极限
XantC
数列极限
定义
\[ \lim_{n\to\infty}a_n=L \]
\[ \iff \forall\epsilon>0, \exists N,\text{s.t.} \;\text{if} \;n>N, \text{then}\;|a_n-L|<\varepsilon \]
极限存在准则
夹逼定理
\[ \exists n_0\in\mathbb{N},\text{s.t.}\; n>n_0, \;x_n\leq y_n\leq z_n, \lim_{n\to\infty}x_n=\lim_{x\to\infty}z_n=a\]
\[ \Rightarrow \lim_{n\to \infty}y_n=a \]
单调有界
\[ 单调有界数列必有极限 \]
柯西极限存在准则
\[ \forall \epsilon >0,\; \exists N\in\mathbb{Z^{*}},\;\text{s.t.}\; m>N,n>N, |x_m-x_n|<\epsilon\]
\[ \iff \lim_{n\to \infty}x_n=a \]
函数极限
定义
\[ \lim_{x\to a}f(x)=L \]
\[ \iff \forall\epsilon>0, \exists\delta, \text{s.t.} \;\text{if} \;0<|x-a|<\delta, \text{then}\;|f(x)-L|<\varepsilon \]
\[ (\delta=\delta(a, \varepsilon)) \]
单边极限
\[ \lim_{x\to a^{+}}f(x)=L \]
\[ \iff \forall\epsilon>0, \exists\delta, \text{s.t.} \;\text{if} \;a<x<a+\delta, \text{then}\;|f(x)-L|<\varepsilon \]
\[ \lim_{x\to a^{-}}f(x)=L \]
\[ \iff \forall\epsilon>0, \exists\delta, \text{s.t.} \;\text{if} \;a-\delta<x<a, \text{then}\;|f(x)-L|<\varepsilon \]
无穷极限
\[ \lim_{x\to a}f(x)=\infty \]
\[ \iff \forall M>0, \exists\delta, \text{s.t.} \;\text{if} \;x\in\mathring{U}(a, \delta), \text{then}\;f(x)>M \]
\[ \lim_{x\to a}f(x)=-\infty \]
\[ \iff \forall N<0, \exists\delta, \text{s.t.}\;\text{if} \;x\in\mathring{U}(a, \delta), \text{then}\;f(x)<N \]
连续性
\[ \lim_{x\to a}f(x)=f(a) \]
\[ \lim_{\Delta x\to 0}\Delta y=0 \]
证正弦连续性
\[ \begin{align*} |\Delta y|&=|\sin(x+\Delta x)-\sin x|\\ &=|2\sin(\frac{\Delta x}{2})\cos(x+\frac{\Delta x}{2})|\\ &\leq |2(\frac{\Delta x}{2})\cdot 1|\\ &=|\Delta x| \end{align*} \]
\[ \Rightarrow 0\leq|\Delta y|\leq |\Delta x| \]
\[ \therefore \lim_{\Delta x\to 0}\Delta y=0 \]
定理
极限存在准则
同数列
局部保号性及推论
\[ \lim_{x\to a}f(x)>0 \Rightarrow f(x)>0 \,(x\in\mathring{U}(a, \delta)) \]
\[ f(x)\geq0\Rightarrow \lim_{x\to a}f(x)\geq0 \]
\[ f(x)\leq g(x) \Rightarrow \lim_{x\to a}f(x)\leq\lim_{x\to a}g(x) \;\bigstar \]
证明保号性
\[ \text{if}\; 0<|x-a|<\delta, \text{then} \;L-\varepsilon<f(x)<L+\varepsilon \]
\[ \varepsilon=L\colon L-\frac{L}{2}<f(x) \]
\[ \Rightarrow f(x)>\frac{L}{2}>0 \]
\[ \therefore f(x)>0\,(x\in\mathring{U}(a,\delta)) \]
夹逼定理
\[ f(x)\leq g(x)\leq h(x),\, \lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L \Rightarrow \lim_{x\to a}g(x)=L \]
证明
\[ \begin{cases} x\in\mathring{U}(a,\delta_1)\Rightarrow L-\varepsilon<f(x)<L+\varepsilon\\ x\in\mathring{U}(a,\delta_2)\Rightarrow L-\varepsilon<h(x)<L+\varepsilon \end{cases} \]
\[ let \;\delta=\min(\delta_1, \delta_2) \]
\[ x\in\mathring{U}(a,\delta)\Rightarrow L-\varepsilon<f(x)\leq g(x)\leq h(x)<L+\varepsilon \]
\[ \Rightarrow L-\varepsilon\leq g(x)\leq L-\varepsilon \]
