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序列极限的性质与夹逼定理

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序列极限的性质与夹逼定理

小象实战讲义 · 人工智能数学基础

在掌握了序列极限的严格定义（ε-N 语言）之后，我们自然要追问：收敛的序列有哪些普遍的性质？这些性质如何帮助我们更高效地计算极限？本节将系统介绍收敛序列的六大核心性质，并重点讲解一个威力强大的计算工具——夹逼定理。学完本节，你将能够运用这些性质，严谨地证明极限的唯一性、有界性，并解决一系列形式复杂的极限计算问题。

💡 核心导读

本节将带你掌握以下关键内容：

收敛序列的六大性质：唯一性、有界性、保序性、四则运算法则、子序列收敛性、单调收敛原理。
性质的应用与证明：学习如何运用 ε-N 语言严谨证明唯一性、有界性和保序性，理解证明背后的直观思想。
极限的计算技巧：掌握处理“多项式比多项式”型极限的通用方法（上下同除最高次幂）。
夹逼定理：理解并证明这个同时给出极限存在性与值的强大定理。
夹逼定理的典型应用：学会处理 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>n</mi><mrow><mn>1</mn><mi mathvariant="normal">/</mi><mi>n</mi></mrow></msup></mrow><annotation encoding="application/x-tex">n^{1/n}</annotation></semantics></math>n1/n 和 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mroot><mrow><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup></mrow><mi>n</mi></mroot></mrow><annotation encoding="application/x-tex">\sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n}</annotation></semantics></math>na1n+a2n+⋯+amn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.28em" viewBox="0 0 400000 1296" preserveAspectRatio="xMinYMin slice"><path d="M263,681c0.7,0,18,39.7,52,119 c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120 c340,-704.7,510.7,-1060.3,512,-1067 l0 -0 c4.7,-7.3,11,-11,19,-11 H40000v40H1012.3 s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232 c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z M1001 80h400000v40h-400000z"/></svg> 型极限的经典套路。

序列极限的基本性质

设 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>x</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{x_n\}</annotation></semantics></math>{xn} 和 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>y</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{y_n\}</annotation></semantics></math>{yn} 为实数序列，且 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>x</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} x_n = a</annotation></semantics></math>limn→∞xn=a, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>y</mi><mi>n</mi></msub><mo>=</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} y_n = b</annotation></semantics></math>limn→∞yn=b。收敛序列具有以下重要性质：

唯一性：若序列收敛，则其极限唯一。即不可能同时收敛到两个不同的值 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>a 和 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>b</mi></mrow><annotation encoding="application/x-tex">b</annotation></semantics></math>b（<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo mathvariant="normal">≠</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">a \neq b</annotation></semantics></math>a=b）。
有界性：收敛序列必有界。即存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>M</mi><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">M > 0</annotation></semantics></math>M>0，使得对一切 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>∈</mo><mi mathvariant="double-struck">N</mi></mrow><annotation encoding="application/x-tex">n \in \mathbb{N}</annotation></semantics></math>n∈N，有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mi mathvariant="normal">∣</mi><mo>≤</mo><mi>M</mi></mrow><annotation encoding="application/x-tex">|x_n| \le M</annotation></semantics></math>∣xn∣≤M。
保序性：若存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>0</mn></msub></mrow><annotation encoding="application/x-tex">N_0</annotation></semantics></math>N0，使得当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>0</mn></msub></mrow><annotation encoding="application/x-tex">n > N_0</annotation></semantics></math>n>N0 时恒有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>y</mi><mi>n</mi></msub></mrow><annotation encoding="application/x-tex">x_n \le y_n</annotation></semantics></math>xn≤yn，则 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>≤</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">a \le b</annotation></semantics></math>a≤b。
四则运算法则：若两序列收敛，则它们的和、差、积、商（分母极限不为零）的序列也收敛，且极限等于对应极限的和、差、积、商。 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mtable rowspacing="0.25em" columnalign="right left" columnspacing="0em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mo stretchy="false">(</mo><msub><mi>x</mi><mi>n</mi></msub><mo>±</mo><msub><mi>y</mi><mi>n</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><mo>±</mo><mi>b</mi><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mo stretchy="false">(</mo><msub><mi>x</mi><mi>n</mi></msub><mo>⋅</mo><msub><mi>y</mi><mi>n</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><mo>⋅</mo><mi>b</mi><mo separator="true">,</mo></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mrow><mo fence="true">(</mo><mfrac><msub><mi>x</mi><mi>n</mi></msub><msub><mi>y</mi><mi>n</mi></msub></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mspace width="1em"/><mo stretchy="false">(</mo><mi>b</mi><mo mathvariant="normal">≠</mo><mn>0</mn><mo stretchy="false">)</mo><mi mathvariant="normal">.</mi></mrow></mstyle></mtd></mtr></mtable><annotation encoding="application/x-tex">\begin{aligned} &\lim_{n \to \infty} (x_n \pm y_n) = a \pm b, \\ &\lim_{n \to \infty} (x_n \cdot y_n) = a \cdot b, \\ &\lim_{n \to \infty} \left(\frac{x_n}{y_n}\right) = \frac{a}{b} \quad (b \neq 0). \end{aligned}</annotation></semantics></math>n→∞lim(xn±yn)=a±b,n→∞lim(xn⋅yn)=a⋅b,n→∞lim(ynxn)=ba(b=0).
子序列的收敛性：若序列收敛于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>a，则它的任一子序列也收敛于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>a。
单调收敛原理：单调有界的实数序列必收敛。这是一个存在性定理，常用于证明极限存在。

