【数学分析笔记】第3章第1节 函数极限（6）
3. 函数极限与连续函数
3.1 函数极限
【例3.1.12】 f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j , b m , b j ≠ 0 , a n , a k ≠ 0 f(x)= \frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}},b_{m},b_{j}\ne 0,a_{n},a_{k}\ne 0 f(x)=bmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk,bm,bj=0,an,ak=0，讨论极限 lim x → ∞ f ( x ) \lim\limits_{x \rightarrow \infty}f(x) x→∞limf(x)与 lim x → 0 f ( x ) \lim\limits_{x \rightarrow 0}f(x) x→0limf(x).
【解】 x → ∞ x\to\infty x→∞：
（1）若 n = m n=m n=m， lim x → ∞ f ( x ) = lim x → ∞ a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = lim x → ∞ a n + a n − 1 x + ⋯ + a k x n − k b n + b n − 1 x + ⋯ + b j x n − j = a n b n ; \lim\limits_{x \rightarrow \infty}f(x)=\lim\limits_{x \rightarrow \infty} \frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\lim\limits_{x \rightarrow \infty} \frac{a_{n}+\frac{a_{n-1}}{x}+\cdots+\frac{a_{k}}{x^{n-k}}}{b_{n}+\frac{b_{n-1}}{x}+\cdots+\frac{b_{j}}{x^{n-j}}}=\frac{a_{n}}{b_{n}} ; x→∞limf(x)=x→∞limbmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=x→∞limbn+xbn−1+⋯+xn−jbjan+xan−1+⋯+xn−kak=bnan;
（2）若 n > m n>m n>m， lim x → ∞ f ( x ) = lim x → ∞ x n ( a n + a n − 1 1 x + ⋯ + a k 1 x n − k x m ( b m + b m − 1 1 x + ⋯ + b j 1 x m − j ) = lim x → ∞ x n − m ( a n + a n − 1 1 x + ⋯ + a k 1 x n − k ( b m + b m − 1 1 x + ⋯ + b j 1 x m − j ) = ∞ \lim\limits_{x \rightarrow \infty}f(x)=\lim\limits_{x \rightarrow \infty} \frac{x^{n}(a_{n}+a_{n-1} \frac{1}{x}+\cdots+a_{k} \frac{1}{x^{n-k}}}{x^{m}(b_{m} +b_{m-1} \frac{1}{x}+\cdots+b_{j} \frac{1}{x^{m-j}})}=\lim\limits_{x \rightarrow \infty}x^{n-m} \frac{(a_{n}+a_{n-1} \frac{1}{x}+\cdots+a_{k} \frac{1}{x^{n-k}}}{(b_{m} +b_{m-1} \frac{1}{x}+\cdots+b_{j} \frac{1}{x^{m-j}})}=\infty x→∞limf(x)=x→∞limxm(bm+bm−1x1+⋯+bjxm−j1)xn(an+an−1x1+⋯+akxn−k1=x→∞limxn−m(bm+bm−1x1+⋯+bjxm−j1)(an+an−1x1+⋯+akxn−k1=∞
（3）若 n < m n<m n<m， lim x → ∞ f ( x ) = lim x → ∞ x n ( a n + a n − 1 1 x + ⋯ + a k 1 x n − k x m ( b m + b m − 1 1 x + ⋯ + b j 1 x m − j ) = lim x → ∞ x n − m ( a n + a n − 1 1 x + ⋯ + a k 1 x n − k ( b m + b m − 1 1 x + ⋯ + b j 1 x m − j ) = lim x → ∞ 1 x m − n ( a n + a n − 1 1 x + ⋯ + a k 1 x n − k ( b m + b m − 1 1 x + ⋯ + b j 1 x m − j ) = 0 \lim\limits_{x \rightarrow \infty}f(x)=\lim\limits_{x \rightarrow \infty} \frac{x^{n}(a_{n}+a_{n-1} \frac{1}{x}+\cdots+a_{k} \frac{1}{x^{n-k}}}{x^{m}(b_{m} +b_{m-1} \frac{1}{x}+\cdots+b_{j} \frac{1}{x^{m-j}})}=\lim\limits_{x \rightarrow \infty}x^{n-m} \frac{(a_{n}+a_{n-1} \frac{1}{x}+\cdots+a_{k} \frac{1}{x^{n-k}}}{(b_{m} +b_{m-1} \frac{1}{x}+\cdots+b_{j} \frac{1}{x^{m-j}})}=\lim\limits_{x \rightarrow \infty}\frac{1}{x^{m-n}} \frac{(a_{n}+a_{n-1} \frac{1}{x}+\cdots+a_{k} \frac{1}{x^{n-k}}}{(b_{m} +b_{m-1} \frac{1}{x}+\cdots+b_{j} \frac{1}{x^{m-j}})}=0 x→∞limf(x)=x→∞limxm(bm+bm−1x1+⋯+bjxm−j1)xn(an+an−1x1+⋯+akxn−k1=x→∞limxn−m(bm+bm−1x1+⋯+bjxm−j1)(an+an−1x1+⋯+akxn−k1=x→∞limxm−n1(bm+bm−1x1+⋯+bjxm−j1)(an+an−1x1+⋯+akxn−k1=0
x → 0 x\to 0 x→0：