\[ \therefore \lim_{x\to a}g(x)=L \]
洛必达法则
\[ \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0 \;\text{or} \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty \]
\[ \exists f^{'}(x),g^{'}(x), \,x\in\mathring{U}(a) \]
\[ \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{'}(x)}{g^{'}(x)} \]
证明
\[ \text{For}\colon f(a)=g(a)=0 \]
\[ \exists \xi \in (x,a)\,\text{or}\,(a,x) \]
\[ \text{s.t.}\, \frac{f^{'}(\xi)}{g^{'}(\xi)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f(x)}{g(x)} \]
\[ x\rightarrow a\Rightarrow \xi \rightarrow a \]
\[ \therefore \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{'}(x)}{g^{'}(x)} \]
\[ \text{For}\colon f(a)=g(a)=\infty \]
\[ \frac{f(x)}{g(x)}=\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}=\frac{0}{0} \]
\[ \Rightarrow \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{(\frac{1}{g(x)})^{'}}{(\frac{1}{f(x)})^{'}}=\lim_{x\to a}\frac{g^{'}(x)f^2(x)}{f^{'}(x)g^2(x)} \]
\[ \therefore \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{'}(x)}{g^{'}(x)} \]
计算
运算法则
\[ \lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x) \tag{1} \]
\[ \lim_{x\to a}[f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x) \tag{2} \]
\[ \lim_{x\to a}[cf(x)]=c\lim_{x\to a}f(x) \tag{3} \]
\[ \lim_{x\to a}[f(x)g(x)]=\lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x) \tag{4} \]
\[ \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\;(\lim_{x\to a}g(x)\neq0) \tag{5} \]
\[ \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))=f(u_0) \tag{6} \;(\lim_{u\to u_0}f(u)=f(u_0)) \]
\[ \lim_{x\to a}[f(x)]^n=[\lim_{x\to a}f(x)]^n \tag{7} \]
\[ \lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a}f(x)} \tag{8} \]
\[ \lim_{x\to a}c=c \tag{9} \]
\[ \lim_{x\to a}x=a \tag{10} \]
\[ \lim_{x\to a}x^n=a^n \tag{11} \]
证明加法运算
\[ \lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\colon \]
\[ \lim_{x\to a}f(x)=L,\; \lim_{x\to a}g(x)=M \]
\[ \begin{cases} 0<|x-a|<\delta_1\Rightarrow|f(x)-L|<\frac{\varepsilon}{2} \\ 0<|x-a|<\delta_2\Rightarrow|g(x)-M|<\frac{\varepsilon}{2} \\ \end{cases} \]
\[ let\;\delta=\min(\delta_1,\delta_2) \]
\[ \because 0<|x-a|<\delta \]
\[ \therefore \begin{cases} |f(x)-L|<\frac{\varepsilon}{2}\\ |g(x)-M|<\frac{\varepsilon}{2} \end{cases}\]
\[ \begin{align*} \Rightarrow |f(x)+g(x)-L-M|&\leq|f(x)-L|+|g(x)-M| \\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ &=\varepsilon \end{align*} \]
\[ \therefore |f(x)+g(x)-L-M|<\varepsilon \]
\[ \therefore \lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x) \]
证明复合函数运算
