1. 等价无穷小($ x\to 0 $)
$$ x-ln(1+x)\sim\frac{1}{2}x^2 $$
$$ 1-\cos^\alpha x\sim\frac{\alpha}{2}x^2 $$
$$ x-\sin(x)\sim\frac{1}{6}x^3, \quad \tan(x)-x\sim\frac{1}{3}x^3 $$
$$ \arcsin(x)-x\sim\frac{1}{6}x^3, \quad x-\arctan(x)\sim\frac{1}{3}x^3 $$
$$ e^x-\sin(x)-\cos(x)\sim x^2 $$
$$ x-\sin(x)\cdot \cos(x)\cdot \cos(2x)\sim\frac{8}{3}x^3 $$
2. 特殊极限
自变量趋于0时
$$ \lim\limits_{x\to 0^+}x^x = 1 $$
$$ \lim\limits_{x\to 0^+}\sin(x)ln(x) = 0 $$
自变量趋于正无穷时
$$ \lim\limits_{x\to+\infty}\frac{ln^a x}{x^b} = 0, (a为任意实数, b>0) $$
$$ \lim\limits_{x\to+\infty}\frac{x^a}{b^x} = 0, (a>0, b>1) $$
3. 做题手法
3-1. 分子(母)有理化
3-2. 有分母，分左右
例题
题目：当$ x\to 1 $时，$ f(x)=\frac{x^2-1}{x-1}e^{\frac{1}{x-1}} $的极限为？
答：① $ \lim\limits_{x\to 1}\frac{x^2-1}{x-1}=\lim\limits_{x\to 1}(x+1)=2 $；
②$ \lim\limits_{x\to 1^-}e^{\frac{1}{x-1}}=0，\quad \lim\limits_{x\to 1^+}e^{\frac{1}{x-1}}=\infty $；
因此，此极限不存在但不是无穷。
3-3. 列与子列问题
例题
题目：极限$ \lim\limits_{x\to 0}\frac{1}{x}\sin\frac{1}{x} $为？
答：① 令$ x_n=\frac{1}{2n\pi}\to 0(n\to \infty) \quad \Rightarrow $
$$
\begin{align}
&\lim\limits_{n\to \infty}\frac{1}{x_n}\sin(\frac{1}{x_n})
\\=&\lim\limits_{n\to \infty}2n\pi \sin(2n\pi)
\\=&0
\end{align}
$$
② 令$ y_n=\frac{1}{2n\pi +\frac{\pi}{2}}\to 0(n\to \infty) \quad \Rightarrow $
$$
\begin{align}
&\lim\limits_{n\to \infty}\frac{1}{y_n}\sin(\frac{1}{y_n})
\\=&\lim\limits_{n\to \infty}(2n\pi +\frac{\pi}{2})\sin(2n\pi +\frac{\pi}{2})
\\=&\infty
\end{align}
$$
因此，此极限不存在但不是无穷。
3-4. 拆分或组合
$$
\begin{align}
&\lim\limits_{x\to 0}\frac{e^x-\sin(x)-\cos(x)}{x^2}
\\=&\lim\limits_{x\to 0}\frac{x-\sin(x)}{x^2}+\lim\limits_{x\to 0}\frac{e^x-\cos(x)-x}{x^2}
\\=&0+\frac{1}{2}\lim\limits_{x\to 0}\frac{e^x-1+\sin(x)}{x}
\\=&\frac{1}{2}\lim\limits_{x\to 0}\frac{2x}{x}
\\=&1
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 0}\frac{xe^x-ln(1+x)}{x^2}
\\=&\lim\limits_{x\to 0}\frac{xe^x-x}{x^2}+\lim\limits_{x\to 0}\frac{x-ln(1+x)}{x^2}
\\=&1+\lim\limits_{x\to 0}\frac{\frac{1}{2}x^2}{x^2}
\\=&\frac{3}{2}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 0}\frac{\tan(\tan(x))-x}{x(1-\sqrt{\cos(x)})}
\\=&\lim\limits_{x\to 0}\frac{(1+\sqrt{\cos(x)})(\tan(\tan(x))-x)}{x(1-\cos(x))}
\\=&4\bigg [ \lim\limits_{x\to 0}\frac{\tan(\tan(x))-x}{x^3}\bigg ]