（1）若 k = j k=j k=j， f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = x k ( a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k ) x j ( b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ) = a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ⇒ lim x → 0 f ( x ) = a k b j f(x)=\frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\frac{x^{k}(a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} )}{x^{j}(b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} )}=\frac{a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} }{b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} }\Rightarrow\lim\limits_{x \rightarrow 0}f(x)=\frac{a^{k}}{b^{j}} f(x)=bmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=xj(bmxm−j+bm−1xm−j−1+⋯+bj)xk(anxn−k+an−1xn−k−1+⋯+ak)=bmxm−j+bm−1xm−j−1+⋯+bjanxn−k+an−1xn−k−1+⋯+ak⇒x→0limf(x)=bjak；
（2）若 k > j k>j k>j， f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = x k ( a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k ) x j ( b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ) = x k − j ( a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k ) ( b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ) ⇒ lim x → 0 f ( x ) = 0 f(x)=\frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\frac{x^{k}(a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} )}{x^{j}(b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} )}=x^{k-j}\frac{(a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} )}{(b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} )}\Rightarrow\lim\limits_{x \rightarrow 0}f(x)=0 f(x)=bmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=xj(bmxm−j+bm−1xm−j−1+⋯+bj)xk(anxn−k+an−1xn−k−1+⋯+ak)=xk−j(bmxm−j+bm−1xm−j−1+⋯+bj)(anxn−k+an−1xn−k−1+⋯+ak)⇒x→0limf(x)=0
（3）若 k < j k<j k<j， f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = x k ( a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k ) x j ( b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ) = 1 x j − k ( a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k ) ( b m x m − j + b m − 1 x m − j − 1 + ⋯ + b j ) ⇒ lim x → 0 f ( x ) = ∞ f(x)=\frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\frac{x^{k}(a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} )}{x^{j}(b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} )}=\frac{1}{x^{j-k}}\frac{(a_{n} x^{n-k}+a_{n-1} x^{n-k-1}+\cdots+a_{k} )}{(b_{m} x^{m-j}+b_{m-1} x^{m-j-1}+\cdots+b_{j} )}\Rightarrow\lim\limits_{x \rightarrow 0}f(x)=\infty f(x)=bmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=xj(bmxm−j+bm−1xm−j−1+⋯+bj)xk(anxn−k+an−1xn−k−1+⋯+ak)=xj−k1(bmxm−j+bm−1xm−j−1+⋯+bj)(anxn−k+an−1xn−k−1+⋯+ak)⇒x→0limf(x)=∞
综上所述：
L = lim x → ∞ a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = { a n b n , n = m , 0 , n < m , ∞ , n > m ; l = lim x → 0 a n x n + a n − 1 x n − 1 + ⋯ + a k x k b m x m + b m − 1 x m − 1 + ⋯ + b j x j = { a k b k , k = j , 0 , k > j , ∞ , k < j . \begin{array}{l} L=\lim\limits_{x \rightarrow \infty} \frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\left\{\begin{array}{ll} \frac{a_{n}}{b_{n}}, & n=m, \\ 0, & n<m, \\ \infty, & n>m ; \end{array}\right. \\ l=\lim\limits_{x \rightarrow 0} \frac{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{k} x^{k}}{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{j} x^{j}}=\left\{\begin{array}{ll} \frac{a_{k}}{b_{k}}, & k=j, \\ 0, & k>j, \\ \infty, & k<j . \end{array}\right. \end{array} L=x→∞limbmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=⎩ ⎨ ⎧bnan,0,∞,n=m,n<m,n>m;l=x→0limbmxm+bm−1xm−1+⋯+bjxjanxn+an−1xn−1+⋯+akxk=⎩ ⎨ ⎧bkak,0,∞,k=j,k>j,k<j.