\[ u=g(x), \lim_{x\to a}g(x)=u_0,\lim_{u\to u_0}f(u)=f(u_0) \]
\[ \Rightarrow \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))=f(u_0)\colon \]
\[ \begin{cases} \text{if}\;0<|x-a|<\delta_0, \text{then}\;|g(x)-u_0|<\eta_0 \\ \text{if}\;0<|u-u_0|<\eta_0, \text{then}\;|f(u)-f(u_0)|<\varepsilon_0 \end{cases} \]
\[ \Rightarrow \text{if}\;0<|x-a|<\delta_0, \text{then}\;|f(u)-f(u_0)|<\varepsilon_0 \]
\[ \Rightarrow \lim_{x\to a}f(u)=f(u_0)=f(\lim_{x\to a}g(x)) \]
\[ \therefore \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))=f(u_0) \]
无穷小
\[ x\to 0\colon\] \[\sin x \sim x, \;\tan x \sim x\]
\[ \arcsin x\sim x,\; 1-\cos x\sim \frac{1}{2}x^2 \]
\[ \sqrt[n]{1+x^m}-1\sim \frac{1}{n}x^m \]
未定式极限
分式
\[ \frac{0}{0}=\frac{0^{'}}{0^{'}},\;\frac{\infty}{\infty}=\frac{\infty^{'}}{\infty^{'}} \]
乘式
\[ 0\cdot\infty=\frac{0}{\frac{1}{\infty}}=\frac{\infty}{\frac{1}{0}} \]
减式
\[ \infty-\infty\colon\text{通分} \]
次方
\[ \infty^0=0^0=1^\infty\colon \lim u^v=\mathrm{e}^{\lim v\ln{u}} \]
重要极限
第一类重要极限
\[ \lim_{x\to 0}\frac{\sin x}{x}=1 \]
证明
\[ \sin x<x<\tan x \]
\[ 1<\frac{x}{\sin x}<\frac{1}{\cos x} \]
\[ \therefore \cos x<\frac{\sin x}{x}<1 \]
\[ \begin{align*} \cos x&=1-2\sin^2(\frac{x}{2}) \\ &>1-2(\frac{x}{2})^2 \\ &=1-\frac{x^2}{2} \end{align*}\]
\[ \sin x<\tan x\Rightarrow \cos x<1 \]
\[ \therefore 1-\frac{x^2}{2}<\cos x<1 \]
\[ \Rightarrow \lim_{x\to 0}\cos x=\lim_{x\to 0}(1-\frac{x^2}{2})=\lim_{x\to 0}1=1 \]
\[ \because \cos x<\frac{\sin x}{x}<1 \]
\[ \therefore \lim_{x\to 0}\frac{\sin x}{x}=\lim_{x \to 0}\cos x=\lim_{x\to 0}1=1 \]
第二类重要极限
\[ \lim_{x\to \infty}(1+\frac{1}{x})^x=\mathrm{e} \]
证明
\[ x_n=(1+\frac{1}{n})^n < y_n=(1+\frac{1}{n})^{n+1} \]
\[ (1+\frac{1}{n+1})^{n+1}\geq(\frac{n(\frac{n+1}{n})+1}{n+1})^{n+1}\geq\frac{n+1}{n}\cdot\frac{n+1}{n}\cdots 1=(1+\frac{1}{n})^n \]
\[ \Rightarrow x_n \uparrow \]
\[ \frac{1}{y_n}=(\frac{n}{n+1})^{n+1}\leq(\frac{(n+1)(\frac{n}{n+1})+1}{n+2})^{n+2}=(\frac{n+1}{n+2})^{n+2}=\frac{1}{y_{n+1}} \]
\[ \Rightarrow y_n \downarrow \]
\[ \therefore 2=x_1\leq x_n < y_n \leq y_1=4 \]
\[ \therefore x_n单调有界 \]
\[ \therefore\lim_{n\to \infty}(1+\frac{1}{n})^n=\mathrm{e} \]
\[ \text{对于}x\in \mathbb{R},\,\lim_{x\to \infty}(1+\frac{1}{x})^x\colon \]
\[ \text{let}\;n\leq x<n+1 \]
\[ (1+\frac{1}{n+1})^n<(1+\frac{1}{x})^x<(1+\frac{1}{n})^{n+1} \]
\[ \lim_{n\to \infty}(1+\frac{1}{n+1})^n=\lim_{n\to \infty}\frac{(1+\frac{1}{n+1})^{n+1}}{1+\frac{1}{n+1}}=\mathrm{e} \]
\[ \lim_{x\to\infty}(1+\frac{1}{n})^{n+1}=\lim_{n\to \infty}[(1+\frac{1}{n})^n\cdot(1+\frac{1}{n})]=\mathrm{e} \]
\[ \therefore \lim_{x\to \infty}(1+\frac{1}{x})^x=\mathrm{e} \]