泰勒公式和展开应用(同阶无穷小求解)

Author:张一极
个人博客:极度空间
公众号:视觉迷航

知识

Tips:函数差值极限为0,不可以证明两个函数极限相等,因为两个函数的极限未必一定存在
不是任意的无穷小都可以比阶
两个无穷小的比值为c不为0,则分子是分母的同阶无穷小
比值为0,则分子是分母的高阶无穷小
$\infty$ ,分子是分母的低阶无穷小
比值为1,分子是分母的等价无穷小
$[b(x)]^k$ =c不为0,则a是b的k阶无穷小

泰勒公式:

$\begin{array}{ll}\sin x=x-\frac{x^{3}}{3 !}+o\left(x^{3}\right), & \text { cos } x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+o\left(x^{1}\right) \\ \arcsin x=x+\frac{x^{3}}{3 !}+o\left(x^{3}\right), & \tan x=x+\frac{x^{3}}{3}+o\left(x^{3}\right) \\ \arctan x=x-\frac{x^{3}}{3}+o\left(x^{2}\right) & \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+o\left(x^{3}\right) \\ \mathrm{e}^{\prime}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+o\left(x^{3}\right), & (1+x)^{*}=1+\alpha \cdot x+\frac{\alpha(a-1)}{2 !} x^{2}+o\left(x^{2}\right)\end{array}$

无穷小运算法则:

\begin{array}{l}O\left(x^{m}\right)+o\left(x^{n}\right)=o\left(x^{l}\right) ,l=min(m,n) \\ o\left(x^{m}\right) \cdot o\left(x^{n}\right)=0\left(x^{m+n}\right)\end{array}

一个非0常数,乘以一个无穷小,不影响阶数.

常用等价无穷小替换:

\begin{array}{l}\sin x \sim x, \quad \text { tan } x \sim x, \quad \text { aresin } x \sim x, \quad \text { arctan } x \sim x, \quad \ln (1+x) \sim x, \quad \mathrm{e}^{x}-1 \sim x \\ \quad a^{\prime}-1 \sim x \ln a, \quad 1-\cos x \sim \frac{1}{2} x^{2}, \quad(1+x)^{a}-1 \sim a \cdot x\end{array}

泰勒公式展开应用:

$\frac{A}{B}$ 型

问:

$\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}$

解:

求#

\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}\\=\lim _{x \rightarrow 0} \frac{x-[x-\frac{1}{6}x^3+o(x^5)]}{x^{3}}\\=\lim _{x \rightarrow 0} \frac{\frac{1}{6} x^{3}+o\left(x^{5}\right)}{x^{1}}-\frac{1}{6}

$A-B$ 型,幂次最低原则:

应该展开到系数不相同的最低次幂

问

$x\to0$ $cosx-e^{-\frac{x^2}{2}}$ $ax^b$ 为等价无穷小,求a,b.

解:

首先根据泰勒展开式:

cosx=1-\frac{x^2}{2}+\frac{1}{24}x^4+o(x^4) \\e^{-\frac{x^2}{2}} = 1-\frac{x^2}{2}+\frac{1}{8}x^4+o(x^4) \\\to\frac{1}{24}x^4-\frac{1}{8}x^4 \\=-\frac{1}{12}x^4+o(x^4)

得:

$a = -\frac{1}{12}\\b = 4$

问

\lim _{x \rightarrow 0} \frac{\arcsin x-\arctan x}{\sin x-\tan x}

解:

根据泰勒展开:

$sinx = x-\frac{x^3}{6}+o(x^2)$

$tanx = x+\frac{x^3}{3}+o(x^3)$

第二项起系数已经不相同,故此时停下,不再展开.

$arcsinx = x+\frac{x^3}{6}+o(x^3)$

$arctanx = x - \frac{x^3}{6}+o(x^3)$

$\lim _{x \rightarrow 0} \frac{\arcsin x-\arctan x}{\sin x-\tan x}$

$lim_{x\to0}\frac{\frac{x^3}{2}}{\frac{x^3}{-2}}$

$-1$

问

$lim_{x\to0}x+sinx$

解:

x+sinx = x-(-sinx) \\x = 1\cdot x \\-sinx = -1\cdot x+\frac{1}{6}x^3+o(x) \\x+sinx=[2x+o(x)]\to2x \\同理(x-sinx)\to\frac{1}{6}x^3 \\\therefore tanx-x\to\frac{1}{3}x^2 \\\therefore sinx-x \to-\frac{1}{6}x^3

3.等价无穷小替换

问

\lim _{x \rightarrow 0} \frac{\sin ^{2} x-x^{2}}{e^{x^{4}}-1}

解:

\\e^{x^4}-1\to x^4 \\sin^2x-x^2 = (sinx+x)(sinx-x) \\\lim _{x \rightarrow 0} \frac{\sin ^{2} x-x^{2}}{e^{x^{4}}-1}\to \lim _{x \rightarrow 0} \frac{(sinx+x)(sinx-x)}{x^{4}} \to 等价无穷小替换得:\lim _{x \rightarrow 0} \frac{(2x)(-\frac{1}{6}x^3)}{x^{4}}=-\frac{1}{3}

4.归结原则

问:

\lim n\to \infty \left(\operatorname{ntan} \frac{1}{n}\right)^{n^{2}}

解:

\lim x\to0 \left(\operatorname{ntan} \frac{1}{n}\right)^{n^{2}} \\=lim_{x\to0}(\frac{tanx}{x})^\frac{1}{x^2} \\幂指函数:lim~u^v = e^{lim(v\cdot lnu)} \\=lim_{x\to0}(\frac{tanx}{x})^\frac{1}{x^2} \\=e^{lim_{x\to0}\frac{1}{x^2}ln\frac{tanx}{x}} \\拆解:lim_{x\to0}ln\frac{tanx}{x}\\ = lim_{x\to0}ln({~1+\frac{tanx}{x}-1})\to lim_{x\to0}(\frac{tanx}{x}-1) \\其中(\frac{tanx}{x}-1)\to0 \\=lim_{x\to0}(\frac{tanx-x}{x}) \\\because tanx-x\to\frac{1}{3}x^3 \\=lim_{x\to0}(\frac{\frac{1}{3}x^3}{x}) \\\therefore lim_{x\to0}e^{lim_{x\to0}\frac{1}{x^2}ln\frac{tanx}{x}} \\=lim_{x\to0}e^{\frac{1}{x^2}\cdot {\frac{\frac{1}{3}x^3}{x}}} \\=e^{\frac{1}{3}}

$n=\frac{1}{n}$ $e^\frac{1}{3}$ .