泰山之管穿石,单极之绠断干。水非石之钻,索非木之锯,渐靡使之然也。

——东汉·班固《汉书·枚乘传》

楔子

前两天看到 blackpenredpen曹老师 的视频,有感触,觉得对于数学的学习而言,刷题中才能真正学会东西。有些东西,看了不一定理解,理解了不一定会应用,用了不一定真正掌握。做题能很好的杂糅这一切,让理解更深入,记忆更加清晰。

文章痕迹

4-3开始进行整理写作

4-6

收集到22个,感觉没得写了。早上偶然看到mdnice群里有人发导数的笔记,然后又翻到了他之前的极限笔记。然而,笔记里第一题我都做的吃力,NND,再看才知晓是Wu老的高数17天。然后找来翻了翻,光极限就讲了57页😮,当然里面方法还是基本方法,但很多都是“稍稍”延伸了几步,却足以让我手足无措。这两天不能放松,好好理理思路。

4-7

要把等价无穷小推导一遍,不然太不好记了。

4-9

利用泰勒展开式推导出所有常见无穷小(4 \to 50)

4-15

累了,多巴胺业已耗尽,五十道,了了之。

🔧2023-7-30

之前换插件导致部分样式出错,今日对文章进行维护。


初出茅庐

这一部分有27道例题,主要介绍极限计算的基本方法。

limn 1n2+1+1n2+2+1n2+n\begin{equation} \lim_{n\to \infty}\ \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\dots \frac{1}{\sqrt{n^2+n}} \end{equation}

🔑Answer

提示:利用夹逼定理(🗣️哪—里—跑—)

1nn2+1bn1nn2+n\frac{1*n}{\sqrt{n^2+1}}\le b_n \le \frac{1*n}{\sqrt{n^2+n}}

limx0ln(1+4x2)sinx2\begin{equation} \lim _{x \rightarrow 0} \frac{\ln \left(1+4 x^{2}\right)}{\sin x^{2}} \end{equation}

Answer

提示:等价无穷小

sin(🥕)ln(1+🥕)\sin(🥕)\sim \ln(1+🥕)

limx0ln(1+4x2)sinx2=4x2x2= 4 \begin{aligned} & \lim _{x \rightarrow 0} \frac{\ln \left(1+4 x^{2}\right)}{\sin x^{2}} \\ &= \frac{4 x^{2}}{x^{2}} \\ &=\ 4 \end{aligned}

limx0x3sin1x3\begin{equation} \lim_{x \rightarrow 0} x^{3} \sin \frac{1}{x^{3}} \end{equation}

🔑Answer

提示:注意等价无穷小的条件

 x3为无穷小,sin1x31无穷小与有界变量的乘积仍为无穷小limx0x3sin1x3=0\ x^3\text{为无穷小},|\sin{\frac{1}{x^3}}|\le 1 \\ {\color{blue} \text{无穷小与有界变量的乘积仍为无穷小}}\\ \therefore \lim _{x \rightarrow 0} x^{3} \sin \frac{1}{x^{3}}= 0

limx0sinxtanxx3\begin{equation} \lim_{x \rightarrow 0} \frac{\sin x-\tan x}{x^{3}} \end{equation}

🔑Answer

提示:拼凑等价无穷小

limx0sinxtanxx3=limx0sinx(11cosx)x3=limx0sinx(cosx1)x3cosx=limx0x(12x2)x3=12\begin{aligned} &lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x^{3}}\\ &=\lim _{x \rightarrow 0} \frac{\sin x\left(1-\frac{1}{\cos x}\right)}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x(\cos x-1)}{x^{3} \cos x} \\ &=\lim _{x \rightarrow 0} \frac{x \cdot\left(-\frac{1}{2} x^{2}\right)}{x^{3}}\\ &=-\frac{1}{2} \end{aligned}

limx0(1+xex)1x\begin{equation} \lim _{x \rightarrow 0}\left(1+x e^{x}\right)^{\frac{1}{x}} \end{equation}

🔑Answer

提示:11^\infty常规思路·第一弹

limx0(1+🥕)1🥕=e(🥕0)\lim_{x\to 0} (1+🥕)^{\frac{1}{🥕}}=e \quad{(🥕\ne0)}

limx0(1tanx1+tanx)1sin2x\begin{equation} \lim _{x \rightarrow 0}\left(\frac{1-\tan x}{1+\tan x}\right)^{\frac{1}{-\sin 2 x}} \end{equation}

