3. 函数极限与连续函数
3.1 函数极限
3.1.1 函数极限的性质
-
【局部有界性】若 lim x → x 0 f ( x ) = A \lim\limits_{x\to x_{0}}f(x)=A x→x0limf(x)=A,则 ∃ δ > 0 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) : m ≤ f ( x ) ≤ M \exists \delta>0,\forall x(0<|x-x_{0}|<\delta):m\le f(x)\le M ∃δ>0,∀x(0<∣x−x0∣<δ):m≤f(x)≤M,其中 M , m M,m M,m是两个固定的实数。
【证】取 m < A < M m<A<M m<A<M,令 g ( x ) = m , h ( x ) = M g(x)=m,h(x)=M g(x)=m,h(x)=M, lim x → x 0 g ( x ) = m , lim x → x 0 h ( x ) = M \lim\limits_{x\to x_{0}}g(x)=m,\lim\limits_{x\to x_{0}}h(x)=M x→x0limg(x)=m,x→x0limh(x)=M,由局部保序性
则 ∃ δ > 0 , ∀ x ( 0 < ∣ x − x 0 < δ ) : m = g ( x ) < f ( x ) < h ( x ) = M \exists\delta>0,\forall x(0<|x-x_{0}<\delta):m=g(x)<f(x)<h(x)=M ∃δ>0,∀x(0<∣x−x0<δ):m=g(x)<f(x)<h(x)=M
若 f ( x ) f(x) f(x)在 x 0 x_{0} x0处有定义,则在 x x x满足 ∣ x − x 0 ∣ < δ |x-x_{0}|<\delta ∣x−x0∣<δ的条件下, min { m , f ( x 0 ) } ≤ f ( x ) ≤ max { M , f ( x 0 ) } \min\{m,f(x_{0})\}\le f(x)\le\max\{M,f(x_{0})\} min{m,f(x0)}≤f(x)≤max{M,f(x0)} -
【夹逼性】【定理3.1.3】若 ∃ R > 0 , ∀ x ( 0 < ∣ x − x 0 ∣ < r ) : g ( x ) ≤ f ( x ) ≤ h ( x ) , lim x → x 0 g ( x ) = lim x → x 0 h ( x ) = A \exists R>0,\forall x(0<|x-x_{0}|<r):g(x)\le f(x)\le h(x),\lim\limits_{x\to x_{0}}g(x)=\lim\limits_{x\to x_{0}}h(x)=A ∃R>0,∀x(0<∣x−x0∣<r):g(x)≤f(x)≤h(x),x→x0limg(x)=x→x0limh(x)=A,则 lim x → x 0 f ( x ) = A \lim\limits_{x\to x_{0}}f(x)=A x→x0limf(x)=A
【证】 ∀ ε > 0 , ∃ δ 1 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) : ∣ g ( x ) − A ∣ < ε ⇒ g ( x ) > A − ε \forall \varepsilon>0,\exists \delta_{1},\forall x(0<|x-x_{0}|<\delta):|g(x)-A|<\varepsilon \Rightarrow g(x)>A-\varepsilon ∀ε>0,∃δ1,∀x(0<∣x−x0∣<δ):∣g(x)−A∣<ε⇒g(x)>A−ε
∃ δ 2 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) : ∣ h ( x ) − A ∣ < ε ⇒ h ( x ) < A + ε \exists \delta_{2},\forall x(0<|x-x_{0}|<\delta):|h(x)-A|<\varepsilon \Rightarrow h(x)<A+\varepsilon ∃δ2,∀x(0<∣x−x0∣<δ):∣h(x)−A∣<ε⇒h(x)<A+ε
取 δ = min { r , δ 1 , δ 2 } , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) : A − ε < g ( x ) ≤ f ( x ) ≤ h ( x ) < A + ε \delta=\min\{r,\delta_{1},\delta_{2}\},\forall x(0<|x-x_{0}|<\delta):A-\varepsilon<g(x)\le f(x)\le h(x)<A+\varepsilon δ=min{r,δ1,δ2},∀x(0<∣x−x0∣<δ):A−ε<g(x)≤f(x)≤h(x)<A+ε
即 ∣ f ( x ) − A ∣ < ε |f(x)-A|<\varepsilon ∣f(x)−A∣<ε