【例3.1.13】证明 lim x → ∞ ( 1 + 1 x ) x = e \lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e} x→∞lim(1+x1)x=e.
【证】先证 lim x → + ∞ ( 1 + 1 x ) x = e \lim\limits_{x \rightarrow+ \infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e} x→+∞lim(1+x1)x=e， ∀ x ≥ 1 , [ x ] ≤ x < [ x ] + 1 \forall x\ge 1,[x]\le x <[x]+1 ∀x≥1,[x]≤x<[x]+1
( 1 + 1 [ x ] + 1 ) [ x ] < ( 1 + 1 x ) x < ( 1 + 1 [ x ] ) [ x ] + 1 (1+\frac{1}{[x]+1})^{[x]}<\left(1+\frac{1}{x}\right)^{x}<(1+\frac{1}{[x]})^{[x]+1} (1+[x]+11)[x]<(1+x1)x<(1+[x]1)[x]+1
由于 [ x ] [x] [x]是取整函数，则 [ x ] [x] [x]相当于数列的 n n n
lim n → ∞ ( 1 + 1 n + 1 ) n = lim n → ∞ ( 1 + 1 n + 1 ) n + 1 ⋅ 1 1 + 1 n + 1 = e \lim\limits_{n \rightarrow \infty}\left(1+\frac{1}{n+1}\right)^{n}=\lim\limits_{n \rightarrow \infty}(1+\frac{1}{n+1})^{n+1}\cdot\frac{1}{1+\frac{1}{n+1}}=\mathrm{e} n→∞lim(1+n+11)n=n→∞lim(1+n+11)n+1⋅1+n+111=e
lim n → ∞ ( 1 + 1 n ) n + 1 = lim n → ∞ ( 1 + 1 n ) n ( 1 + 1 n ) = e \lim \limits_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n+1}=\lim \limits_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}(1+\frac{1}{n})=\mathrm{e} n→∞lim(1+n1)n+1=n→∞lim(1+n1)n(1+n1)=e
由夹逼法可知 lim x → + ∞ ( 1 + 1 x ) x = e \lim\limits_{x \rightarrow+ \infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e} x→+∞lim(1+x1)x=e
设 x → − ∞ x\to-\infty x→−∞，令 y = − x y=-x y=−x，则 y → + ∞ y\to+\infty y→+∞
lim x → − ∞ ( 1 + 1 x ) x = lim y → + ∞ ( 1 − 1 y ) − y = lim y → + ∞ ( y − 1 y ) − y = lim y → + ∞ ( y y − 1 ) y = lim y → + ∞ ( 1 + 1 y − 1 ) y = lim y → + ∞ ( 1 + 1 y − 1 ) y − 1 ( 1 + 1 y − 1 ) = e \lim\limits_{x \rightarrow- \infty}\left(1+\frac{1}{x}\right)^{x}=\lim\limits_{y \rightarrow+ \infty}\left(1-\frac{1}{y}\right)^{-y}=\lim\limits_{y \rightarrow+ \infty}\left(\frac{y-1}{y}\right)^{-y}=\lim\limits_{y \rightarrow+ \infty}(\frac{y}{y-1})^{y}=\lim\limits_{y \rightarrow+ \infty}(1+\frac{1}{y-1})^{y}=\lim\limits_{y \rightarrow+ \infty}(1+\frac{1}{y-1})^{y-1}(1+\frac{1}{y-1})=\mathrm{e} x→−∞lim(1+x1)x=y→+∞lim(1−y1)−y=y→+∞lim(yy−1)−y=y→+∞lim(y−1y)y=y→+∞lim(1+y−11)y=y→+∞lim(1+y−11)y−1(1+y−11)=e
综上所述 lim x → ∞ ( 1 + 1 x ) x = e \lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e} x→∞lim(1+x1)x=e