🔑Answer

提示:11^\infty常规思路·第二弹

limx0(1tanx1+tanx)1sin2x=limx0(1+(1tanx1+tanx1))11tanx1+tanx1×2tanx1+tanx1sin2x=limx0e2tanx1+tanx1sin2x=limx0e2x1+tanx12x=limx0e11+tanx=e\begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{1-\tan x}{1+\tan x}\right)^{\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}\left(1+(\frac{1-\tan x}{1+\tan x}-1)\right)^{\frac{1}{\frac{1-\tan x}{1+\tan x}-1}\times \frac{-2\tan x}{1+\tan x}\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{-2\tan x}{1+\tan x}\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{-2 x}{1+\tan x}\frac{1}{- 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{1}{1+\tan x}}\\ &=e \end{aligned}

limn(n2+nn)\begin{equation} \lim_{n\to \infty} (\sqrt{n^2+n}-n) \end{equation}

From Deyi

提示:常规思路:平方差拼凑

limn(n2+nn)=limn(n2+nn1)=limn(n2+nn2n2+n+n)=limn(nn2+n+n)=limn(11+1n+1)=12\begin{aligned} &\lim_{n\to \infty} (\sqrt{n^2+n}-n)\\ &=\lim_{n\to \infty} (\frac{\sqrt{n^2+n}-n}{1})\\ &=\lim_{n\to \infty} (\frac{n^2+n-n^2}{\sqrt{n^2+n}+n)}\\ &=\lim_{n\to \infty} (\frac{n}{\sqrt{n^2+n}+n)}\\ &=\lim_{n\to \infty} (\frac{1}{\sqrt{1+\frac{1}{n}}+1})\\ &=\frac{1}{2}\\ \end{aligned}

limxx(x+2x3)\begin{equation} \lim_{x\to \infty} \sqrt{x}(\sqrt{x+2}-\sqrt{x-3}) \end{equation}

From Deyi

提示:常规思路:平方差拼凑

limxx(x+2x3)=limxx(x+2x3)1=limxx×5x+2+x3)=limx51+2/x+13/x)=52\begin{aligned} &\lim_{x\to \infty} \sqrt{x}(\sqrt{x+2}-\sqrt{x-3})\\ &=\lim_{x\to \infty} \frac{\sqrt{x}(\sqrt{x+2}-\sqrt{x-3})}{1}\\ &=\lim_{x\to \infty} \frac{\sqrt{x}\times 5}{\sqrt{x+2}+\sqrt{x-3})}\\ &=\lim_{x\to \infty} \frac{ 5}{\sqrt{1+2/x}+\sqrt{1-3/x})}\\ &=\frac{5}{2} \end{aligned}

limx0(1x1ex1)\begin{equation} \lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) \end{equation}

1987卷三

提示:凑乘积形式,等价无穷小+洛必达

limx0(1x1ex1)=limx0(ex1x(ex1)x)=limx0(ex1xx2)=limx0(ex12x)=limx0(ex2)=12\begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1-x}{(e^x-1)x}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1-x}{x^2}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1}{2x}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x}{2}\right)\\ &=\frac{1}{2} \end{aligned}

limx0(xsinx)ex21x31\begin{equation} \lim_{x\to 0} \frac{(x-\sin x) e^{-x^2}}{\sqrt{1-x^3}-1} \end{equation}

🔑Answer

提示:等价无穷小替换

(1+🥕)a1a🥕(1+🥕)^a-1 \sim a🥕

limx0(xsinx)ex21x31=limx0ex2×limx0xsinx1/2×(x3)=2×limx0xsinxx3洛必达法则=2×limx01cosx3x2=2×limx01/2×x23x2=13\begin{aligned} &\lim_{x\to 0} \frac{(x-\sin x) e^{-x^2}}{\sqrt{1-x^3}-1}\\ &= \lim_{x\to 0} e^{-x^2}\times \lim_{x\to 0} \frac{x-\sin x}{1/2\times (-x^3)}\\ &= -2\times \lim_{x\to 0} \frac{x-\sin x}{x^3}\qquad \text{洛必达法则}\\ &= -2\times \lim_{x\to 0} \frac{1-\cos x}{3x^2}\\ &= -2\times \lim_{x\to 0} \frac{1/2 \times x^2}{3x^2}\\ &=- \frac{1}{3} \end{aligned}

limx0xsin2xx+sin5x\begin{equation} \lim_{x\to 0} \frac{x-\sin 2x}{x + \sin 5x} \end{equation}

🔑Answer

提示:加减法使用等价无穷小的条件

\becausex0x \rightarrow 0 时, sin2x2x,sin5x5x\sin 2 x \sim 2 x, \sin 5 x \sim 5 x, 且 limx0sin2xx\lim _{x \rightarrow 0} \frac{\sin 2 x}{x} =21,limx0sin5xx=51=2 \neq 1, \lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=5 \neq-1,