下面我们选择其中几个性质进行证明，以加深对 ε-N 语言的理解。

唯一性的证明

证明：（反证法）假设序列 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>x</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{x_n\}</annotation></semantics></math>{xn} 同时收敛到两个不同的极限 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>a 和 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>b</mi></mrow><annotation encoding="application/x-tex">b</annotation></semantics></math>b，且 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>></mo><mi>b</mi></mrow><annotation encoding="application/x-tex">a > b</annotation></semantics></math>a>b。取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\varepsilon = \frac{a - b}{2} > 0</annotation></semantics></math>ε=2a−b>0。根据极限定义：

存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>1</mn></msub></mrow><annotation encoding="application/x-tex">N_1</annotation></semantics></math>N1，当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>1</mn></msub></mrow><annotation encoding="application/x-tex">n > N_1</annotation></semantics></math>n>N1 时，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">|x_n - a| < \varepsilon</annotation></semantics></math>∣xn−a∣<ε。
存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">N_2</annotation></semantics></math>N2，当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">n > N_2</annotation></semantics></math>n>N2 时，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>b</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">|x_n - b| < \varepsilon</annotation></semantics></math>∣xn−b∣<ε。取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><msub><mi>N</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>N</mi><mn>2</mn></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">N = \max\{N_1, N_2\}</annotation></semantics></math>N=max{N1,N2}，则当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mi>N</mi></mrow><annotation encoding="application/x-tex">n > N</annotation></semantics></math>n>N 时，上述两式同时成立。由此可得： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>></mo><mi>a</mi><mo>−</mo><mi>ε</mi><mo>=</mo><mi>a</mi><mo>−</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo separator="true">,</mo><mspace width="1em"/><msub><mi>x</mi><mi>n</mi></msub><mo><</mo><mi>b</mi><mo>+</mo><mi>ε</mi><mo>=</mo><mi>b</mi><mo>+</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">x_n > a - \varepsilon = a - \frac{a-b}{2} = \frac{a+b}{2}, \quad x_n </semantics></math>xn>a−ε=a−2a−b=2a+b,xn<b+ε=b+2a−b=2a+b. 这导致 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub></mrow><annotation encoding="application/x-tex">x_n</annotation></semantics></math>xn 既大于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mn>2</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{a+b}{2}</annotation></semantics></math>2a+b 又小于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mn>2</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{a+b}{2}</annotation></semantics></math>2a+b，矛盾。故假设不成立，极限唯一。

