两个重要极限定理:
lim
x
→
0
sin
x
x
=
1
(1)
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \tag{1}
x→0limxsinx=1(1)
和
lim
x
→
∞
(
1
+
1
x
)
x
=
e
(2)
\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e \tag{2}
x→∞lim(1+x1)x=e(2)
引理(夹逼定理)
定义一:
如果数列
{
X
n
}
\lbrace X_n \rbrace
{Xn},
{
Y
n
}
\lbrace Y_n \rbrace
{Yn} 及
{
Z
n
}
\lbrace Z_n \rbrace
{Zn} ,满足下列条件:
(1) 当
n
>
N
0
n > N_0
n>N0 时,其中
N
0
∈
N
∗
N_0 \in N^*
N0∈N∗ ,有
Y
n
≤
X
n
≤
Z
n
Y_n \le X_n \le Z_n
Yn≤Xn≤Zn,
(2)
{
Y
n
}
\lbrace Y_n\rbrace
{Yn},
{
Z
n
}
\lbrace Z_n \rbrace
{Zn} 有相同的极限
a
a
a,设
−
∞
<
a
<
+
∞
- \infty < a < + \infty
−∞<a<+∞,则,数列
{
X
n
}
\lbrace X_n \rbrace
{Xn} 的极限存在,且
lim
n
→
∞
X
n
=
a
\lim_{n \rightarrow \infty} X_n = a
n→∞limXn=a
定义二:
F
(
x
)
F(x)
F(x) 与
G
(
x
)
G(x)
G(x) 在
X
0
X_0
X0 连续且存在相同的极限
A
A
A,即
x
→
X
0
x \rightarrow X_0
x→X0 时,
lim
F
(
x
)
=
lim
G
(
x
)
=
A
\lim F(x) = \lim G(x) = A
limF(x)=limG(x)=A,则
若有函数在
f
(
x
)
f(x)
f(x) 在
X
0
X_0
X0 的某领域内恒有
F
(
x
)
≤
f
(
x
)
≤
G
(
x
)
F(x) \le f(x) \le G(x)
F(x)≤f(x)≤G(x) ,则当
X
X
X 趋近
X
0
X_0
X0, 有
lim
F
(
x
)
≤
lim
f
(
x
)
≤
l
i
m
G
(
x
)
\lim F(x) \le \lim f(x) \le lim G(x)
limF(x)≤limf(x)≤limG(x)
即
A
≤
l
i
m
f
(
x
)
≤
A
A \le lim f(x) \le A
A≤limf(x)≤A
故
lim
(
X
0
)
=
A
\lim(X_0) = A
lim(X0)=A
简单地说:函数
A
>
B
A>B
A>B,函数
B
>
C
B>C
B>C,函数
A
A
A的极限是
X
X
X,函数
C
C
C 的极限也是
X
X
X ,那么函数
B
B
B 的极限就一定是
X
X
X,这个就是夹逼定理。
定理 1 证明:
![在这里插入图片描述](https://img-blog.csdnimg.cn/20200703213016286.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xpbmd0aWFueXVsb25n,size_16,color_FFFFFF,t_70#pic_center)
如上图,对于弧
A
C
⌢
\mathop{AC}\limits^{\frown}
AC⌢ ,由于半径
1
1
1,所以,弧
A
C
⌢
\mathop{AC}\limits^{\frown}
AC⌢ 长
x
x
x。图片很直观地看出
sin
x
≤
x
≤
tan
x
\sin x \le x \le \tan x
sinx≤x≤tanx,并在
x
→
0
x \rightarrow 0
x→0的时候,他们都"相等"。这个是几何直观的,如果我们假设化曲为直是可行的。
所以,
由上述公式,
sin
x
≤
x
≤
t
a
n
x
⟺
1
≤
x
sin
x
≤
tan
x
sin
x
⟺
1
≤
x
sin
x
≤
1
cos
x
\sin x \le x \le tan x \iff 1 \le \frac{x}{\sin x} \le \frac{\tan x}{\sin x} \iff 1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}
sinx≤x≤tanx⟺1≤sinxx≤sinxtanx⟺1≤sinxx≤cosx1
由上式取倒数得:
cos
x
≤
sin
x
x
≤
1
\cos x \le \frac{\sin x}{x} \le 1
cosx≤xsinx≤1
因为,
lim
x
→
0
cos
x
=
1
\lim_{x \rightarrow 0} \cos x = 1
x→0limcosx=1
所以,
lim
x
→
0
sin
x
x
=
1
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1
x→0limxsinx=1
定理 1,得证。
定理2,证明:
首先,证明此极限存在;
构造数列
x
n
=
(
1
+
1
n
)
n
x_n = (1 + \frac{1}{n})^n
xn=(1+n1)n
根据二项式定理,进行展开:
x
n
=
C
n
0
1
n
(
1
n
)
0
+
C
n
1
1
n
−
1
(
1
n
)
1
+
C
n
2
1
n
−
2
(
1
n
)
2
+
⋯
+
n
(
n
−
1
)
(
n
−
2
)
⋯
1
n
!