满足等价无穷小替换对加减法成立的条件,

limx0xsin2xx+sin5x=limx0x2xx+5x=16\therefore \lim _{x \rightarrow 0} \frac{x-\sin 2 x}{x+\sin 5 x}=\lim _{x \rightarrow 0} \frac{x-2 x}{x+5 x}=-\frac{1}{6}

limx(cos1x)x2\begin{equation} \lim _{x \rightarrow \infty}\left(\cos \frac{1}{x}\right)^{x^{2}} \end{equation}

🔑Answer

提示:11^{\infty}型·第三弹

limx(cos1x)x2=limx[1+(1cosx1)]11cosx1×(1cosx1)×x2=elimx(cos1x1)x2=elimxcos1x1(1x)2=elimx12(1x)2(1x)2=e12=1e\begin{aligned} &\displaystyle \lim _{x \rightarrow \infty}\left(\cos \frac{1}{x}\right)^{x^{2}}\\ &=\displaystyle\lim_{x\to \infty}[1+(\frac{1}{\cos x}-1)]^{\frac{1}{\frac{1}{\cos x}-1}\times (\frac{1}{\cos x}-1)\times x^2}\\ &=e^{\displaystyle\lim_{x\to \infty}\left(\cos \frac{1}{x}-1\right) x^{2}}\\ &=e^{\displaystyle\lim_{x\to \infty} \frac{\cos \frac{1}{x}-1}{\left(\frac{1}{x}\right)^{2}}}\\ &=e^{\displaystyle\lim_{x\to \infty} \frac{-\frac{1}{2}\left(\frac{1}{x}\right)^{2}}{\left(-\frac{1}{x}\right)^{2}}}\\ &=e^{^{-\frac{1}{2}}}\\ &=\frac{1}{\sqrt{e}} \end{aligned}

limx(x+a)x+a(x+b)x+b(x+a+b)2x+a+b\begin{equation} \lim _{x \rightarrow \infty} \frac{(x+a)^{x+a}(x+b)^{x+b}}{(x+a+b)^{2 x+a+b}} \end{equation}

From Zhongxiang-17Days

提示:11^{\infty}型·第四弹

limx(x+a)x+a(x+b)x+b(x+a+b)2x+a+b=limx(x+a)x+a(x+a+b)x+a(x+b)x+b(x+a+b)x+b=limx1(1+bx+a)x+a1(1+ax+b)x+b=1eb1ea=e(a+b)\begin{aligned} &\lim _{x \rightarrow \infty} \frac{(x+a)^{x+a}(x+b)^{x+b}}{(x+a+b)^{2 x+a+b}}\\ &=\lim _{x \rightarrow \infty} \frac{(x+a)^{x+a}}{(x+a+b)^{x+a}} \cdot \frac{(x+b)^{x+b}}{(x+a+b)^{x+b}} \\ &=\lim _{x \rightarrow \infty} \frac{1}{\left(1+\frac{b}{x+a}\right)^{x+a}} \cdot \frac{1}{\left(1+\frac{a}{x+b}\right)^{x+b}} \\ &=\frac{1}{\mathrm{e}^{b}} \cdot \frac{1}{\mathrm{e}^{a}} \\ &=\mathrm{e}^{-(a+b)} \end{aligned}

limx0[ln(x+1+x2)x]11cosx\begin{equation} \lim _{x \rightarrow 0}\left[\frac{\ln (x+\sqrt{1+x^{2}})}{x}\right]^{\frac{1}{1-\cos x}} \end{equation}

From Zhongxiang-17Days

提示:11^{\infty}型·第五弹,常规思路

limx0[ln(x+1+x2)x]11cosx=limx0[1+ln(x+1+x2)xx]11cosx=limx0eln(x+1+x2)xx(1cosx)幂化简:limx0ln(x+1+x2)xx12x2=limx011+x2132x2=limx012x232x2=13原式=e13\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{x}\right]^{\frac{1}{1-\cos x}}\\ &=\lim _{x \rightarrow 0}\left[1+\frac{\ln \left(x+\sqrt{1+x^{2}}\right)-x}{x}\right]^{\frac{1}{1-\cos x}} \\ &=\lim _{x \rightarrow 0} e^{\frac{\ln (x + \sqrt{1+x^2})-x}{x(1-\cos x )}}\\ \text{幂化简:}&\color{blue}\lim _{x \rightarrow 0} \frac{\ln \left(x+\sqrt{1+x^{2}}\right)-x}{x \cdot \frac{1}{2} x^{2}} \\ &\color{blue}=\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{1+x^{2}}}-1}{\frac{3}{2} x^{2}} \\ &\color{blue}=\lim _{x \rightarrow 0} \frac{-\frac{1}{2} x^{2}}{\frac{3}{2} x^{2}}\\ &\color{blue}=-\frac{1}{3}\\ \text{原式}&=e^{-\frac{1}{3}}\\ \end{aligned}

limn(π2arctann)1lnn\begin{equation} \lim _{n \rightarrow \infty}\left(\frac{\pi}{2}-\arctan n\right)^{\frac{1}{\ln n}} \end{equation}