有界性的证明

证明：因为 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>x</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} x_n = a</annotation></semantics></math>limn→∞xn=a，取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>ε</mi><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\varepsilon = 1</annotation></semantics></math>ε=1，则存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi></mrow><annotation encoding="application/x-tex">N</annotation></semantics></math>N，使得当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mi>N</mi></mrow><annotation encoding="application/x-tex">n > N</annotation></semantics></math>n>N 时，有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">|x_n - a| < 1</annotation></semantics></math>∣xn−a∣<1。由三角不等式，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mi mathvariant="normal">∣</mi><mo>=</mo><mi mathvariant="normal">∣</mi><mo stretchy="false">(</mo><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mo stretchy="false">)</mo><mo>+</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo>≤</mo><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo>+</mo><mi mathvariant="normal">∣</mi><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mn>1</mn><mo>+</mo><mi mathvariant="normal">∣</mi><mi>a</mi><mi mathvariant="normal">∣</mi></mrow><annotation encoding="application/x-tex">|x_n| = |(x_n - a) + a| \le |x_n - a| + |a| < 1 + |a|</annotation></semantics></math>∣xn∣=∣(xn−a)+a∣≤∣xn−a∣+∣a∣<1+∣a∣。对于前 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi></mrow><annotation encoding="application/x-tex">N</annotation></semantics></math>N 项 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>x</mi><mn>2</mn></msub><mo separator="true">,</mo><mo>…</mo><mo separator="true">,</mo><msub><mi>x</mi><mi>N</mi></msub></mrow><annotation encoding="application/x-tex">x_1, x_2, \dots, x_N</annotation></semantics></math>x1,x2,…,xN，这是有限个数，必存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>M</mi><mn>0</mn></msub><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mn>1</mn></msub><mi mathvariant="normal">∣</mi><mo separator="true">,</mo><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mn>2</mn></msub><mi mathvariant="normal">∣</mi><mo separator="true">,</mo><mo>…</mo><mo separator="true">,</mo><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>N</mi></msub><mi mathvariant="normal">∣</mi><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">M_0 = \max\{|x_1|, |x_2|, \dots, |x_N|\}</annotation></semantics></math>M0=max{∣x1∣,∣x2∣,…,∣xN∣}。取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>M</mi><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><msub><mi>M</mi><mn>0</mn></msub><mo separator="true">,</mo><mn>1</mn><mo>+</mo><mi mathvariant="normal">∣</mi><mi>a</mi><mi mathvariant="normal">∣</mi><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">M = \max\{M_0, 1 + |a|\}</annotation></semantics></math>M=max{M0,1+∣a∣}，则对一切 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>∈</mo><mi mathvariant="double-struck">N</mi></mrow><annotation encoding="application/x-tex">n \in \mathbb{N}</annotation></semantics></math>n∈N，都有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mi mathvariant="normal">∣</mi><mo>≤</mo><mi>M</mi></mrow><annotation encoding="application/x-tex">|x_n| \le M</annotation></semantics></math>∣xn∣≤M。故序列有界。

保序性的证明

证明：（反证法）假设结论不成立，即 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>></mo><mi>b</mi></mrow><annotation encoding="application/x-tex">a > b</annotation></semantics></math>a>b。取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\varepsilon = \frac{a - b}{2} > 0</annotation></semantics></math>ε=2a−b>0。根据极限定义：

存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>1</mn></msub></mrow><annotation encoding="application/x-tex">N_1</annotation></semantics></math>N1，当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>1</mn></msub></mrow><annotation encoding="application/x-tex">n > N_1</annotation></semantics></math>n>N1 时，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">|x_n - a| < \varepsilon</annotation></semantics></math>∣xn−a∣<ε，即 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>></mo><mi>a</mi><mo>−</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">x_n > a - \varepsilon</annotation></semantics></math>xn>a−ε。
存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">N_2</annotation></semantics></math>N2，当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">n > N_2</annotation></semantics></math>n>N2 时，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>y</mi><mi>n</mi></msub><mo>−</mo><mi>b</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">|y_n - b| < \varepsilon</annotation></semantics></math>∣yn−b∣<ε，即 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>y</mi><mi>n</mi></msub><mo><</mo><mi>b</mi><mo>+</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">y_n </semantics></math>yn<b+ε。取 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><msub><mi>N</mi><mn>0</mn></msub><mo separator="true">,</mo><msub><mi>N</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>N</mi><mn>2</mn></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">N = \max\{N_0, N_1, N_2\}</annotation></semantics></math>N=max{N0,N1,N2}，则当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mi>N</mi></mrow><annotation encoding="application/x-tex">n > N</annotation></semantics></math>n>N 时，有 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>></mo><mi>a</mi><mo>−</mo><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mi>b</mi><mo>+</mo><mi>ε</mi><mo>></mo><msub><mi>y</mi><mi>n</mi></msub><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">x_n > a - \varepsilon = \frac{a+b}{2} = b + \varepsilon > y_n.</annotation></semantics></math>xn>a−ε=2a+b=b+ε>yn. 这与条件“当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>0</mn></msub></mrow><annotation encoding="application/x-tex">n > N_0</annotation></semantics></math>n>N0 时 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>y</mi><mi>n</mi></msub></mrow><annotation encoding="application/x-tex">x_n \le y_n</annotation></semantics></math>xn≤yn”矛盾。故假设错误，必有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>≤</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">a \le b</annotation></semantics></math>a≤b。

极限的计算：四则运算的应用

利用极限的四则运算法则，我们可以计算许多常见序列的极限。

例1：计算极限 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mfrac><mrow><mn>3</mn><msup><mi>n</mi><mn>2</mn></msup><mo>+</mo><mn>4</mn><mi>n</mi><mo>−</mo><mn>100</mn></mrow><mrow><mn>4</mn><msup><mi>n</mi><mn>2</mn></msup><mo>−</mo><mn>5</mn><mi>n</mi><mo>+</mo><mn>10</mn></mrow></mfrac></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \frac{3n^2 + 4n - 100}{4n^2 - 5n + 10}</annotation></semantics></math>limn→∞4n2−5n+103n2+4n−100。

