本文主要是介绍05:极限-无穷小,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
1.无穷小的概念及比较
定义: lim x → x 0 f ( x ) = 0 , 则称 f ( x ) 是 x = x 0 时的无穷小 定义: \lim_{x \to x_0} f(x) =0,则称f(x)是x=x_0时的无穷小 定义:x→x0limf(x)=0,则称f(x)是x=x0时的无穷小
性质 | Value |
---|---|
① | 有限个无穷小相加还是无穷小 |
② | 有界变量乘以无穷小还是无穷小 |
导管 | $1 |
有界函数:
有界 | 类型 |
---|---|
① | sin△及其反函数 |
② | cos△及其反函数 |
③ | arctan△ |
③ | arccot△ |
例1: 求 lim n → x 0 x ⋅ s i n 1 x 极限 求 \lim_{n \to x_0} x·sin \frac{1}{x} 极限 求n→x0limx⋅sinx1极限
解: ∵ s i n 1 x 为有界函数,有界 ⋅ 0 = 0 , ∴ lim n → x 0 x ⋅ s i n 1 x = 0 解:∵sin \frac{1}{x} 为有界函数,有界·0=0, ∴\lim_{n \to x_0} x·sin \frac{1}{x}=0 解:∵sinx1为有界函数,有界⋅0=0,∴n→x0limx⋅sinx1=0
1.2无穷小的阶的比较
x->x0时,α(x)->0, β(x)->0
lim x → x 0 α ( x ) β ( x ) = { 0 α ( x ) 是 β ( x ) 的高阶无穷小,记作: α ( x ) = o β ( x ) 1 α ( x ) 是 β ( x ) 的等价无穷小,记作: α ( x ) ∽ β ( x ) A ( A ≠ 0 ) α ( x ) 是 β ( x ) 的同阶无穷小 ∞ , α(x)是β(x)的低阶无穷小 \lim_{x \to x_0} \frac{α(x)}{β(x)} = \begin{cases} 0 & α(x)是β(x)的高阶无穷小,记作:α(x)=oβ(x) \\[2ex] 1 & α(x)是β(x)的等价无穷小,记作:α(x)\backsimβ(x) \\[2ex] A(A≠0) & α(x)是β(x)的同阶无穷小 \\[2ex] \infty, & \text{α(x)是β(x)的低阶无穷小} \\ \end{cases} x→x0limβ(x)α(x)=⎩ ⎨ ⎧01A(A=0)∞,α(x)是β(x)的高阶无穷小,记作:α(x)=oβ(x)α(x)是β(x)的等价无穷小,记作:α(x)∽β(x)α(x)是β(x)的同阶无穷小α(x)是β(x)的低阶无穷小
例 1 : x = 1 , α ( x ) = x 2 − 1 , β ( x ) = 2 x 2 − x − 1 lim x → 1 x 2 − 1 2 x 2 − x − 1 = lim x → 1 ( x − 1 ) ( x + 1 ) ( x − 1 ) ( 2 x + 1 ) = lim x → 1 x + 1 2 x + 1 = 2 3 ∴ α ( x ) 是 β ( x ) 的同阶无穷小 \begin{align*} 例1: &x=1,α(x)=x^2-1, β(x)=2x^2-x-1 \\ & \lim_{x \to 1} \frac{x^2-1}{2x^2-x-1}\\ &= \lim_{x \to 1} \frac{(x-1)(x+1)}{(x-1)(2x+1)} \\ & = \lim_{x \to 1} \frac{x+1}{2x+1} \\ & ={\frac{2}{3}} \\ & ∴α(x)是β(x)的同阶无穷小 \\ \end{align*} 例1:x=1,α(x)=x2−1,β(x)=2x2−x−1x→1lim2x2−x−1x2−1=x→1lim(x−1)(2x+1)(x−1)(x+1)=x→1lim2x+1x+1=32∴α(x)是β(x)的同阶无穷小
1.3 等价无穷小的替换(9类)只针对 0 0 型 \frac{0}{0}型 00型
前提: △ → 0 △\rightarrow0 △→0
项目 | 函数类型 |
---|---|
① | sin△=△ |
② | arcsin△=△ |
③ | tan△=△ |
④ | arctan△=△ |
⑤ | e △ − 1 = △ e^△-1=△ e△−1=△ |
⑥ | ln ( 1 + △ ) = △ \ln(1+△)=△ ln(1+△)=△ |
⑦ | 1-cos△= 1 2 △ 2 \frac{1}{2}△^2 21△2 |
⑧ | a △ − 1 = △ ⋅ ln a a^△-1=△·\ln a a△−1=△⋅lna |
⑨ | ( 1 + △ ) a − 1 = a ⋅ △ (1+△)^a-1=a·△ (1+△)a−1=a⋅△ |
⑩ | tan △ − cos △ = 1 2 △ 3 \tan△-\cos△=\frac{1}{2}△^3 tan△−cos△=21△3 |
例 1 : lim x → 0 1 + x 2 − 1 1 − c o s x = lim x → 0 1 + x 2 − 1 1 2 x 2 = lim x → 0 ( 1 + x 2 ) 1 2 − 1 1 2 x 2 = lim x → 0 1 2 x 2 1 2 x 2 = 1 \begin{align*} 例1: & \lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{1-cosx}\\ &= \lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{ \frac{1}{2}x^2} \\ & = \lim_{x \to 0} \frac{(1+x^2)^\frac{1}{2}-1}{ \frac{1}{2}x^2}\\ & =\lim_{x \to 0} \frac{\frac{1}{2}x^2}{\frac{1}{2}x^2} \\ & =1\\ \end{align*} 例1:x→0lim1−cosx1+x2−1=x→0lim21x21+x2−1=x→0lim21x2(1+x2)21−1=x→0lim21x221x2=1
