍为单调有界，}\\ & \text{则极限仍存在。}\end{array}$$
【小结】1、递推公式极限的基本思路：首先证明数列{an}单调有界，从而得到该数列极限存在；然后设lim⁡n→∞an=A，在等式两边同时取极限，得到方程A=f(A)，解方程即可得出极限.（2）分析和证明数列单调有界的基本思路：①分析：先求出极限，从极限与数列前几项的大小关系分析数列的单调性，进而确定有上界还是有下界，以及上界或下界各是多少；②证明：先证有界性，再证单调性；在证明有界性时一般会用到数学归纳法. \begin{aligned} &\text{【小结】1、递推公式极限的基本思路：首先证明数列}\{a_n\}\text{单调有界，从而得到该数列} \\ &\text{极限存在；然后设}\lim_{n \to \infty}a_n = A，\text{在等式两边同时取极限，得到方程}A = f(A)，\text{解方程} \\ &\text{即可得出极限.} \\ & \\ &\text{（2）分析和证明数列单调有界的基本思路：} \\ &①\text{分析：先求出极限，从极限与数列前几项的大小关系分析数列的单调性，进而确定有} \\ &\text{上界还是有下界，以及上界或下界各是多少；} \\ &②\text{证明：先证有界性，再证单调性；在证明有界性时一般会用到数学归纳法.} \end{aligned} ​【小结】1、递推公式极限的基本思路：首先证明数列{an​}单调有界，从而得到该数列极限存在；然后设n→∞lim​an​=A，在等式两边同时取极限，得到方程A=f(A)，解方程即可得出极限.（2）分析和证明数列单调有界的基本思路：①分析：先求出极限，从极限与数列前几项的大小关系分析数列的单调性，进而确定有上界还是有下界，以及上界或下界各是多少；②证明：先证有界性，再证单调性；在证明有界性时一般会用到数学归纳法.​
$$\begin{array}{rl}& \text{2、证明单调性时常用的不等式：}\\ & （1）\text{当}x\in \left(0,\frac{\pi }{2}\right)\text{时，}\sin x<x<\tan x\\ & \\ & （2）\text{当}x>0\text{时，}{e}^x-1>x>\ln (1+x)\\ & \\ & （3）\text{均值不等式：设}a,b>0，则a+b\ge 2\sqrt{ab}\end{array}$$

$$\begin{array}{rl}& \text{分析：有}e\text{猜}0\text{，有}n\text{猜1}\\ & A{e}^A=e^A-1,A=0\\ & \text{拉格朗日}\\ & x_n{e}^{x_{n+1}}=e^{x_n}-e^0\stackrel{0<{\xi }_n<x_n}{=}(x_n-0)e^{{\xi }_n}\\ & <x_n{e}^{x_n}\end{array}$$
证：先证有界性，x1>0，假设当n=k时，xn>0，则xn+1=ln⁡exn−1xn>ln⁡1=0，故由数学归纳法可知，{xn}有下界.再证单调性，xnexn+1=exn−1=exn−e0=(xn−0)eξn (0<ξn<xn)<xnexn,故xn+1<xn即{xn}单调递减.由单调有界收敛准则可知，{xn}收敛，设lim⁡n→∞xn=α，则αeα=eα−1，故α=0，即lim⁡n→∞xn=0 \begin{aligned} &\text{证：先证有界性，}x_1>0，\text{假设当}n=k\text{时，}x_n>0， \\ &\text{则}x_{n+1} = \ln\frac{e^{x_n}-1}{x_n} > \ln1=0，\text{故由数学归纳法} \\ &\text{可知，}\{x_n\}\text{有下界}. \\ & \\ &\text{再证单调性，}x_n e^{x_{n+1}} = e^{x_n}-1 = e^{x_n}-e^0 = (x_n-0)e^{\xi_n}\ (0<\xi_n<x_n) \\ &< x_n e^{x_n},\text{故}x_{n+1}<x_n \\ &\text{即}\{x_n\}\text{单调递减}. \\ & \\ &\text{由单调有界收敛准则可知，}\{x_n\}\text{收敛，设}\lim_{n \to \infty}x_n=\alpha，则 \\ &\alpha e^\alpha = e^\alpha - 1，\text{故}\alpha=0，\text{即}\lim_{n \to \infty}x_n=0 \end{aligned} ​证：先证有界性，x1​>0，假设当n=k时，xn​>0，则xn+1​=lnxn​exn​−1​>ln1=0，故由数学归纳法可知，{xn​}有下界.再证单调性，xn​exn+1​=exn​−1=exn​−e0=(xn​−0)eξn​ (0<ξn​<xn​)<xn​exn​,故xn+1​<xn​即{xn​}单调递减.由单调有界收敛准则可知，{xn​}收敛，设n→∞lim​xn​=α，则αeα=eα−1，故α=0，即n→∞lim​xn​=0​
$$\begin{array}{rl}& 小结：用拉格朗日中值定理的情况\\ & e^{x_n}-1=e^{x_n}-e^0\\ & \ln x_n=\ln x_n-\ln 1\end{array}$$