极限100/2

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。

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

🔑Answer

提示：利用夹逼定理（🗣️哪—里—跑—）

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

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

Answer

提示：等价无穷小

$$\sin(🥕)\sim \ln(1+🥕)$$

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

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

🔑Answer

提示：注意等价无穷小的条件

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

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

🔑Answer

提示：拼凑等价无穷小

$$\begin{aligned} &lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x^{3}}\\ &=\lim _{x \rightarrow 0} \frac{\sin x\left(1-\frac{1}{\cos x}\right)}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x(\cos x-1)}{x^{3} \cos x} \\ &=\lim _{x \rightarrow 0} \frac{x \cdot\left(-\frac{1}{2} x^{2}\right)}{x^{3}}\\ &=-\frac{1}{2} \end{aligned}$$

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

🔑Answer

提示：$1^\infty$常规思路·第一弹

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

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

🔑Answer

提示：$1^\infty$常规思路·第二弹

$$\begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{1-\tan x}{1+\tan x}\right)^{\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}\left(1+(\frac{1-\tan x}{1+\tan x}-1)\right)^{\frac{1}{\frac{1-\tan x}{1+\tan x}-1}\times \frac{-2\tan x}{1+\tan x}\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{-2\tan x}{1+\tan x}\frac{1}{-\sin 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{-2 x}{1+\tan x}\frac{1}{- 2 x}}\\ &=\lim _{x \rightarrow 0}e^{ \frac{1}{1+\tan x}}\\ &=e \end{aligned}$$

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

From Deyi

提示：常规思路：平方差拼凑

$$\begin{aligned} &\lim_{n\to \infty} (\sqrt{n^2+n}-n)\\ &=\lim_{n\to \infty} (\frac{\sqrt{n^2+n}-n}{1})\\ &=\lim_{n\to \infty} (\frac{n^2+n-n^2}{\sqrt{n^2+n}+n)}\\ &=\lim_{n\to \infty} (\frac{n}{\sqrt{n^2+n}+n)}\\ &=\lim_{n\to \infty} (\frac{1}{\sqrt{1+\frac{1}{n}}+1})\\ &=\frac{1}{2}\\ \end{aligned}$$

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

From Deyi

提示：常规思路：平方差拼凑

$$\begin{aligned} &\lim_{x\to \infty} \sqrt{x}(\sqrt{x+2}-\sqrt{x-3})\\ &=\lim_{x\to \infty} \frac{\sqrt{x}(\sqrt{x+2}-\sqrt{x-3})}{1}\\ &=\lim_{x\to \infty} \frac{\sqrt{x}\times 5}{\sqrt{x+2}+\sqrt{x-3})}\\ &=\lim_{x\to \infty} \frac{ 5}{\sqrt{1+2/x}+\sqrt{1-3/x})}\\ &=\frac{5}{2} \end{aligned}$$

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

1987卷三

提示：凑乘积形式，等价无穷小+洛必达

$$\begin{aligned} &\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1-x}{(e^x-1)x}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1-x}{x^2}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x-1}{2x}\right)\\ &=\lim _{x \rightarrow 0}\left(\frac{e^x}{2}\right)\\ &=\frac{1}{2} \end{aligned}$$

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

🔑Answer

提示：等价无穷小替换

$$(1+🥕)^a-1 \sim a🥕$$

$$\begin{aligned} &\lim_{x\to 0} \frac{(x-\sin x) e^{-x^2}}{\sqrt{1-x^3}-1}\\ &= \lim_{x\to 0} e^{-x^2}\times \lim_{x\to 0} \frac{x-\sin x}{1/2\times (-x^3)}\\ &= -2\times \lim_{x\to 0} \frac{x-\sin x}{x^3}\qquad \text{洛必达法则}\\ &= -2\times \lim_{x\to 0} \frac{1-\cos x}{3x^2}\\ &= -2\times \lim_{x\to 0} \frac{1/2 \times x^2}{3x^2}\\ &=- \frac{1}{3} \end{aligned}$$

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

🔑Answer

提示：加减法使用等价无穷小的条件

$\because$ 当 $x \rightarrow 0$ 时, $\sin 2 x \sim 2 x, \sin 5 x \sim 5 x$, 且 $\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}$ $=2 \neq 1, \lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=5 \neq-1$,