【注】本题还能得到这个结果： lim x → ∞ ( 1 − 1 x ) x = 1 e \lim\limits_{x \rightarrow \infty}(1-\frac{1}{x})^{x}=\frac{1}{e} x→∞lim(1−x1)x=e1
3.1.5 函数极限的Cauchy（柯西）收敛原理
回忆数列极限的柯西收敛原理，数列 lim n → ∞ x n \lim\limits_{n\to\infty}x_{n} n→∞limxn收敛 ⇔ , ∀ ε > 0 , ∃ N , ∀ n , m > N : ∣ x m − x n ∣ < ε \Leftrightarrow,\forall \varepsilon>0,\exists N,\forall n,m>N:|x_{m}-x_{n}|<\varepsilon ⇔,∀ε>0,∃N,∀n,m>N:∣xm−xn∣<ε
对函数的柯西收敛原理（以 x → + ∞ x\to +\infty x→+∞为例）
lim x → + ∞ f ( x ) \lim\limits_{x\to +\infty}f(x) x→+∞limf(x)存在且有限（收敛） ⇔ ∀ ε > 0 , ∃ X , ∀ x ′ , x ′ ′ > X : ∣ f ( x ′ ) − f ( x ′ ′ ) ∣ < ε \Leftrightarrow\forall\varepsilon>0,\exists X,\forall x',x''> X:|f(x')-f(x'')|<\varepsilon ⇔∀ε>0,∃X,∀x′,x′′>X:∣f(x′)−f(x′′)∣<ε
【证】先证必要性，设 lim x → + ∞ f ( x ) = A \lim\limits_{x\to +\infty}f(x)=A x→+∞limf(x)=A，则 ∀ ε > 0 , ∃ X > 0 , ∀ x ′ , x ′ ′ > X : ∣ f ( x ′ ) − A ∣ < ε 2 , ∣ f ( x ′ ′ ) − A ∣ < ε 2 \forall \varepsilon>0,\exists X>0,\forall x',x''>X:|f(x')-A|<\frac{\varepsilon}{2},|f(x'')-A|<\frac{\varepsilon}{2} ∀ε>0,∃X>0,∀x′,x′′>X:∣f(x′)−A∣<2ε,∣f(x′′)−A∣<2ε
所以 ∣ f ( x ′ ) − f ( x ′ ′ ) ∣ = ∣ f ( x ′ ) − A + A − f ( x ′ ′ ) ∣ ≤ ∣ f ( x ′ ) − A ∣ + ∣ A − f ( x ′ ′ ) ∣ = ∣ f ( x ′ ) − A ∣ + ∣ f ( x ′ ′ ) − A ∣ < ε 2 + ε 2 = ε |f(x')-f(x'')|=|f(x')-A+A-f(x'')|\le|f(x')-A|+|A-f(x'')|=|f(x')-A|+|f(x'')-A|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon ∣f(x′)−f(x′′)∣=∣f(x′)−A+A−f(x′′)∣≤∣f(x′)−A∣+∣A−f(x′′)∣=∣f(x′)−A∣+∣f(x′′)−A∣<2ε+2ε=ε
再证充分性， ∀ ε > 0 , ∃ X > 0 , ∀ x ′ , x ′ ′ > X : ∣ f ( x ′ ) − f ( x ′ ′ ) ∣ < ε \forall\varepsilon>0,\exists X>0,\forall x',x''>X:|f(x')-f(x'')|<\varepsilon ∀ε>0,∃X>0,∀x′,x′′>X:∣f(x′)−f(x′′)∣<ε
任取 { x n } , lim n → ∞ x n = + ∞ \{x_{n}\},\lim\limits_{n\to\infty}x_{n}=+\infty {xn},n→∞limxn=+∞，则对上述 X > 0 , ∃ N , ∀ m , n > N : x n > X ， x m > X X>0, \exists N, \forall m,n>N: x_{n}>X，x_{m}>X X>0,∃N,∀m,n>N:xn>X，xm>X
则有 ∣ f ( x n ) − f ( x m ) ∣ < ε |f(x_{n})-f(x_{m})|<\varepsilon ∣f(xn)−f(xm)∣<ε
{ f ( x n ) } \{f(x_{n})\} {f(xn)}是基本数列，必定收敛
由对应的Heine（海涅）定理（任取的 { x n } \{x_{n}\} {xn}）可知， lim x → + ∞ f ( x ) \lim\limits_{x\to +\infty}f(x) x→+∞limf(x)存在且有限。
今后，在建立反常积分的收敛性判别法则等方面，函数极限的Cauchy收敛原理将发挥重要的作用。