所以 lim x → x 0 f ( x ) = A \lim\limits_{x\to x_{0}}f(x)=A x→x0limf(x)=A
【例3.1.4】证明
lim
x
→
0
sin
x
x
=
1
\lim\limits_{x\to 0}\frac{\sin x}{x}=1
x→0limxsinx=1
【证】作如下四分之一单位圆:
S
△
O
A
B
<
扇形
O
A
B
的面积
<
S
△
O
B
C
S_{\triangle O A B}<扇形OAB的面积<S_{\triangle O B C}
S△OAB<扇形OAB的面积<S△OBC
S
△
O
A
B
=
1
2
sin
x
S_{\triangle O A B}=\frac{1}{2}\sin x
S△OAB=21sinx
扇形
O
A
B
的面积
=
1
2
x
扇形OAB的面积=\frac{1}{2}x
扇形OAB的面积=21x(扇形面积公式
1
2
α
r
2
\frac{1}{2}\alpha r^{2}
21αr2,其中
α
\alpha
α为顶角,
r
r
r为半径)
S
△
O
B
C
=
1
2
tan
x
S_{\triangle O B C}=\frac{1}{2}\tan x
S△OBC=21tanx
则
1
2
sin
x
<
1
2
x
<
1
2
tan
x
\frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x
21sinx<21x<21tanx
亦即
sin
x
<
x
<
tan
x
,
0
<
x
<
π
2
\sin x <x <\tan x,0<x<\frac{\pi}{2}
sinx<x<tanx,0<x<2π
sin
x
x
<
x
x
=
1
\frac{\sin x}{x}<\frac{x}{x}=1
xsinx<xx=1
x
<
tan
x
=
sin
x
cos
x
⇒
sin
x
x
>
cos
x
x<\tan x=\frac{\sin x}{\cos x}\Rightarrow\frac{\sin x}{x}>\cos x
x<tanx=cosxsinx⇒xsinx>cosx
则
cos
x
<
sin
x
x
<
1
,
0
<
x
<
π
2
\cos x<\frac{\sin x}{x}<1,0<x<\frac{\pi}{2}
cosx<xsinx<1,0<x<2π
由于
sin
x
x
\frac{\sin x}{x}
xsinx是偶函数,则不等式区间可以对称过去,即
cos
x
<
sin
x
x
<
1
,
0
<
∣
x
∣
<
π
2
\cos x<\frac{\sin x}{x}<1,0<|x|<\frac{\pi}{2}
cosx<xsinx<1,0<∣x∣<2π
现在证明
lim
x
→
0
cos
x
=
1
\lim\limits_{x\to 0}\cos x=1
x→0limcosx=1
∀
ε
>
0
,
∃
δ
∗
,
∀
x
(
0
<
∣
x
−
0
∣
=
∣
x
∣
<
δ
)
:
∣
cos
x
−
1
∣
=
2
sin
2
x
2
<
2
×
x
2
4
=
x
2
2
\forall \varepsilon>0,\exists \delta^{*},\forall x(0<|x-0|=|x|<\delta):|\cos x -1|=2\sin^{2}\frac{x}{2}<2\times\frac{x^{2}}{4}=\frac{x^{2}}{2}
∀ε>0,∃δ∗,∀x(0<∣x−0∣=∣x∣<δ):∣cosx−1∣=2sin22x<2×4x2=2x2
取
δ
∗
=
2
ε
\delta^{*}=\sqrt{2\varepsilon}
δ∗=2ε
则
lim
x
→
0
cos
x
=
1
\lim\limits_{x\to 0}\cos x=1
x→0limcosx=1
又
lim
x
→
0
1
=
1
\lim\limits_{x\to 0}1=1
x→0lim1=1
由夹逼性定理,
lim
x
→
0
sin
x
x
=
1
\lim\limits_{x\to 0}\frac{\sin x}{x}=1
x→0limxsinx=1
【注】
lim
n
→
∞
n
sin
π
n
=
lim
n
→
∞
π
sin
π
n
π
n
=
π
⇏
lim
x
→
0
sin
x
x
=
1
\lim\limits_{n\to\infty}n\sin \frac{\pi}{n}=\lim\limits_{n\to\infty}\pi\frac{\sin\frac{\pi}{n}}{\frac{\pi}{n}}=\pi\not\Rightarrow\lim\limits_{x\to 0}\frac{\sin x}{x}=1
n→∞limnsinnπ=n→∞limπnπsinnπ=π⇒x→0limxsinx=1
x
∈
(
−
π
2
,
π
2
)
\
{
0
}
,
∃
n
∈
N
+
x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \backslash\{0\},\exists n\in\mathbb{N}^{+}