满足等价无穷小替换对加减法成立的条件,

$$\therefore \lim _{x \rightarrow 0} \frac{x-\sin 2 x}{x+\sin 5 x}=\lim _{x \rightarrow 0} \frac{x-2 x}{x+5 x}=-\frac{1}{6}$$

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

🔑Answer

提示：$1^{\infty}$型·第三弹

$$\begin{aligned} &\displaystyle \lim _{x \rightarrow \infty}\left(\cos \frac{1}{x}\right)^{x^{2}}\\ &=\displaystyle\lim_{x\to \infty}[1+(\frac{1}{\cos x}-1)]^{\frac{1}{\frac{1}{\cos x}-1}\times (\frac{1}{\cos x}-1)\times x^2}\\ &=e^{\displaystyle\lim_{x\to \infty}\left(\cos \frac{1}{x}-1\right) x^{2}}\\ &=e^{\displaystyle\lim_{x\to \infty} \frac{\cos \frac{1}{x}-1}{\left(\frac{1}{x}\right)^{2}}}\\ &=e^{\displaystyle\lim_{x\to \infty} \frac{-\frac{1}{2}\left(\frac{1}{x}\right)^{2}}{\left(-\frac{1}{x}\right)^{2}}}\\ &=e^{^{-\frac{1}{2}}}\\ &=\frac{1}{\sqrt{e}} \end{aligned}$$

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

From Zhongxiang-17Days

提示：$1^{\infty}$型·第四弹

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

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

From Zhongxiang-17Days

提示：$1^{\infty}$型·第五弹，常规思路

$$\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{x}\right]^{\frac{1}{1-\cos x}}\\ &=\lim _{x \rightarrow 0}\left[1+\frac{\ln \left(x+\sqrt{1+x^{2}}\right)-x}{x}\right]^{\frac{1}{1-\cos x}} \\ &=\lim _{x \rightarrow 0} e^{\frac{\ln (x + \sqrt{1+x^2})-x}{x(1-\cos x )}}\\ \text{幂化简：}&\color{blue}\lim _{x \rightarrow 0} \frac{\ln \left(x+\sqrt{1+x^{2}}\right)-x}{x \cdot \frac{1}{2} x^{2}} \\ &\color{blue}=\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{1+x^{2}}}-1}{\frac{3}{2} x^{2}} \\ &\color{blue}=\lim _{x \rightarrow 0} \frac{-\frac{1}{2} x^{2}}{\frac{3}{2} x^{2}}\\ &\color{blue}=-\frac{1}{3}\\ \text{原式}&=e^{-\frac{1}{3}}\\ \end{aligned}$$

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

From Zhongxiang-17Days

提示：$1^{\infty}$型·变体，思路还是一样的，幂运算变乘法

$$\begin{aligned} &\lim _{n \rightarrow \infty}\left(\frac{\pi}{2}-\arctan n\right)^{\frac{1}{\ln n}}\\ &=\lim _{n \rightarrow \infty}e^{\frac{\ln (\pi /2 - \arctan x)}{\ln x}}\\ \text{幂化简：}&\color{blue}\lim _{n \rightarrow \infty} \frac{\ln (\pi /2 - \arctan x)}{\ln x}\\ &\color{blue}=\lim _{x \rightarrow+\infty} \frac{\frac{1}{\frac{\pi}{2}-\arctan x} \cdot\left(-\frac{1}{1+x^{2}}\right)}{\frac{1}{x}}\\ &\color{blue}=-\lim _{x \rightarrow+\infty} \frac{\frac{1}{x}}{\frac{\pi}{2}-\arctan x} \\ &\color{blue}=-\lim _{x \rightarrow+\infty} \frac{-\frac{1}{x^{2}}}{-\frac{1}{1+x^{2}}} \\ &\color{blue}=-1\\ \text{原式}&=e^{-1}\\ \end{aligned}$$