From Zhongxiang-17Days

提示:11^{\infty}型·变体,思路还是一样的,幂运算变乘法

limn(π2arctann)1lnn=limneln(π/2arctanx)lnx幂化简:limnln(π/2arctanx)lnx=limx+1π2arctanx(11+x2)1x=limx+1xπ2arctanx=limx+1x211+x2=1原式=e1\begin{aligned} &\lim _{n \rightarrow \infty}\left(\frac{\pi}{2}-\arctan n\right)^{\frac{1}{\ln n}}\\ &=\lim _{n \rightarrow \infty}e^{\frac{\ln (\pi /2 - \arctan x)}{\ln x}}\\ \text{幂化简:}&\color{blue}\lim _{n \rightarrow \infty} \frac{\ln (\pi /2 - \arctan x)}{\ln x}\\ &\color{blue}=\lim _{x \rightarrow+\infty} \frac{\frac{1}{\frac{\pi}{2}-\arctan x} \cdot\left(-\frac{1}{1+x^{2}}\right)}{\frac{1}{x}}\\ &\color{blue}=-\lim _{x \rightarrow+\infty} \frac{\frac{1}{x}}{\frac{\pi}{2}-\arctan x} \\ &\color{blue}=-\lim _{x \rightarrow+\infty} \frac{-\frac{1}{x^{2}}}{-\frac{1}{1+x^{2}}} \\ &\color{blue}=-1\\ \text{原式}&=e^{-1}\\ \end{aligned}

limx01cos2x12xsin2xx2(ex21)\begin{equation} \lim_{x\to 0} \frac{1-\cos^2x-\frac{1}{2} x \sin 2x}{x^2(e^{x^2}-1)} \end{equation}

🔑Answer

提示:多次使用洛必达法则

limx01cos2x12xsin2xx2(ex21)=limx01cos2x12xsin2xx2×x2上下都趋近于0,使用洛必达法则=limx02cosxsinx12sin2xcos2x×x4x3=limx012sin2xcos2x×x4x3上下都趋近于0,使用洛必达法则=limx0cos2xcos2x+sin2x×2x12x2=limx02x×2x12x2=13\begin{aligned} &\displaystyle \lim_{x\to 0} \frac{1-\cos^2x-\frac{1}{2} x \sin 2x}{x^2(e^{x^2}-1)}\\ &=\displaystyle \lim_{x\to 0} \frac{1-\cos^2x-\frac{1}{2} x \sin 2x}{x^2\times x^2}\qquad \text{上下都趋近于0,使用洛必达法则}\\ &=\displaystyle \lim_{x\to 0} \frac{2\cos x\sin x-\frac{1}{2}\sin 2x-\cos 2x\times x}{4x^3}\\ &=\displaystyle \lim_{x\to 0} \frac{\frac{1}{2}\sin 2x-\cos 2x\times x}{4x^3}\qquad \text{上下都趋近于0,使用洛必达法则}\\ &=\displaystyle \lim_{x\to 0} \frac{\cos 2x-\cos 2x+\sin 2x\times 2 x}{12x^2}\\ &=\displaystyle \lim_{x\to 0} \frac{ 2x\times 2 x}{12x^2}\\ &=\frac{1}{3} \end{aligned}

limx0(1x21sin2x)\begin{equation} \lim _{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2x}) \end{equation}

🔑Answer

提示:多次使用洛必达法则

limx0(1x21sin2x)=limx0sin2xx2x2sin2x上下都趋近于0,使用洛必达法则=limx0sin2x2x4x3上下都趋近于0,使用洛必达法则=limx02cos2x212x2=limx0cos2x16x2=limx012(2x)26x2=13\begin{aligned} &\displaystyle\lim _{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2x})\\ &=\displaystyle\lim _{x\to 0}\frac{\sin^2x-x^2}{x^2 \sin^2x}\qquad \text{上下都趋近于0,使用洛必达法则}\\ &=\displaystyle\lim _{x\to 0}\frac{\sin 2x-2x}{4x^3}\qquad \text{上下都趋近于0,使用洛必达法则} \\ &=\displaystyle\lim _{x\to 0}\frac{2\cos 2x-2}{12x^2} \\ &=\displaystyle\lim _{x\to 0}\frac{\cos 2x-1}{6x^2} \\ &=\displaystyle\lim _{x\to 0}\frac{\frac{1}{2} (2x)^2}{6x^2}\\ &=\frac{1}{3} \end{aligned}

limx0(cosxcos2x)1x2\begin{equation} \lim _{x \rightarrow 0} (\frac{\cos x}{\cos 2x})^{\frac{1}{x^2}} \end{equation}