\\=&4\bigg [ \lim\limits_{x\to 0}\frac{\tan(\tan(x))-\tan(x)}{x^3}+\lim\limits_{x\to 0}\frac{\tan(x)-x}{x^3}\bigg ]
\\=&4\bigg [ \lim\limits_{x\to 0}\frac{\tan(\tan(x))-\tan(x)}{\tan^3(x)}+\lim\limits_{x\to 0}\frac{\frac{1}{3}x^3}{x^3}\bigg ]
\\=&4\bigg [ \lim\limits_{t\to 0}\frac{\tan(t)-t}{t^3}+\frac{1}{3}\bigg ]
\\=&\frac{8}{3}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 0^+}\frac{e^{\tan(x)}-e^x}{x^3}
\\=&\lim\limits_{x\to 0^+}\frac{e^x(e^{\tan(x)-x}-1)}{x^3}
\\=&\lim\limits_{x\to 0^+}\frac{\tan(x)-x}{x^3}
\\=&\frac{1}{3}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 0^+}\frac{ln(\frac{\arcsin(x)}{x})}{x^2}
\\=&\lim\limits_{x\to 0^+}\frac{ln(1+\frac{\arcsin(x)-x}{x})}{x^2}
\\=&\lim\limits_{x\to 0^+}\frac{\arcsin(x)-x}{x^3}
\\=&\frac{1}{6}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 1}\frac{\sin^2\pi x}{(x-1)ln(x)}
\\=&\lim\limits_{x\to 1}\frac{\sin^2[\pi +\pi (x-1)]}{(x-1)ln[1+(x-1)]}
\\=&\lim\limits_{x\to 1}\frac{\sin^2\pi (x-1)}{(x-1)^2}
\\=&\pi ^2
\end{align}
$$
4. 其他公式
4-1. 结合定积分
分子次数齐，分母次数齐，分母多一次时，有如下关系式成立：
$$ \lim\limits_{n\to+\infty}\frac{1}{n}\sum_{i=1}^nf(\frac{i-1}{n})=\lim\limits_{n\to+\infty}\frac{1}{n}\sum_{i=1}^nf(\frac{i}{n})=\int_0^1 f(x)dx $$
4-2. 华里士公式(点火公式)
$ \int_0^{\frac{\pi}{2}}\sin^{2n}xdx=\int_0^{\frac{\pi}{2}}\cos^{2n}xdx = \frac{1}{2}\cdot \frac{3}{4}\cdot ··· \cdot \frac{2n-1}{2n}\cdot \frac{\pi}{2} $
$ \quad $
$ \int_0^{\frac{\pi}{2}}\sin^{2n+1}xdx=\int_0^{\frac{\pi}{2}}\cos^{2n+1}xdx =1\cdot \frac{2}{3}\cdot \frac{4}{5} ··· \cdot \frac{2n}{2n+1} $
4-1及4-2例题
$$
\begin{align}
&\lim\limits_{n\to \infty}\bigg (\frac{\sin^{2}(\frac{\pi}{n})}{n}+\frac{\sin^{2}(\frac{2 \pi}{n})}{n}+ ··· +\frac{\sin^{2}(\frac{n \pi}{n})}{n}\bigg )
\\=&\lim\limits_{x\to \infty}\frac{1}{n}\sum_{i=1}^n \sin^{2}(\frac{i \pi}{n})
\\=&\int_0^1 \sin^2(\pi x)dx
\\=&\frac{1}{\pi}\int_0^1 \sin^2(\pi x)d\pi x
\\=&\frac{1}{\pi}\int_0^\pi \sin^2(x)dx
\\=&\frac{2}{\pi}\int_0^{\frac{\pi}{2}} \sin^2(x)dx
\\=&\frac{2}{\pi} × \frac{1}{2} × \frac{\pi}{2}
\\=&\frac{1}{2}
\end{align}
$$
4-3. 变上下限积分求导公式
$$
\frac{d}{dx}\int_{\varphi _1(x)}^{\varphi _2(x)}f(t)dt=f[\varphi _2(x)]\cdot \varphi _2^{'}(x)-f[\varphi _1(x)]\cdot \varphi _1^{'}(x)