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

🔑Answer

提示：多次使用洛必达法则

$$\begin{aligned} &\displaystyle \lim_{x\to 0} \frac{1-\cos^2x-\frac{1}{2} x \sin 2x}{x^2(e^{x^2}-1)}\\ &=\displaystyle \lim_{x\to 0} \frac{1-\cos^2x-\frac{1}{2} x \sin 2x}{x^2\times x^2}\qquad \text{上下都趋近于0，使用洛必达法则}\\ &=\displaystyle \lim_{x\to 0} \frac{2\cos x\sin x-\frac{1}{2}\sin 2x-\cos 2x\times x}{4x^3}\\ &=\displaystyle \lim_{x\to 0} \frac{\frac{1}{2}\sin 2x-\cos 2x\times x}{4x^3}\qquad \text{上下都趋近于0，使用洛必达法则}\\ &=\displaystyle \lim_{x\to 0} \frac{\cos 2x-\cos 2x+\sin 2x\times 2 x}{12x^2}\\ &=\displaystyle \lim_{x\to 0} \frac{ 2x\times 2 x}{12x^2}\\ &=\frac{1}{3} \end{aligned}$$

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

🔑Answer

提示：多次使用洛必达法则

$$\begin{aligned} &\displaystyle\lim _{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2x})\\ &=\displaystyle\lim _{x\to 0}\frac{\sin^2x-x^2}{x^2 \sin^2x}\qquad \text{上下都趋近于0，使用洛必达法则}\\ &=\displaystyle\lim _{x\to 0}\frac{\sin 2x-2x}{4x^3}\qquad \text{上下都趋近于0，使用洛必达法则} \\ &=\displaystyle\lim _{x\to 0}\frac{2\cos 2x-2}{12x^2} \\ &=\displaystyle\lim _{x\to 0}\frac{\cos 2x-1}{6x^2} \\ &=\displaystyle\lim _{x\to 0}\frac{\frac{1}{2} (2x)^2}{6x^2}\\ &=\frac{1}{3} \end{aligned}$$

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

🔑Answer

提示：等价无穷小之 加不反

$$\begin{aligned} &\lim _{x \rightarrow 0} (\frac{\cos x}{\cos 2x})^{\frac{1}{x^2}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\left(\frac{\cos x}{\cos 2 x}-1\right)}{x^{2}}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\cos x-\cos 2x}{x^2 \cos 2 x}}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{\cos x-\cos 2x}{x^2 }}\\ &=\mathrm{e}^{\displaystyle \lim_{x\to 0} \frac{-\sin x+2 \sin 2 x}{2 x}}\qquad \because \lim_{x\to 0}\frac{-\sin x}{2\sin 2x}\ne -1\\ &=\mathrm{e}^{\frac{3}{2}} \end{aligned}$$

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

🔑Answer

提示：等价无穷小+洛必达法则

$$\begin{aligned} &\lim _{x \rightarrow 0} \frac{[\sin x-\sin (\sin x)] \cdot \sin x}{x^{4}} \\ &=\lim _{x \rightarrow 0} \frac{\cos x-\cos (\sin x) \cdot \cos x}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \cos x \times \lim _{x \rightarrow 0} \frac{1-\cos (\sin x)}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sin ^{2} x}{3 x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{1 x^{2}}{6 x^{2}} \\ &=\frac{1}{6} \end{aligned}$$

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

1987卷一

提示：洛必达+等价无穷小,看见积分，必定洛必达。

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{1}{b x-\sin x} \int_{0}^{x} \frac{t^{2}}{\sqrt{a+t^{2}}} d t&=1\\ \lim _{x \rightarrow 0}\frac{ \frac{x^{2}}{\sqrt{a+x^{2}}} }{ {b -\cos x}}&=1\\ \lim _{x \rightarrow 0} \frac{1}{\sqrt{a+x^{2}}}\times \lim _{x \rightarrow 0}\frac{ x^{2} }{ {b -\cos x}}&=1\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{ x^{2} }{ {b -\cos x}}&=1\qquad \text{🥝}\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{1}{b} \frac{ x^{2} }{ {1 -\frac{1}{b}\cos x}}&=1\\ \frac{1}{\sqrt{a}}\times \lim _{x \rightarrow 0}\frac{1}{b} \frac{ x^{2} }{ \frac{1}{2b}x^2-\frac{1}{b}+1}&=1\\ \frac{1}{\sqrt{a}}\times2&=1\\ a&=4\\ (a=4\to\text{🥝})\Longrightarrow b&=1 \end{aligned}$$

$$\begin{equation} \lim _{x \rightarrow 0} \frac{\int_{0}^{x}(x-t) \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right)\left(1-\sqrt{1-x^{2}}\right)} \end{equation}$$