x∈(−2π,2π)\{0},∃n∈N+使得
π
n
+
1
<
∣
x
∣
≤
π
n
\frac{\pi}{n+1}<|x|\le\frac{\pi}{n}
n+1π<∣x∣≤nπ(找到这样一个
n
n
n,保证
(
0
,
π
2
]
=
⋃
n
=
2
∞
(
π
n
+
1
,
π
n
]
\left(0, \frac{\pi}{2}\right]=\bigcup\limits_{n=2}^{\infty}\left(\frac{\pi}{n+1}, \frac{\pi}{n}\right]
(0,2π]=n=2⋃∞(n+1π,nπ],将
(
0
,
π
2
]
(0,\frac{\pi}{2}]
(0,2π]划分成无限个小区间,左开右闭这样既不重叠也不遗漏,即
∣
x
∣
⊂
(
0
,
π
2
)
⊂
(
0
,
π
2
]
|x|\subset(0,\frac{\pi}{2})\subset(0,\frac{\pi}{2}]
∣x∣⊂(0,2π)⊂(0,2π],保证不等式两边都是无穷小量,和
x
→
0
x\to 0
x→0一致,即
n
→
∞
n\to \infty
n→∞,
π
n
+
1
→
0
,
π
n
→
0
\frac{\pi}{n+1}\to 0,\frac{\pi}{n}\to0
n+1π→0,nπ→0)
x
≥
π
n
+
1
,
x
≤
−
π
n
+
1
,
−
π
n
<
x
<
π
n
x\ge\frac{\pi}{n+1},x\le-\frac{\pi}{n+1},-\frac{\pi}{n}<x<\frac{\pi}{n}
x≥n+1π,x≤−n+1π,−nπ<x<nπ
n
n
+
1
⋅
sin
π
n
+
1
π
n
+
1
=
π
n
+
1
π
n
⋅
sin
π
n
+
1
π
n
+
1
=
sin
π
n
+
1
π
n
<
sin
x
x
<
sin
π
n
π
n
+
1
=
sin
π
n
π
n
⋅
π
n
π
n
+
1
=
sin
π
n
π
n
⋅
n
+
1
n
\frac{n}{n+1}\cdot\frac{\sin \frac{\pi}{n+1}}{\frac{\pi}{n+1}}=\frac{\frac{\pi}{n+1}}{\frac{\pi}{n}}\cdot\frac{\sin \frac{\pi}{n+1}}{\frac{\pi}{n+1}}=\frac{\sin \frac{\pi}{n+1}}{\frac{\pi}{n}}<\frac{\sin x}{x}<\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n+1}}=\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}\cdot\frac{\frac{\pi}{n}}{\frac{\pi}{n+1}}=\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}\cdot\frac{n+1}{n}
n+1n⋅n+1πsinn+1π=nπn+1π⋅n+1πsinn+1π=nπsinn+1π<xsinx<n+1πsinnπ=nπsinnπ⋅n+1πnπ=nπsinnπ⋅nn+1
由于
lim
n
→
∞
n
n
+
1
⋅
sin
π
n
+
1
π
n
+
1
=
1
=
lim
n
→
∞
sin
π
n
π
n
⋅
n
+
1
n
=
1
\lim\limits_{n\to\infty}\frac{n}{n+1}\cdot\frac{\sin \frac{\pi}{n+1}}{\frac{\pi}{n+1}}=1=\lim\limits_{n\to\infty}\frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}\cdot\frac{n+1}{n}=1
n→∞limn+1n⋅n+1πsinn+1π=1=n→∞limnπsinnπ⋅nn+1=1
由夹逼性可知
lim
x
→
0
sin
x
x
=
1
\lim\limits_{x\to 0}\frac{\sin x}{x}=1
x→0limxsinx=1
3.1.2 函数极限的四则运算
【定理3.1.4】设 lim x → x 0 f ( x ) = A , lim x → x 0 g ( x ) = B \lim\limits_{x\to x_{0}}f(x)=A,\lim\limits_{x\to x_{0}}g(x)=B x→x0limf(x)=A,x→x0limg(x)=B,则:
- lim x → x 0 ( α f ( x ) + β g ( x ) ) = α A + β B \lim\limits _{x \rightarrow x_{0}}(\alpha f(x)+\beta g(x))=\alpha A+\beta B x→x0lim(αf(x)+βg(x))=αA+βB( α , β \alpha,\beta α,β是常数)