$$
$$
特别地，\frac{d}{dx}\int_0^{\varphi(x)}f(t)dt=f[\varphi (x)]\cdot \varphi^{'}(x)
$$
$$
\begin{align}
&\lim\limits_{x\to 0}\frac{\int_0^x(e^t-1-t)^2}{x^2(x-\arctan (x))}
\\=&3\lim\limits_{x\to 0}\frac{\int_0^x(e^t-1-t)^2}{x^5}
\\=&\frac{3}{5}\lim\limits_{x\to 0}\frac{(e^x-1-x)^2}{x^4}
\\=&\frac{3}{10}\lim\limits_{x\to 0}\frac{(e^x-1-x)(e^x-1)}{x^3}
\\=&\frac{3}{10}\lim\limits_{x\to 0}\frac{e^x-1-x}{x^2}
\\=&\frac{3}{20}\lim\limits_{x\to 0}\frac{e^x-1}{x}
\\=&\frac{3}{20}
\end{align}
$$
4-4. 三角函数公式
$$
\begin{align}
\tan(\frac{\pi}{2}+\alpha)=-\tan(\frac{\pi}{2}-\alpha)&=-\cot(\alpha)=-\frac{1}{\tan(\alpha)}
\\ \tan(\alpha \pm \beta)&=\frac{\tan(\alpha)\pm \tan(\beta)}{1\mp \tan(\alpha)\cdot \tan(\beta)}
\\ \sec^2(x)-&\tan^2(x)=1
\\ \quad
\\ 自己尝试解一下这道题：&\lim\limits_{x\to 1}\tan(\frac{\pi}{4}x)^{\tan(\frac{\pi}{2}x)}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{x\to 1}\tan(\frac{\pi}{4}x)^{\tan(\frac{\pi}{2}x)}
\\&=\lim\limits_{x\to 1}\Bigg\{ \bigg [ 1+(\tan(\frac{\pi}{4}x-1)) \bigg ]^{\frac{1}{\tan(\frac{\pi}{4}x-1)}} \Bigg \}^{\tan(\frac{\pi}{2}x)(\tan(\frac{\pi}{4}-1))}
\\&=e^{\lim\limits_{x\to 1}\tan[\frac{\pi}{2}+\frac{\pi}{2}(x-1)]\{ \tan[\frac{\pi}{4}+\frac{\pi}{4}(x-1)] \}}
\\&\overset{x-1=t}{=}e^{\lim\limits_{t\to 0}\tan(\frac{\pi}{2}+\frac{\pi}{2}t)\tan[(\frac{\pi}{4}+\frac{\pi}{4}t)-1]}
\\&=e^{-\lim\limits_{t\to 0}\frac{1}{\tan(\frac{\pi}{2}t)}\bigg( \frac
{1+\tan(\frac{\pi}{4}t)}{1-\tan(\frac{\pi}{4}t)}-1 \bigg)}
\\&=e^{-\lim\limits_{t\to 0}\frac{1}{1-\tan(\frac{\pi}{4}t)}\cdot \frac{2\tan(\frac{\pi}{4}t)}{\frac{\pi}{2}t}}
\\&=e^{-1}
\end{align}
$$
4-5. 麦克劳林展开式
麦克劳林级数是泰勒级数在$ x=0 $处展开的特殊形式，其表达式为：
$$ f(x)=\sum_{n=0}^{\infty} \frac{1}{n!}f^{(n)}(0)x^n,\quad (|x|<r) $$
常见的函数展开形式如下：
$$ e^x=1+x+\frac{x^2}{2!}+···+\frac{x^n}{n!}+o(x^n) $$
$$ sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+···+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+2}) $$
$$ cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+···+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1}) $$
$$ ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+···+(-1)^n\frac{x^{n+1}}{n+1}+o(x^{n+1}) $$
$$ \frac{1}{1-x}=1+x+x^2+···+x^n+o(x^n) $$