解：这是 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mi mathvariant="normal">∞</mi><mi mathvariant="normal">∞</mi></mfrac></mrow><annotation encoding="application/x-tex">\frac{\infty}{\infty}</annotation></semantics></math>∞∞ 型未定式。处理多项式比值极限的通用方法是分子分母同除以 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math>n 的最高次幂（此处为 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>n</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">n^2</annotation></semantics></math>n2）。 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mtable rowspacing="0.25em" columnalign="right left" columnspacing="0em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mrow><mn>3</mn><msup><mi>n</mi><mn>2</mn></msup><mo>+</mo><mn>4</mn><mi>n</mi><mo>−</mo><mn>100</mn></mrow><mrow><mn>4</mn><msup><mi>n</mi><mn>2</mn></msup><mo>−</mo><mn>5</mn><mi>n</mi><mo>+</mo><mn>10</mn></mrow></mfrac></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mrow><mn>3</mn><mo>+</mo><mfrac><mn>4</mn><mi>n</mi></mfrac><mo>−</mo><mfrac><mn>100</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow><mrow><mn>4</mn><mo>−</mo><mfrac><mn>5</mn><mi>n</mi></mfrac><mo>+</mo><mfrac><mn>10</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow></mfrac></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mn>3</mn><mo>+</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mn>4</mn><mi>n</mi></mfrac><mo>−</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mn>100</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mn>4</mn><mo>−</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mn>5</mn><mi>n</mi></mfrac><mo>+</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mn>10</mn><msup><mi>n</mi><mn>2</mn></msup></mfrac></mrow></mfrac></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mfrac><mrow><mn>3</mn><mo>+</mo><mn>0</mn><mo>−</mo><mn>0</mn></mrow><mrow><mn>4</mn><mo>−</mo><mn>0</mn><mo>+</mo><mn>0</mn></mrow></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi mathvariant="normal">.</mi></mrow></mstyle></mtd></mtr></mtable><annotation encoding="application/x-tex">\begin{aligned} \lim_{n \to \infty} \frac{3n^2 + 4n - 100}{4n^2 - 5n + 10} &= \lim_{n \to \infty} \frac{3 + \frac{4}{n} - \frac{100}{n^2}}{4 - \frac{5}{n} + \frac{10}{n^2}} \\ &= \frac{\lim_{n \to \infty}3 + \lim_{n \to \infty}\frac{4}{n} - \lim_{n \to \infty}\frac{100}{n^2}}{\lim_{n \to \infty}4 - \lim_{n \to \infty}\frac{5}{n} + \lim_{n \to \infty}\frac{10}{n^2}} \\ &= \frac{3 + 0 - 0}{4 - 0 + 0} = \frac{3}{4}. \end{aligned}</annotation></semantics></math>n→∞lim4n2−5n+103n2+4n−100=n→∞lim4−n5+n2103+n4−n2100=limn→∞4−limn→∞n5+limn→∞n210limn→∞3+limn→∞n4−limn→∞n2100=4−0+03+0−0=43.

例2：计算极限 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msubsup><mo>∑</mo><mrow><mi>k</mi><mo>=</mo><mn>0</mn></mrow><mi>n</mi></msubsup><msup><mi>q</mi><mi>k</mi></msup></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sum_{k=0}^{n} q^k</annotation></semantics></math>limn→∞∑k=0nqk，其中 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><mi>q</mi><mi mathvariant="normal">∣</mi><mo><</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">|q| < 1</annotation></semantics></math>∣q∣<1。

解：这是等比数列求和。等比数列前 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">n+1</annotation></semantics></math>n+1 项和公式为 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>S</mi><mi>n</mi></msub><mo>=</mo><mfrac><mrow><mn>1</mn><mo>−</mo><msup><mi>q</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>q</mi></mrow></mfrac></mrow><annotation encoding="application/x-tex">S_n = \frac{1 - q^{n+1}}{1 - q}</annotation></semantics></math>Sn=1−q1−qn+1（首项为1）。 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><munderover><mo>∑</mo><mrow><mi>k</mi><mo>=</mo><mn>0</mn></mrow><mi>n</mi></munderover><msup><mi>q</mi><mi>k</mi></msup><mo>=</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mrow><mn>1</mn><mo>−</mo><msup><mi>q</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>q</mi></mrow></mfrac><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sum_{k=0}^{n} q^k = \lim_{n \to \infty} \frac{1 - q^{n+1}}{1 - q}.</annotation></semantics></math>n→∞limk=0∑nqk=n→∞lim1−q1−qn+1. 由于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><mi>q</mi><mi mathvariant="normal">∣</mi><mo><</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">|q| < 1</annotation></semantics></math>∣q∣<1，有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msup><mi>q</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msup><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} q^{n+1} = 0</annotation></semantics></math>limn→∞qn+1=0。根据四则运算法则： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mfrac><mrow><mn>1</mn><mo>−</mo><msup><mi>q</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>q</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn><mo>−</mo><mn>0</mn></mrow><mrow><mn>1</mn><mo>−</mo><mi>q</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>−</mo><mi>q</mi></mrow></mfrac><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \frac{1 - q^{n+1}}{1 - q} = \frac{1 - 0}{1 - q} = \frac{1}{1 - q}.</annotation></semantics></math>n→∞lim1−q1−qn+1=1−q1−0=1−q1.