🔑Answer

提示:等价无穷小之 加不反

limx0(cosxcos2x)1x2=elimx0(cosxcos2x1)x2=elimx0cosxcos2xx2cos2x=elimx0cosxcos2xx2=elimx0sinx+2sin2x2xlimx0sinx2sin2x1=e32\begin{aligned} &\lim _{x \rightarrow 0} (\frac{\cos x}{\cos 2x})^{\frac{1}{x^2}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\left(\frac{\cos x}{\cos 2 x}-1\right)}{x^{2}}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\cos x-\cos 2x}{x^2 \cos 2 x}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\cos x-\cos 2x}{x^2 }}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{-\sin x+2 \sin 2 x}{2 x}}\qquad \because \lim_{x\to 0}\frac{-\sin x}{2\sin 2x}\ne -1\\ &=\mathrm{e}^{\frac{3}{2}} \end{aligned}

limx0[sinxsin(sinx)]sinxx4\begin{equation} \lim _{x \rightarrow 0} \frac{[\sin x-\sin (\sin x)] \cdot \sin x}{x^{4}} \end{equation}

🔑Answer

提示:等价无穷小+洛必达法则

limx0[sinxsin(sinx)]sinxx4=limx0cosxcos(sinx)cosx3x2=limx0cosx×limx01cos(sinx)3x2=limx012sin2x3x2=limx01x26x2=16\begin{aligned} &\lim _{x \rightarrow 0} \frac{[\sin x-\sin (\sin x)] \cdot \sin x}{x^{4}} \\ &=\lim _{x \rightarrow 0} \frac{\cos x-\cos (\sin x) \cdot \cos x}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \cos x \times \lim _{x \rightarrow 0} \frac{1-\cos (\sin x)}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sin ^{2} x}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{1 x^{2}}{6 x^{2}} \\ &=\frac{1}{6} \end{aligned}

求正的常数a与b,使下式成立:limx01bxsinx0xt2a+t2dt=1\begin{equation} \text{求正的常数a与b,使下式成立:} \lim _{x \rightarrow 0} \frac{1}{b x-\sin x} \int_{0}^{x} \frac{t^{2}}{\sqrt{a+t^{2}}} d t=1 \end{equation}

1987卷一

提示:洛必达+等价无穷小,看见积分,必定洛必达。

limx01bxsinx0xt2a+t2dt=1limx0x2a+x2bcosx=1limx01a+x2×limx0x2bcosx=11a×limx0x2bcosx=1🥝1a×limx01bx211bcosx=11a×limx01bx212bx21b+1=11a×2=1a=4(a=4🥝)b=1\begin{aligned} \lim _{x \rightarrow 0} \frac{1}{b x-\sin x} \int_{0}^{x} \frac{t^{2}}{\sqrt{a+t^{2}}} d t&=1\\ \lim _{x \rightarrow 0}\frac{ \frac{x^{2}}{\sqrt{a+x^{2}}} }{ {b -\cos x}}&=1\\ \lim _{x \rightarrow 0} \frac{1}{\sqrt{a+x^{2}}}\times \lim _{x \rightarrow 0}\frac{ x^{2} }{ {b -\cos x}}&=1\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{ x^{2} }{ {b -\cos x}}&=1\qquad \text{🥝}\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{1}{b} \frac{ x^{2} }{ {1 -\frac{1}{b}\cos x}}&=1\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{1}{b} \frac{ x^{2} }{ \frac{1}{2b}x^2-\frac{1}{b}+1}&=1\\ \frac{1}{\sqrt{a}}\times2&=1\\ a&=4\\ (a=4\to\text{🥝})\Longrightarrow b&=1 \end{aligned}

limx00x(xt)sint2 dt(x2+x3)(11x2)\begin{equation} \lim _{x \rightarrow 0} \frac{\int_{0}^{x}(x-t) \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right)\left(1-\sqrt{1-x^{2}}\right)} \end{equation}