- lim x → x 0 ( f ( x ) g ( x ) ) = A B \lim \limits_{x \rightarrow x_{0}}(f(x) g(x))=A B x→x0lim(f(x)g(x))=AB
-
lim
x
→
x
0
f
(
x
)
g
(
x
)
=
A
B
(
B
≠
0
)
\lim\limits _{x \rightarrow x_{0}} \frac{f(x)}{g(x)}=\frac{A}{B}(B\ne 0)
x→x0limg(x)f(x)=BA(B=0)
【证】由 lim x → x 0 f ( x ) = A \lim\limits_{x\to x_{0}}f(x)=A x→x0limf(x)=A,则 ∃ δ 0 > 0 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ 0 ) : ∣ f ( x ) ∣ ≤ X \exists\delta_{0}>0,\forall x(0<|x-x_{0}|<\delta_{0}):|f(x)|\le X ∃δ0>0,∀x(0<∣x−x0∣<δ0):∣f(x)∣≤X(有界的条件后面能用到)
∀ ε > 0 , ∃ δ 1 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ 1 ) : ∣ f ( x ) − A ∣ < ε \forall\varepsilon>0,\exists\delta_{1},\forall x(0<|x-x_{0}|<\delta_{1}):|f(x)-A|<\varepsilon ∀ε>0,∃δ1,∀x(0<∣x−x0∣<δ1):∣f(x)−A∣<ε
由于 lim x → x 0 g ( x ) = B \lim\limits_{x\to x_{0}}g(x)=B x→x0limg(x)=B
则 ∃ δ 2 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ 2 ) : ∣ g ( x ) − B ∣ < ε \exists\delta_{2},\forall x(0<|x-x_{0}|<\delta_{2}):|g(x)-B|<\varepsilon ∃δ2,∀x(0<∣x−x0∣<δ2):∣g(x)−B∣<ε
取 δ = min { δ 1 , δ 2 } , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) \delta=\min\{\delta_{1},\delta_{2}\},\forall x(0<|x-x_{0}|<\delta) δ=min{δ1,δ2},∀x(0<∣x−x0∣<δ)
(1) ∣ ( α f ( x ) + β g ( x ) ) − ( α A + β B ) ∣ = ∣ α ( f ( x ) − A ) + β ( g ( x ) − B ) ∣ ≤ ∣ α ( f ( x ) − A ) ∣ + ∣ β ( g ( x ) − B ) ∣ < ( ∣ α ∣ + ∣ β ∣ ) ε |(\alpha f(x)+\beta g(x))-(\alpha A+\beta B)|=|\alpha(f(x)-A)+\beta(g(x)-B)|\le|\alpha(f(x)-A)|+|\beta(g(x)-B)|<(|\alpha|+|\beta|)\varepsilon ∣(αf(x)+βg(x))−(αA+βB)∣=∣α(f(x)−A)+β(g(x)−B)∣≤∣α(f(x)−A)∣+∣β(g(x)−B)∣<(∣α∣+∣β∣)ε
所以 lim x → x 0 ( α f ( x ) + β g ( x ) ) = α A + β B \lim\limits _{x \rightarrow x_{0}}(\alpha f(x)+\beta g(x))=\alpha A+\beta B x→x0lim(αf(x)+βg(x))=αA+βB
(2) ∣ f ( x ) g ( x ) − A B ∣ = ∣ f ( x ) g ( x ) − B f ( x ) + B f ( x ) − A B ∣ = ∣ f ( x ) ( g ( x ) − B ) + B ( f ( x ) − A ) ∣ ≤ ∣ f ( x ) ∣ ⋅ ∣ g ( x ) − B ∣ + ∣ B ∣ ⋅ ∣ f ( x ) − A ∣ < ( ∣ X ∣ + ∣ B ∣ ) ε |f(x)g(x)-AB|=|f(x)g(x)-Bf(x)+Bf(x)-AB|=|f(x)(g(x)-B)+B(f(x)-A)|\le|f(x)|\cdot|g(x)-B|+|B|\cdot|f(x)-A|<(|X|+|B|)\varepsilon ∣f(x)g(x)−AB∣=∣f(x)g(x)−Bf(x)+Bf(x)−AB∣=∣f(x)(g(x)−B)+B(f(x)−A)∣≤∣f(x)∣⋅∣g(x)−B∣+∣B∣⋅∣f(x)−A∣<(∣X∣+∣B∣)ε