夹逼定理及其应用

夹逼定理是计算极限和证明极限存在的利器。

定理内容与证明

定理（夹逼定理）：设三个序列 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>x</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{x_n\}</annotation></semantics></math>{xn}, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>y</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{y_n\}</annotation></semantics></math>{yn}, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>z</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{z_n\}</annotation></semantics></math>{zn} 满足：存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>N</mi><mn>0</mn></msub><mo>∈</mo><mi mathvariant="double-struck">N</mi></mrow><annotation encoding="application/x-tex">N_0 \in \mathbb{N}</annotation></semantics></math>N0∈N，使得当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><msub><mi>N</mi><mn>0</mn></msub></mrow><annotation encoding="application/x-tex">n > N_0</annotation></semantics></math>n>N0 时，有 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>z</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>y</mi><mi>n</mi></msub><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">x_n \le z_n \le y_n.</annotation></semantics></math>xn≤zn≤yn. 如果 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>x</mi><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>y</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = a</annotation></semantics></math>limn→∞xn=limn→∞yn=a，那么序列 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">{</mo><msub><mi>z</mi><mi>n</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">\{z_n\}</annotation></semantics></math>{zn} 也收敛，且 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>z</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} z_n = a</annotation></semantics></math>limn→∞zn=a。

证明：对于任意 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>ε</mi><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\varepsilon > 0</annotation></semantics></math>ε>0，因为 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>x</mi><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>y</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = a</annotation></semantics></math>limn→∞xn=limn→∞yn=a，存在 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi></mrow><annotation encoding="application/x-tex">N</annotation></semantics></math>N（不妨设 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>N</mi><mo>></mo><msub><mi>N</mi><mn>0</mn></msub></mrow><annotation encoding="application/x-tex">N > N_0</annotation></semantics></math>N>N0），使得当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mi>N</mi></mrow><annotation encoding="application/x-tex">n > N</annotation></semantics></math>n>N 时，同时有 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>x</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi><mspace width="1em"/><mtext>和</mtext><mspace width="1em"/><mi mathvariant="normal">∣</mi><msub><mi>y</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">|x_n - a| < \varepsilon \quad \text{和} \quad |y_n - a| < \varepsilon.</annotation></semantics></math>∣xn−a∣<ε和∣yn−a∣<ε. 这意味着 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>−</mo><mi>ε</mi><mo><</mo><msub><mi>x</mi><mi>n</mi></msub></mrow><annotation encoding="application/x-tex">a - \varepsilon < x_n</annotation></semantics></math>a−ε<xn 且 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>y</mi><mi>n</mi></msub><mo><</mo><mi>a</mi><mo>+</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">y_n < a + \varepsilon</annotation></semantics></math>yn<a+ε。结合条件 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>x</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>z</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>y</mi><mi>n</mi></msub></mrow><annotation encoding="application/x-tex">x_n \le z_n \le y_n</annotation></semantics></math>xn≤zn≤yn，得到当 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mi>N</mi></mrow><annotation encoding="application/x-tex">n > N</annotation></semantics></math>n>N 时， <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>a</mi><mo>−</mo><mi>ε</mi><mo><</mo><msub><mi>x</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>z</mi><mi>n</mi></msub><mo>≤</mo><msub><mi>y</mi><mi>n</mi></msub><mo><</mo><mi>a</mi><mo>+</mo><mi>ε</mi><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">a - \varepsilon < x_n \le z_n \le y_n < a + \varepsilon.</annotation></semantics></math>a−ε<xn≤zn≤yn<a+ε. 即 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><msub><mi>z</mi><mi>n</mi></msub><mo>−</mo><mi>a</mi><mi mathvariant="normal">∣</mi><mo><</mo><mi>ε</mi></mrow><annotation encoding="application/x-tex">|z_n - a| < \varepsilon</annotation></semantics></math>∣zn−a∣<ε。由极限定义，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>z</mi><mi>n</mi></msub><mo>=</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} z_n = a</annotation></semantics></math>limn→∞zn=a。