ShowTime

提示:看见积分,必定洛必达,但也要注意顺序。

limx00x(xt)sint2 dt(x2+x3)(11x2)=limx0x0xsint2 dt0xtsint2 dt(x2+x3)12x2=limx0x0xsint2 dt0xtsint2 dt12x4快使用洛必达,哼哼哈嘿!=limx00xsint2 dt+xsinx2xsinx22x3=limx0sinx26x2=16\begin{aligned} &\lim _{x \rightarrow 0} \frac{\int_{0}^{x}(x-t) \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right)\left(1-\sqrt{1-x^{2}}\right)}\\ &=\lim _{x \rightarrow 0} \frac{x \int_{0}^{x} \sin t^{2} \mathrm{~d} t-\int_{0}^{x} t \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right) \cdot \frac{1}{2} x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{ x \int_{0}^{x} \sin t^{2} \mathrm{~d} t- \int_{0}^{x} t \sin t^{2} \mathrm{~d} t}{\frac{1}{2} x^{4}} \quad \text{快使用洛必达,哼哼哈嘿!} \\ &=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t^{2} \mathrm{~d} t+x \sin x^{2}-x \sin x^{2}}{2 x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x^{2}}{6 x^{2}}\\&=\frac{1}{6} \end{aligned}

limx0[1ex11ln(1+x)]\begin{equation} \lim _{x \rightarrow 0}\left[\frac{1}{e^{x}-1}-\frac{1}{\ln (1+x)}\right] \end{equation}

🔑Answer

提示:等价无穷小+洛必达法则

limx0[1ex11ln(1+x)]=limx0ln(1+x)ex+1(ex1)ln(1+x)=limx0ln(1+x)ex+1xx=limx011+xex2x上下都趋近于0,使用洛必达法则=12limx01ex(1+x)xlimx011+x=12limx0(1ex(1+x))x=12limx0(ex(2+x))=1.\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{1}{\mathrm{e}^{x}-1}-\frac{1}{\ln (1+x)}\right]\\&=\lim _{x \rightarrow 0} \frac{\ln (1+x)-\mathrm{e}^{x}+1}{\left(\mathrm{e}^{x}-1\right) \ln (1+x)}\\&=\lim _{x \rightarrow 0} \frac{\ln (1+x)-\mathrm{e}^{x}+1}{x \cdot x} \\& = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}-\mathrm{e}^{x}}{2 x}\qquad \text{上下都趋近于0,使用洛必达法则} \\ &=\frac{1}{2} \lim _{x \rightarrow 0} \frac{1-\mathrm{e}^{x}(1+x)}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{1+x}\\&=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(1-\mathrm{e}^{x}(1+x)\right)^{\prime}}{x^{\prime}}\\&=-\frac{1}{2} \lim _{x \rightarrow 0}\left(\mathrm{e}^{x}(2+x)\right)\\&=-1 . \end{aligned}

limx0x2sin1xx+sinx\begin{equation} \lim _{x \rightarrow 0} \frac{x^{2} \sin \frac{1}{x}}{x+\sin x} \end{equation}

From Zhongxiang-17Days

提示:请不要用洛必达!!

简单分析一下,可以发现这是一个00\frac{0}{0}型极限,要是你手痒痒直接用了,就会陷入绝境: 原式 =limx02xsin1xcos1x1+cosx\text { 原式 }=\lim _{x \rightarrow 0} \frac{2 x \sin \frac{1}{x}-\cos \frac{1}{x}}{1+\cos x}

实际上这道题的意义就是告诉你要注意洛必达的条件:求导后的极限存在才能用洛必达

 正解:原式 =limx0xsin1x1+sinxx=02=0\text { 正解:原式 }=\lim _{x \rightarrow 0} \frac{x \sin \frac{1}{x}}{1+\frac{\sin x}{x}}=\frac{0}{2}=0


拉格朗日中值定理专区

如果函数 f(x)f(x) 在闭区间 [a,b][a, b] 上连续,在开区间 (a,b)(a, b) 上可㝵,那么在开区间 (a,b)(a, b) 内至少存在一点 ξ\xi 使得 :

f(ξ)=f(b)f(a)baf^{\prime}(\xi)=\frac{f(b)-f(a)}{b-a}

这个方法神奇的紧!

limx0cos(sinx)cosxarcsin4x\begin{equation} \lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{\arcsin ^{4} x} \end{equation}

From Jin

提示:

limx0cos(sinx)cosxarcsin4x=limx0sinξ(sinxx)x4\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{\arcsin ^{4} x}=\lim _{x \rightarrow 0} \frac{-\sin \xi \cdot(\sin x-x)}{x^{4}}

ξ\xi介于sinx\sin xxx之间,由夹逼定理可以得到limx0sinξx=1\lim_{x\to 0}\frac{\sin \xi}{x}=1

limx0cos(sinx)cosxarcsin4x=limx0(sinxx)x3=limx01cosx3x2=16\begin{aligned} &\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{\arcsin ^{4} x}\\ &=\lim _{x \rightarrow 0} \frac{-(\sin x-x)}{x^{3}}\\ &=\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x^{2}}&\\ &=\frac{1}{6}\\ \end{aligned}