(3) lim x → x 0 g ( x ) = B ≠ 0 \lim\limits_{x\to x_{0}}g(x)=B\ne 0 x→x0limg(x)=B=0,则 ∃ δ ∗ > 0 , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ∗ \exists\delta_{*}>0,\forall x(0<|x-x_{0}|<\delta_{*} ∃δ∗>0,∀x(0<∣x−x0∣<δ∗,由局部保序性的推论1可知 ∣ g ( x ) ∣ > ∣ B ∣ 2 |g(x)|>\frac{|B|}{2} ∣g(x)∣>2∣B∣
取 δ = min ( δ ∗ , δ 1 , δ 2 ) , ∀ x ( 0 < ∣ x − x 0 ∣ < δ ) : ∣ f ( x ) g ( x ) − A B ∣ = ∣ B f ( x ) − A g ( x ) ∣ ∣ B g ( x ) ∣ = ∣ B f ( x ) − A B + A B − A g ( x ) ∣ ∣ B g ( x ) ∣ = ∣ B ( f ( x ) − A ) + A ( B − g ( x ) ) ∣ ∣ B g ( x ) ∣ ≤ ∣ B ∣ ⋅ ∣ f ( x ) − A ∣ + ∣ A ∣ ⋅ ∣ B − g ( x ) ∣ ∣ B g ( x ) ∣ = ∣ B ∣ ⋅ ∣ f ( x ) − A ∣ + ∣ A ∣ ⋅ ∣ g ( x ) − B ∣ ∣ B ∣ ⋅ ∣ g ( x ) ∣ < ∣ B ∣ ⋅ ε + ∣ A ∣ ⋅ ε ∣ B ∣ ⋅ ∣ B ∣ 2 = 2 ( ∣ B ∣ + ∣ A ∣ ) ε ∣ B ∣ 2 \delta=\min \left(\delta_{*}, \delta_{1}, \delta_{2}\right), \forall x\left(0<\left|x-x_{0}\right|<\delta\right):|\frac{f(x)}{g(x)}-\frac{A}{B}|=\frac{|Bf(x)-Ag(x)|}{|Bg(x)|}=\frac{|Bf(x)-AB+AB-Ag(x)|}{|Bg(x)|}=\frac{|B(f(x)-A)+A(B-g(x))|}{|Bg(x)|}\le\frac{|B|\cdot|f(x)-A|+|A|\cdot|B-g(x)|}{|Bg(x)|}=\frac{|B|\cdot|f(x)-A|+|A|\cdot|g(x)-B|}{|B|\cdot|g(x)|}<\frac{|B|\cdot\varepsilon+|A|\cdot\varepsilon}{|B|\cdot\frac{|B|}{2}}=\frac{2(|B|+|A|)\varepsilon}{|B|^{2}} δ=min(δ∗,δ1,δ2),∀x(0<∣x−x0∣<δ):∣g(x)f(x)−BA∣=∣Bg(x)∣∣Bf(x)−Ag(x)∣=∣Bg(x)∣∣Bf(x)−AB+AB−Ag(x)∣=∣Bg(x)∣∣B(f(x)−A)+A(B−g(x))∣≤∣Bg(x)∣∣B∣⋅∣f(x)−A∣+∣A∣⋅∣B−g(x)∣=∣B∣⋅∣g(x)∣∣B∣⋅∣f(x)−A∣+∣A∣⋅∣g(x)−B∣<∣B∣⋅2∣B∣∣B∣⋅ε+∣A∣⋅ε=∣B∣22(∣B∣+∣A∣)ε
【例】求 lim x → 0 sin α x x , α ≠ 0 \lim\limits_{x \rightarrow 0} \frac{\sin \alpha x}{x},\alpha\ne 0 x→0limxsinαx,α=0
【解】 lim x → 0 sin α x x = lim x → 0 α ( sin α x α x ) = α \lim\limits_{x \rightarrow 0} \frac{\sin \alpha x}{x}=\lim\limits_{x \rightarrow 0} \alpha\left(\frac{\sin \alpha x}{\alpha x}\right)=\alpha x→0limxsinαx=x→0limα(αxsinαx)=α
【例】求
lim
x
→
0
sin
α
x
sin
β
x
,
α
,
β
≠
0
\lim\limits_{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x},\alpha,\beta\ne 0
x→0limsinβxsinαx,α,β=0
【解】
lim
x
→
0
sin
α
x
sin
β
x
=
lim
x
→
0
sin
α
x
α
x
sin
β
x
β
x
⋅
α
β
=
α
β
\lim\limits_{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x}=\lim\limits_{x \rightarrow 0}\frac{\frac{\sin \alpha x}{\alpha x}}{\frac{\sin \beta x}{\beta x}}\cdot\frac{\alpha}{\beta}=\frac{\alpha}{\beta}
x→0limsinβxsinαx=x→0limβxsinβxαxsinαx⋅βα=βα