典型应用示例

例3：证明 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mroot><mi>n</mi><mi>n</mi></mroot><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{n} = 1</annotation></semantics></math>limn→∞nn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>=1。

分析：显然 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mroot><mi>n</mi><mi>n</mi></mroot><mo>></mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\sqrt[n]{n} > 1</annotation></semantics></math>nn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>>1（<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>></mo><mn>1</mn></mrow><annotation encoding="application/x-tex">n>1</annotation></semantics></math>n>1）。令 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>h</mi><mi>n</mi></msub><mo>=</mo><mroot><mi>n</mi><mi>n</mi></mroot><mo>−</mo><mn>1</mn><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">h_n = \sqrt[n]{n} - 1 > 0</annotation></semantics></math>hn=nn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>−1>0，则 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>=</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msub><mi>h</mi><mi>n</mi></msub><msup><mo stretchy="false">)</mo><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">n = (1+h_n)^n</annotation></semantics></math>n=(1+hn)n。利用二项式定理展开： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>n</mi><mo>=</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msub><mi>h</mi><mi>n</mi></msub><msup><mo stretchy="false">)</mo><mi>n</mi></msup><mo>=</mo><munderover><mo>∑</mo><mrow><mi>k</mi><mo>=</mo><mn>0</mn></mrow><mi>n</mi></munderover><mrow><mo fence="true">(</mo><mfrac linethickness="0px"><mi>n</mi><mi>k</mi></mfrac><mo fence="true">)</mo></mrow><msubsup><mi>h</mi><mi>n</mi><mi>k</mi></msubsup><mo>≥</mo><mrow><mo fence="true">(</mo><mfrac linethickness="0px"><mi>n</mi><mn>2</mn></mfrac><mo fence="true">)</mo></mrow><msubsup><mi>h</mi><mi>n</mi><mn>2</mn></msubsup><mo>=</mo><mfrac><mrow><mi>n</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mn>2</mn></mfrac><msubsup><mi>h</mi><mi>n</mi><mn>2</mn></msubsup><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">n = (1+h_n)^n = \sum_{k=0}^{n} \binom{n}{k} h_n^k \ge \binom{n}{2} h_n^2 = \frac{n(n-1)}{2} h_n^2.</annotation></semantics></math>n=(1+hn)n=k=0∑n(kn)hnk≥(2n)hn2=2n(n−1)hn2. 这里我们只保留了展开式中的 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>k</mi><mo>=</mo><mn>2</mn></mrow><annotation encoding="application/x-tex">k=2</annotation></semantics></math>k=2 项（因为所有项非负）。由此可得： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msubsup><mi>h</mi><mi>n</mi><mn>2</mn></msubsup><mo>≤</mo><mfrac><mn>2</mn><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></mfrac><mspace width="1em"/><mo>⇒</mo><mspace width="1em"/><mn>0</mn><mo><</mo><msub><mi>h</mi><mi>n</mi></msub><mo>≤</mo><msqrt><mfrac><mn>2</mn><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></mfrac></msqrt><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">h_n^2 \le \frac{2}{n-1} \quad \Rightarrow \quad 0 < h_n \le \sqrt{\frac{2}{n-1}}.</annotation></semantics></math>hn2≤n−12⇒0<hn≤n−12<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="2.48em" viewBox="0 0 400000 2592" preserveAspectRatio="xMinYMin slice"><path d="M424,2478 c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514 c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20 s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13,-13s76,-122,76,-122s77,-121,77,-121 s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081 l0 -0c4,-6.7,10,-10,18,-10 H400000 v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185 c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80 h400000v40h-400000z"/></svg>. 由于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msqrt><mfrac><mn>2</mn><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></mfrac></msqrt><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt{\frac{2}{n-1}} = 0</annotation></semantics></math>limn→∞n−12<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.88em" viewBox="0 0 400000 1944" preserveAspectRatio="xMinYMin slice"><path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z"/></svg>=0，根据夹逼定理，<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><msub><mi>h</mi><mi>n</mi></msub><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} h_n = 0</annotation></semantics></math>limn→∞hn=0。因此， <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mroot><mi>n</mi><mi>n</mi></mroot><mo>=</mo><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msub><mi>h</mi><mi>n</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mn>1.</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{n} = \lim_{n \to \infty} (1 + h_n) = 1.</annotation></semantics></math>n→∞limnn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>=n→∞lim(1+hn)=1.