驾轻就熟

limx0[ax(1x2a2)ln(1+ax)](a0)\begin{equation} \lim _{x \rightarrow 0}\left[\frac{a}{x}-\left(\frac{1}{x^{2}}-a^{2}\right) \ln (1+a x)\right](a \neq 0) \end{equation}

1997数三

提示:等价无穷小推论

limx0[ax(1x2a2)ln(1+ax)]=limx0[axln(1+ax)x2]+limx0a2ln(1+ax)=limx0axln(1+ax)x2=limx012(ax)2x2(🥕ln(1+🥕)12🥕2)=a22.\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{a}{x}-\left(\frac{1}{x^{2}}-a^{2}\right) \ln (1+a x)\right]\\ &=\lim _{x \rightarrow 0}\left[\frac{a}{x}-\frac{\ln (1+a x)}{x^{2}}\right]+\lim _{x \rightarrow 0} a^{2} \ln (1+a x) \\ &=\lim _{x \rightarrow 0} \frac{a x-\ln (1+a x)}{x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{\frac{1}{2}(a x)^{2}}{x^{2}} {\color{red}\quad\left(🥕-\ln (1+🥕) \sim \frac{1}{2} 🥕^{2}\right)} \\ &=\frac{a^{2}}{2} . \end{aligned}

limx+[(nx)n+1]n+1(x+1)(x2+2)(xn+n)\begin{equation} \lim _{x \rightarrow+\infty} \frac{\sqrt{\left[(n x)^{n}+1\right]^{n+1}}}{(x+1)\left(x^{2}+2\right) \cdots\left(x^{n}+n\right)} \end{equation}

From Zhongxiang-17Days

提示:找到相同的项,消消消,提取出来x,凑出来可以计算的无穷小。

limx+[(nx)n+1]n+1(x+1)(x2+2)(xn+n)=limx+(nn+1xn)n+12(1+1x)(1+2x2)(1+nxn)=nn(n+1)2\begin{aligned} &\lim _{x \rightarrow+\infty} \frac{\sqrt{\left[(n x)^{n}+1\right]^{n+1}}}{(x+1)\left(x^{2}+2\right) \cdots\left(x^{n}+n\right)}\\ &=\lim _{x \rightarrow+\infty} \frac{\left(n^{n}+\frac{1}{x^{n}}\right)^{\frac{n+1}{2}}}{\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x^{2}}\right) \cdots\left(1+\frac{n}{x^{n}}\right)}\\ &=n^{\frac{n(n+1)}{2}} \end{aligned}

limx0x2(ex1)1+tanx1+x\begin{equation} \lim_{x\to 0} \frac{x^2(e^x-1)}{\sqrt{1+\tan x}-\sqrt{1+ x}} \end{equation}

From Deyi

提示:泰勒展开式

tanx=n=1B2n(4)n(14n)(2n)!x2n1=x+13x3+215x5+17315x7+622835x9+1382155925x11+218446081075x13+929569638512875x15+x(1,1)\tan x=\sum_{n=1}^{\infty} \frac{B_{2 n}(-4)^{n}\left(1-4^{n}\right)}{(2 n) !} x^{2 n-1}=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7}+\frac{62}{2835} x^{9}+\frac{1382}{155925} x^{11}+\frac{21844}{6081075} x^{13}+\frac{929569}{638512875} x^{15}+\cdots x \in(-1,1)

limx0x2(ex1)1+tanx1+x=limx0x3×(1+tanx+1+x)tanxx=limx0x3×2tanxx=limx0x3×213x3=6\begin{aligned} &\lim_{x\to 0} \frac{x^2(e^x-1)}{\sqrt{1+\tan x}-\sqrt{1+ x}}\\ &=\lim_{x\to 0} \frac{x^3\times(\sqrt{1+\tan x}+\sqrt{1+ x})}{\tan x-x}\\ &=\lim_{x\to 0} \frac{x^3\times2}{\tan x-x}\\ &=\lim_{x\to 0} \frac{x^3\times2}{\frac{1}{3}x^3}\\ &=6 \end{aligned}

limx04x2+x1+x+1x2+sinx\begin{equation} \lim_{x\to 0} \frac{\sqrt{4x^2+x-1}+x+1}{\sqrt{x^2+\sin x}} \end{equation}