ShowTime

提示：看见积分，必定洛必达，但也要注意顺序。

$$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\int_{0}^{x}(x-t) \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right)\left(1-\sqrt{1-x^{2}}\right)}\\ &=\lim _{x \rightarrow 0} \frac{x \int_{0}^{x} \sin t^{2} \mathrm{~d} t-\int_{0}^{x} t \sin t^{2} \mathrm{~d} t}{\left(x^{2}+x^{3}\right) \cdot \frac{1}{2} x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{ x \int_{0}^{x} \sin t^{2} \mathrm{~d} t- \int_{0}^{x} t \sin t^{2} \mathrm{~d} t}{\frac{1}{2} x^{4}} \quad \text{快使用洛必达，哼哼哈嘿！} \\ &=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t^{2} \mathrm{~d} t+x \sin x^{2}-x \sin x^{2}}{2 x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x^{2}}{6 x^{2}}\\&=\frac{1}{6} \end{aligned}$$

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

🔑Answer

提示：等价无穷小+洛必达法则

$$\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{1}{\mathrm{e}^{x}-1}-\frac{1}{\ln (1+x)}\right]\\&=\lim _{x \rightarrow 0} \frac{\ln (1+x)-\mathrm{e}^{x}+1}{\left(\mathrm{e}^{x}-1\right) \ln (1+x)}\\&=\lim _{x \rightarrow 0} \frac{\ln (1+x)-\mathrm{e}^{x}+1}{x \cdot x} \\& = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}-\mathrm{e}^{x}}{2 x}\qquad \text{上下都趋近于0，使用洛必达法则} \\ &=\frac{1}{2} \lim _{x \rightarrow 0} \frac{1-\mathrm{e}^{x}(1+x)}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{1+x}\\&=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(1-\mathrm{e}^{x}(1+x)\right)^{\prime}}{x^{\prime}}\\&=-\frac{1}{2} \lim _{x \rightarrow 0}\left(\mathrm{e}^{x}(2+x)\right)\\&=-1 . \end{aligned}$$

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

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$\frac{0}{0}$型极限，要是你手痒痒直接用了，就会陷入绝境：$\text { 原式 }=\lim _{x \rightarrow 0} \frac{2 x \sin \frac{1}{x}-\cos \frac{1}{x}}{1+\cos x}$

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。

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

---

拉格朗日中值定理专区

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

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

这个方法神奇的紧！

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

From Jin

提示：

$$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{\arcsin ^{4} x}=\lim _{x \rightarrow 0} \frac{-\sin \xi \cdot(\sin x-x)}{x^{4}}$$

$\xi$介于$\sin x$ 与$x$之间，由夹逼定理可以得到$\lim_{x\to 0}\frac{\sin \xi}{x}=1$

$$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{\arcsin ^{4} x}\\ &=\lim _{x \rightarrow 0} \frac{-(\sin x-x)}{x^{3}}\\ &=\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x^{2}}&\\ &=\frac{1}{6}\\ \end{aligned}$$

驾轻就熟

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

1997数三

提示：等价无穷小推论

$$\begin{aligned} &\lim _{x \rightarrow 0}\left[\frac{a}{x}-\left(\frac{1}{x^{2}}-a^{2}\right) \ln (1+a x)\right]\\ &=\lim _{x \rightarrow 0}\left[\frac{a}{x}-\frac{\ln (1+a x)}{x^{2}}\right]+\lim _{x \rightarrow 0} a^{2} \ln (1+a x) \\ &=\lim _{x \rightarrow 0} \frac{a x-\ln (1+a x)}{x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{\frac{1}{2}(a x)^{2}}{x^{2}} {\color{red}\quad\left(🥕-\ln (1+🥕) \sim \frac{1}{2} 🥕^{2}\right)} \\ &=\frac{a^{2}}{2} . \end{aligned}$$

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

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。

$$\begin{aligned} &\lim _{x \rightarrow+\infty} \frac{\sqrt{\left[(n x)^{n}+1\right]^{n+1}}}{(x+1)\left(x^{2}+2\right) \cdots\left(x^{n}+n\right)}\\ &=\lim _{x \rightarrow+\infty} \frac{\left(n^{n}+\frac{1}{x^{n}}\right)^{\frac{n+1}{2}}}{\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x^{2}}\right) \cdots\left(1+\frac{n}{x^{n}}\right)}\\ &=n^{\frac{n(n+1)}{2}} \end{aligned}$$