例4：设 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>a</mi><mn>2</mn></msub><mo separator="true">,</mo><mo>…</mo><mo separator="true">,</mo><msub><mi>a</mi><mi>m</mi></msub></mrow><annotation encoding="application/x-tex">a_1, a_2, \dots, a_m</annotation></semantics></math>a1,a2,…,am 为正数，求极限 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mroot><mrow><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup></mrow><mi>n</mi></mroot></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n}</annotation></semantics></math>limn→∞na1n+a2n+⋯+amn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.28em" viewBox="0 0 400000 1296" preserveAspectRatio="xMinYMin slice"><path d="M263,681c0.7,0,18,39.7,52,119 c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120 c340,-704.7,510.7,-1060.3,512,-1067 l0 -0 c4.7,-7.3,11,-11,19,-11 H40000v40H1012.3 s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232 c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z M1001 80h400000v40h-400000z"/></svg>。

解：令 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><msub><mi>a</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>a</mi><mn>2</mn></msub><mo separator="true">,</mo><mo>…</mo><mo separator="true">,</mo><msub><mi>a</mi><mi>m</mi></msub><mo stretchy="false">}</mo></mrow><annotation encoding="application/x-tex">A = \max\{a_1, a_2, \dots, a_m\}</annotation></semantics></math>A=max{a1,a2,…,am}。显然 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">A > 0</annotation></semantics></math>A>0。对每个 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">a_i</annotation></semantics></math>ai，有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msubsup><mi>a</mi><mi>i</mi><mi>n</mi></msubsup><mo>≤</mo><msup><mi>A</mi><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">a_i^n \le A^n</annotation></semantics></math>ain≤An，因此 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msubsup><mi>a</mi><mi>k</mi><mi>n</mi></msubsup><mo>≤</mo><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup><mo>≤</mo><mi>m</mi><msup><mi>A</mi><mi>n</mi></msup><mo separator="true">,</mo><mspace width="1em"/><mtext>其中 </mtext><msub><mi>a</mi><mi>k</mi></msub><mo>=</mo><mi>A</mi><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">a_k^n \le a_1^n + a_2^n + \cdots + a_m^n \le m A^n, \quad \text{其中 } a_k = A.</annotation></semantics></math>akn≤a1n+a2n+⋯+amn≤mAn,其中 ak=A. 特别地，有 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>A</mi><mi>n</mi></msup><mo>≤</mo><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup><mo>≤</mo><mi>m</mi><msup><mi>A</mi><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">A^n \le a_1^n + a_2^n + \cdots + a_m^n \le m A^n</annotation></semantics></math>An≤a1n+a2n+⋯+amn≤mAn。两边同时开 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math>n 次方： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>A</mi><mo>≤</mo><mroot><mrow><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup></mrow><mi>n</mi></mroot><mo>≤</mo><mroot><mi>m</mi><mi>n</mi></mroot><mo>⋅</mo><mi>A</mi><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">A \le \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} \le \sqrt[n]{m} \cdot A.</annotation></semantics></math>A≤na1n+a2n+⋯+amn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.28em" viewBox="0 0 400000 1296" preserveAspectRatio="xMinYMin slice"><path d="M263,681c0.7,0,18,39.7,52,119 c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120 c340,-704.7,510.7,-1060.3,512,-1067 l0 -0 c4.7,-7.3,11,-11,19,-11 H40000v40H1012.3 s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232 c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z M1001 80h400000v40h-400000z"/></svg>≤nm<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>⋅A. 由于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mroot><mi>m</mi><mi>n</mi></mroot><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{m} = 1</annotation></semantics></math>limn→∞nm<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>=1（可类似例3证明），根据夹逼定理： <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></munder><mroot><mrow><msubsup><mi>a</mi><mn>1</mn><mi>n</mi></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mi>n</mi></msubsup><mo>+</mo><mo>⋯</mo><mo>+</mo><msubsup><mi>a</mi><mi>m</mi><mi>n</mi></msubsup></mrow><mi>n</mi></mroot><mo>=</mo><mi>A</mi><mo>=</mo><mi>max</mi><mo>⁡</mo><mo stretchy="false">{</mo><msub><mi>a</mi><mn>1</mn></msub><mo separator="true">,</mo><msub><mi>a</mi><mn>2</mn></msub><mo separator="true">,</mo><mo>…</mo><mo separator="true">,</mo><msub><mi>a</mi><mi>m</mi></msub><mo stretchy="false">}</mo><mi mathvariant="normal">.</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = A = \max\{a_1, a_2, \dots, a_m\}.</annotation></semantics></math>n→∞limna1n+a2n+⋯+amn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.28em" viewBox="0 0 400000 1296" preserveAspectRatio="xMinYMin slice"><path d="M263,681c0.7,0,18,39.7,52,119 c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120 c340,-704.7,510.7,-1060.3,512,-1067 l0 -0 c4.7,-7.3,11,-11,19,-11 H40000v40H1012.3 s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232 c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z M1001 80h400000v40h-400000z"/></svg>=A=max{a1,a2,…,am}.