From Deyi

提示:除以出现次数最多的,最讨厌的那一项。

limx04x2+x1+x+1x2+sinx=limx4+1x1x2+1+1x1+sinxx2=4+11=3\begin{aligned} &\lim_{x\to 0} \frac{\sqrt{4x^2+x-1}+x+1}{\sqrt{x^2+\sin x}}\\ &=\lim _{x \rightarrow \infty} \frac{\sqrt{4+\frac{1}{x}-\frac{1}{x^{2}}}+1+\frac{1}{x}}{\sqrt{1+\frac{\sin x}{x^{2}}}}\\ &=\frac{\sqrt{4}+1}{1}\\ &=3 \\ \end{aligned}

limx00xtln(1+tsint)dt1cosx2\begin{equation} \lim _{x \rightarrow 0} \frac{\int_{0}^{x} t \ln (1+t \sin t) \mathrm{d} t}{1-\cos x^{2}} \end{equation}

2016年考研数一

提示:判断极限类型+等价无穷小+洛必达法则

limx00xtln(1+tsint)dt1cosx2=limx00xtln(1+tsint)x42=limx0xln(1+xsinx)2x3=limx0x(xsinx)2x3=12\begin{aligned} &\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x} t \ln (1+t \sin t) \mathrm{d} t}{1-\cos x^{2}}\\ &=\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x} t \ln (1+t \sin t)}{ \frac{x^4}{2}} \\ &=\displaystyle \lim _{x \rightarrow 0} \frac{x \ln (1+x \sin x)}{2 x^{3}}\\ &=\displaystyle\lim _{x \rightarrow 0} \frac{x(x \sin x)}{2 x^{3}}\\ &=\frac{1}{2} \end{aligned}

limx+1x(t2(e1t1)t)dtx2ln(1+1x)\begin{equation} \lim _{x \rightarrow+\infty} \frac{\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t}{x^{2} \ln \left(1+\frac{1}{x}\right)} \end{equation}

2014年考研数一

提示:判断极限类型+等价无穷小+洛必达法则

limx+1x(t2(e1t1)t)dtx2ln(1+1x)=limx+1x(t2(e1t1)t)dtx21x=limx+(1x(t2(e1t1)t)dt)(x)=limx+(x2(e1x1)x)令 t=1x=1xlimt0+et1tt2上下都趋近于0,使用洛必达法则=limt0+et12t=12\begin{aligned} &\lim _{x \rightarrow+\infty} \frac{\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t}{x^{2} \ln \left(1+\frac{1}{x}\right)}\\ &=\lim _{x \rightarrow+\infty} \frac{\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t}{x^{2} \cdot \frac{1}{x}}\\&=\lim _{x \rightarrow+\infty} \frac{\left(\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t\right)^{\prime}}{(x)^{\prime}} \\ &=\lim _{x \rightarrow+\infty}\left(x^{2}\left(\mathrm{e}^{\frac{1}{x}}-1\right)-x\right) \qquad \text{令}\ t=\frac{1}{x} \\ &=\frac{1}{x} \lim _{t \rightarrow 0^{+}} \frac{\mathrm{e}^{t}-1-t}{t^{2}}\qquad \text{上下都趋近于0,使用洛必达法则} \\ &=\lim _{t \rightarrow 0^{+}} \frac{\mathrm{e}^{t}-1}{2 t}\\ &=\frac{1}{2} \end{aligned}

limnnn2+n+1n(n+1)n(5n1)\begin{equation} \lim _{n \rightarrow \infty} \frac{n^{\frac{n^{2}+n+1}{n}}}{(n+1)^{n}}(\sqrt[n]{5}-1) \end{equation}

From Zhongxiang-17Days

提示:对数字敏感+对公式敏感+剥洋葱法

a🥕1🥕lna\frac{ a^{🥕}-1}{🥕}\sim \ln a

limnnn2+n+1n(n+1)n(5n1)=limnnn+1nn(n+1)n(5n1)=limnnn(n+1)n511/n=limn1(1+1n)nln5=ln5e\begin{aligned} & \lim _{n \rightarrow \infty} \frac{n^{\frac{n^{2}+n+1}{n}}}{(n+1)^{n}}(\sqrt[n]{5}-1) \\ =& \lim _{n \rightarrow \infty} \frac{n^{n+1} \cdot \sqrt[n]{n}}{(n+1)^{n}}(\sqrt[n]{5}-1) \\ =& \lim _{n \rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} \frac{\sqrt{5}-1}{1 / n} \\ =& \lim _{n \rightarrow \infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}} \cdot \ln 5 \\ =& \frac{\ln 5}{e} \end{aligned}

limx01ln(1+x2)1sin2x\begin{equation} \lim_{x\to 0}\frac{1}{\ln(1+x^2)}-\frac{1}{\sin^2x} \end{equation}

From Zhongxiang-17Days

提示:对平方差敏感!