例 2 : lim x → 0 tan x − cos x x 3 = lim x → 0 tan x ⋅ ( 1 − cos x ) x 3 = lim x → 0 1 2 x 3 x 3 = 1 2 \begin{align*} 例2: & \lim_{x \to 0} \frac{\tan x-\cos x}{x^3}\\ &= \lim_{x \to 0} \frac{\tan x·(1- \cos x)}{x^3}\\ & = \lim_{x \to 0} \frac{\frac{1}{2}x^3}{x^3}\\ & = \frac{1}{2}\\ \end{align*} 例2:x→0limx3tanx−cosx=x→0limx3tanx⋅(1−cosx)=x→0limx321x3=21
例 3 : lim x → 0 2 − 1 + cos x sin 2 x = lim x → 0 ( 2 − 1 + cos x ) ⋅ ( 2 + 1 + cos x ) ( x 2 ) ⋅ ( 2 + 1 + cos x ) = lim x → 0 2 − ( 1 + cos x ) x 2 ⋅ ( 2 + 1 + cos x ) = lim x → 0 1 − cos x x 2 ⋅ ( 2 + 1 + cos x ) = lim x → 0 1 2 x 2 x 2 ⋅ ( 2 + 1 + cos x ) = lim x → 0 1 2 2 + 1 + cos x = 2 8 \begin{align*} 例3: & \lim_{x \to 0} \frac{\sqrt 2-\sqrt {1+\cos x}}{\sin^2 x}\\ &= \lim_{x \to 0} \frac{(\sqrt 2-\sqrt {1+\cos x})·{(\sqrt 2+\sqrt {1+\cos x})}}{ (x^2)·(\sqrt 2+\sqrt {1+\cos x})}\\ & = \lim_{x \to 0} \frac{2-(1+\cos x)}{ x^2·(\sqrt 2+\sqrt {1+\cos x})}\\ & = \lim_{x \to 0} \frac{1-\cos x}{ x^2·(\sqrt 2+\sqrt {1+\cos x})}\\ & = \lim_{x \to 0} \frac{\frac{1}{2}x^2}{ x^2·(\sqrt 2+\sqrt {1+\cos x})}\\ & = \lim_{x \to 0} \frac{\frac{1}{2}}{ \sqrt 2+\sqrt {1+\cos x}}\\ & = \frac{\sqrt 2}{8}\\ \end{align*} 例3:x→0limsin2x2−1+cosx=x→0lim(x2)⋅(2+1+cosx)(2−1+cosx)⋅(2+1+cosx)=x→0limx2⋅(2+1+cosx)2−(1+cosx)=x→0limx2⋅(2+1+cosx)1−cosx=x→0limx2⋅(2+1+cosx)21x2=x→0lim2+1+cosx21=82
例 4 : lim x → 0 ( 1 + t a n x n − 1 ) ⋅ ( 1 + x − 1 ) 2 x ⋅ sin x = lim x → 0 1 n tan x ⋅ 1 2 x 2 x ⋅ sin x = lim x → 0 1 n x ⋅ 1 2 x 2 x 2 = lim x → 0 1 2 n x 2 2 x 2 = 1 4 n \begin{align*} 例4: & \lim_{x \to 0} \frac{(\sqrt [n]{1+\ tan x}-1)·(\sqrt {1+x}-1)}{2x·\sin x}\\ &= \lim_{x \to 0} \frac{\frac{1}{n}\tan x·\frac{1}{2}x}{2x·\sin x}\\ & =\lim_{x \to 0} \frac{\frac{1}{n} x·\frac{1}{2}x}{2x^2}\\ & = \lim_{x \to 0} \frac{\frac{1}{2n} {x^2}}{2x^2}\\ & = \frac{1}{4n} \end{align*} 例4:x→0lim2x⋅sinx(n1+ tanx−1)⋅(1+x−1)=x→0lim2x⋅sinxn1tanx⋅21x=x→0lim2x2n1x⋅21x=x→0lim2x22n1x2=4n1
例 5 : lim x → 0 e tan x − e sin x x 3 = lim x → 0 e sin x ( e tan x − sin x − 1 ) x 3 = lim x → 0 e sin x ⋅ ( tan x − sin x ) x 3 = lim x → 0 e sin x ⋅ 1 2 x 3 x 3 = lim x → 0 1 2 e sin x = 1 2 思路: ①对于底或指数相同的,提取后一项 ② e tan x − sin x − 1 要联想到 e △ − 1 = △ \begin{align*} 例5: & \lim_{x \to 0} \frac{e^{\tan x}-e^{\sin x}}{x^3}\\ &= \lim_{x \to 0} \frac{e^{\sin x}(e^{\tan x-\sin x}-1)}{x^3}\\ & = \lim_{x \to 0} \frac{e^{\sin x}·(\tan x-\sin x)}{x^3}\\ & = \lim_{x \to 0} \frac{e^{\sin x}·\frac{1}{2}x^3}{x^3}\\ & =\lim_{x \to 0} \frac{1}{2}e^{\sin x} \\ & =\frac{1}{2} \\ 思路:& ①对于底或指数相同的,提取后一项 \\ & ② e^{\tan x-\sin x} -1要联想到 e^△-1=△ \end{align*} 例5:思路:x→0limx3etanx−esinx=x→0limx3esinx(etanx−sinx−1)=x→0limx3esinx⋅(tanx−sinx)=x→0limx3esinx⋅21x3=x→0lim21esinx=21①对于底或指数相同的,提取后一项②etanx−sinx−1要联想到e△−1=△
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