import numpy as np

# 示例：数值验证夹逼定理的应用
# 例3: 验证 lim_{n->∞} n^(1/n) = 1
n_vals = np.array([10, 50, 100, 500, 1000, 5000])
limits_n = np.power(n_vals, 1/n_vals)
print("例3: lim n^(1/n)")
for n, val in zip(n_vals, limits_n):
    print(f"  n={n:4d}, n^(1/n) = {val:.6f}")

# 例4: 验证 lim_{n->∞} (a1^n + a2^n + a3^n)^(1/n) = max(a1, a2, a3)
a = np.array([2.0, 3.0, 5.0])
A = np.max(a)
print(f"\n例4: a = {a}, 理论极限 max(a) = {A}")
for n in [10, 20, 30, 40, 50]:
    sum_powers = np.sum(np.power(a, n))
    limit_val = np.power(sum_powers, 1/n)
    print(f"  n={n:2d}, 计算值 = {limit_val:.6f}, 误差 = {abs(limit_val - A):.6f}")

📝 动手练一练

计算极限：求 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mfrac><mrow><mn>2</mn><msup><mi>n</mi><mn>3</mn></msup><mo>−</mo><mi>n</mi><mo>+</mo><mn>7</mn></mrow><mrow><mn>5</mn><msup><mi>n</mi><mn>3</mn></msup><mo>+</mo><mn>3</mn><msup><mi>n</mi><mn>2</mn></msup><mo>−</mo><mn>2</mn></mrow></mfrac></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \frac{2n^3 - n + 7}{5n^3 + 3n^2 - 2}</annotation></semantics></math>limn→∞5n3+3n2−22n3−n+7。 参考答案：分子分母同除以 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>n</mi><mn>3</mn></msup></mrow><annotation encoding="application/x-tex">n^3</annotation></semantics></math>n3，极限为 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mn>2</mn><mn>5</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{2}{5}</annotation></semantics></math>52。
应用夹逼定理：设 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo><</mo><mi>a</mi><mo><</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">0 < a < b</annotation></semantics></math>0<a<b，证明 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mroot><mrow><msup><mi>a</mi><mi>n</mi></msup><mo>+</mo><msup><mi>b</mi><mi>n</mi></msup></mrow><mi>n</mi></mroot><mo>=</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{a^n + b^n} = b</annotation></semantics></math>limn→∞nan+bn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>=b。 参考答案：由于 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>b</mi><mi>n</mi></msup><mo><</mo><msup><mi>a</mi><mi>n</mi></msup><mo>+</mo><msup><mi>b</mi><mi>n</mi></msup><mo><</mo><mn>2</mn><msup><mi>b</mi><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">b^n < a^n + b^n < 2b^n</annotation></semantics></math>bn<an+bn<2bn，两边开 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math>n 次方得 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>b</mi><mo><</mo><mroot><mrow><msup><mi>a</mi><mi>n</mi></msup><mo>+</mo><msup><mi>b</mi><mi>n</mi></msup></mrow><mi>n</mi></mroot><mo><</mo><mroot><mn>2</mn><mi>n</mi></mroot><mo>⋅</mo><mi>b</mi></mrow><annotation encoding="application/x-tex">b < \sqrt[n]{a^n + b^n} < \sqrt[n]{2} \cdot b</annotation></semantics></math>b<nan+bn<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg><n2<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>⋅b，利用 <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>n</mi><mo>→</mo><mi mathvariant="normal">∞</mi></mrow></msub><mroot><mn>2</mn><mi>n</mi></mroot><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\lim_{n \to \infty} \sqrt[n]{2} = 1</annotation></semantics></math>limn→∞n2<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>=1 和夹逼定理即得结论。

本章小结

本节深入探讨了收敛序列的核心性质，并引入了强大的夹逼定理。

性质是基石：唯一性、有界性、保序性、四则运算法则等是分析序列行为的基础，其证明过程是运用 ε-N 语言的绝佳练习。
计算有套路：对于有理分式极限，上下同除最高次幂是标准方法；夹逼定理是处理形式复杂或不易直接求极限问题的关键工具。
存在性与值：单调收敛原理保证了单调有界序列极限的存在性；夹逼定理在给出极限值的同时，也证明了极限的存在。

行动清单：

默写性质：尝试默写收敛序列的六大性质，并理解其直观含义。
动手证明：选择保序性或有界性，独立用 ε-N 语言完成一次证明。
代码验证：运行讲义中的 Python 代码，改变参数，观察数值结果如何逼近理论极限，加深对夹逼定理的理解。

— 小象教研组

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