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

From Deyi

提示：泰勒展开式

$$\tan x=\sum_{n=1}^{\infty} \frac{B_{2 n}(-4)^{n}\left(1-4^{n}\right)}{(2 n) !} x^{2 n-1}=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7}+\frac{62}{2835} x^{9}+\frac{1382}{155925} x^{11}+\frac{21844}{6081075} x^{13}+\frac{929569}{638512875} x^{15}+\cdots x \in(-1,1)$$

$$\begin{aligned} &\lim_{x\to 0} \frac{x^2(e^x-1)}{\sqrt{1+\tan x}-\sqrt{1+ x}}\\ &=\lim_{x\to 0} \frac{x^3\times(\sqrt{1+\tan x}+\sqrt{1+ x})}{\tan x-x}\\ &=\lim_{x\to 0} \frac{x^3\times2}{\tan x-x}\\ &=\lim_{x\to 0} \frac{x^3\times2}{\frac{1}{3}x^3}\\ &=6 \end{aligned}$$

$$\begin{equation} \lim_{x\to 0} \frac{\sqrt{4x^2+x-1}+x+1}{\sqrt{x^2+\sin x}} \end{equation}$$

From Deyi

提示：除以出现次数最多的，最讨厌的那一项。

$$\begin{aligned} &\lim_{x\to 0} \frac{\sqrt{4x^2+x-1}+x+1}{\sqrt{x^2+\sin x}}\\ &=\lim _{x \rightarrow \infty} \frac{\sqrt{4+\frac{1}{x}-\frac{1}{x^{2}}}+1+\frac{1}{x}}{\sqrt{1+\frac{\sin x}{x^{2}}}}\\ &=\frac{\sqrt{4}+1}{1}\\ &=3 \\ \end{aligned}$$

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

2016年考研数一

提示：判断极限类型+等价无穷小+洛必达法则

$$\begin{aligned} &\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x} t \ln (1+t \sin t) \mathrm{d} t}{1-\cos x^{2}}\\ &=\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x} t \ln (1+t \sin t)}{ \frac{x^4}{2}} \\ &=\displaystyle \lim _{x \rightarrow 0} \frac{x \ln (1+x \sin x)}{2 x^{3}}\\ &=\displaystyle\lim _{x \rightarrow 0} \frac{x(x \sin x)}{2 x^{3}}\\ &=\frac{1}{2} \end{aligned}$$

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

2014年考研数一

提示：判断极限类型+等价无穷小+洛必达法则

$$\begin{aligned} &\lim _{x \rightarrow+\infty} \frac{\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t}{x^{2} \ln \left(1+\frac{1}{x}\right)}\\ &=\lim _{x \rightarrow+\infty} \frac{\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t}{x^{2} \cdot \frac{1}{x}}\\&=\lim _{x \rightarrow+\infty} \frac{\left(\int_{1}^{x}\left(t^{2}\left(\mathrm{e}^{\frac{1}{t}}-1\right)-t\right) \mathrm{d} t\right)^{\prime}}{(x)^{\prime}} \\ &=\lim _{x \rightarrow+\infty}\left(x^{2}\left(\mathrm{e}^{\frac{1}{x}}-1\right)-x\right) \qquad \text{令}\ t=\frac{1}{x} \\ &=\frac{1}{x} \lim _{t \rightarrow 0^{+}} \frac{\mathrm{e}^{t}-1-t}{t^{2}}\qquad \text{上下都趋近于0，使用洛必达法则} \\ &=\lim _{t \rightarrow 0^{+}} \frac{\mathrm{e}^{t}-1}{2 t}\\ &=\frac{1}{2} \end{aligned}$$

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

From Zhongxiang-17Days

提示：对数字敏感+对公式敏感+剥洋葱法

$$\frac{ a^{🥕}-1}{🥕}\sim \ln a$$

$$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{n^{\frac{n^{2}+n+1}{n}}}{(n+1)^{n}}(\sqrt[n]{5}-1) \\ =& \lim _{n \rightarrow \infty} \frac{n^{n+1} \cdot \sqrt[n]{n}}{(n+1)^{n}}(\sqrt[n]{5}-1) \\ =& \lim _{n \rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} \frac{\sqrt{5}-1}{1 / n} \\ =& \lim _{n \rightarrow \infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}} \cdot \ln 5 \\ =& \frac{\ln 5}{e} \end{aligned}$$

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