$$ (1+x)^{\alpha}=1+\alpha x+\frac{\alpha (\alpha -1)}{2!}x^2+···+\frac{\alpha (\alpha -1)···(\alpha -n+1)}{n!}x^n+o(x^n) $$
4-6. 拉格朗日中值定理
无穷大乘以同值函数相减可以用拉格朗日中值定理求极限
$$
\begin{align}
&\lim\limits_{x\to +\infty}x^2\bigg [\arctan(\frac{1}{2x-1})-\arctan(\frac{1}{2x+1})\bigg ]
\\&\infty \cdot (同值函数相减)
\\&令 f(t)=\arctan(t)，则f^{'}(t)=\frac{1}{1+t^2}
\\&\arctan(\frac{1}{2x-1})-\arctan(\frac{1}{2x+1})
\\&=\frac{1}{1+\xi ^2}×(\frac{1}{2x-1}-\frac{1}{2x+1})
\\&=\frac{1}{1+\xi ^2}×\frac{2}{4x^2-1}\qquad (\frac{1}{2x+1}<\xi<\frac{1}{2x-1})
\\&\quad
\\原式&=\lim\limits_{x\to +\infty}x^2\frac{1}{1+\xi ^2}\frac{2}{4x^2-1}
\\&=\frac{1}{2}×1
\\&=\frac{1}{2}
\end{align}
$$
4-7. 夹逼定理
n项相加，分子和分母次数不齐，可以用夹逼定理求极限
例一：
$$
\lim\limits_{n\to \infty}\bigg ( \frac{1}{\sqrt{4n^2+1}}+\frac{1}{\sqrt{4n^2+2}}+···+\frac{1}{\sqrt{4n^2+n}} \bigg )
$$
例二：
$$
\lim\limits_{n\to \infty}\bigg ( \frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+···+\frac{n}{n^2+n+n} \bigg )
$$
$$
\begin{align}
&\lim\limits_{n\to \infty}(\frac{1}{\sqrt{4n^2+1}}+\frac{1}{\sqrt{4n^2+2}}+···+\frac{1}{\sqrt{4n^2+n}})
\\ (·· &·)\triangleq b_n
\\ \because & \frac{1}{\sqrt{4n^2+n}}\le \frac{1}{\sqrt{4n^2+i}} \le \frac{1}{\sqrt{4n^2+1}}
\\ \therefore & \frac{n}{\sqrt{4n^2+n}} \le b_n \le \frac{n}{\sqrt{4n^2+1}}
\\ \because & \lim\limits_{n\to \infty}左=\lim\limits_{n\to \infty}\frac{1}{\sqrt{4+\frac{1}{n}}}=\frac{1}{2}
\\ \quad & \lim\limits_{n\to \infty}右=\lim\limits_{n\to \infty}\frac{1}{\sqrt{4+\frac{1}{n^2}}}=\frac{1}{2}
\\ \therefore & 原式=\frac{1}{2}
\end{align}
$$
$$
\begin{align}
&\lim\limits_{n\to \infty}\bigg ( \frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+···+\frac{n}{n^2+n+n} \bigg )
\\ (·· &·)\triangleq b_n
\\ \because & \frac{i}{n^2+n+n}\le \frac{i}{n^2+n+i} \le \frac{i}{n^2+n+1}
\\ \therefore & \frac{\frac{n}{2}(n+1)}{n^2+n+n} \le b_n \le \frac{\frac{n}{2}(n+1)}{n^2+n+1}
\\ \because & \lim\limits_{n\to \infty}左=\frac{1}{2}\lim\limits_{n\to \infty}\frac{n^2+n}{n^2+n+n}=\frac{1}{2}
\\ \quad & \lim\limits_{n\to \infty}右=\frac{1}{2}\lim\limits_{n\to \infty}\frac{n^2+n}{n^2+n+1}=\frac{1}{2}
\\ \therefore & 原式=\frac{1}{2}
\end{align}
$$
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