1
0
(
1
n
)
n
=
1
+
1
+
1
2
!
(
1
−
1
n
)
+
1
3
!
(
1
−
1
n
)
(
1
−
2
n
)
+
⋯
+
1
n
!
(
1
−
1
n
)
(
1
−
2
n
)
⋯
(
1
−
n
−
1
n
)
<
2
+
1
2
!
+
1
3
!
+
⋯
1
n
!
<
2
+
1
2
+
1
2
2
+
1
2
3
+
⋯
+
1
2
n
−
1
=
3
−
1
2
n
−
1
<
3
x_n = C_n^01^n(\frac{1}{n})^0 + C_n^11^{n-1}({\frac{1}{n}})^1 + C_n^21^{n-2}({\frac{1}{n}})^2 + \cdots + \frac{n(n-1)(n-2)\cdots1}{n!}1^0(\frac{1}{n})^n \\ = 1 + 1 + \frac{1}{2!}(1 - \frac{1}{n}) + \frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n}) + \cdots + \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n}) \\ < 2 + \frac{1}{2!} +\frac{1}{3!} + \cdots \frac{1}{n!} \\ < 2 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{n-1}} = 3 - \frac{1}{2^{n-1}} < 3
xn=Cn01n(n1)0+Cn11n−1(n1)1+Cn21n−2(n1)2+⋯+n!n(n−1)(n−2)⋯110(n1)n=1+1+2!1(1−n1)+3!1(1−n1)(1−n2)+⋯+n!1(1−n1)(1−n2)⋯(1−nn−1)<2+2!1+3!1+⋯n!1<2+21+221+231+⋯+2n−11=3−2n−11<3
而对于
x
n
+
1
=
(
1
+
1
n
+
1
)
n
+
1
=
2
+
1
2
!
(
1
−
1
n
)
+
⋯
+
1
n
!
(
1
−
1
n
)
(
1
−
2
n
)
⋯
(
1
−
n
−
1
n
)
+
1
(
n
+
1
)
!
(
1
−
1
n
+
1
)
(
1
−
2
n
+
1
)
)
⋯
(
1
−
n
n
+
1
)
x_{n+1} = (1 + \frac{1}{n+1})^{n+1} \\ = 2 + \frac{1}{2!}(1 - \frac{1}{n}) + \cdots + \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n}) + \frac{1}{(n+1)!}(1-\frac{1}{n+1})(1-\frac{2}{n+1}))\cdots(1- \frac{n}{n+1})
xn+1=(1+n+11)n+1=2+2!1(1−n1)+⋯+n!1(1−n1)(1−n2)⋯(1−nn−1)+(n+1)!1(1−n+11)(1−n+12))⋯(1−n+1n)
所以
x
n
+
1
−
x
n
=
1
(
n
+
1
)
!
(
1
−
1
n
+
1
)
(
1
−
2
n
+
1
)
)
⋯
(
1
−
n
n
+
1
)
>
0
x_{n+1}-x_n = \frac{1}{(n+1)!}(1-\frac{1}{n+1})(1-\frac{2}{n+1}))\cdots(1- \frac{n}{n+1}) > 0
xn+1−xn=(n+1)!1(1−n+11)(1−n+12))⋯(1−n+1n)>0
故,
x
n
+
1
>
x
n
x_{n+1} > x_n
xn+1>xn
该序列为单调递增序列,存在极限,记此极限为
e
e
e。
对于实数
x
x
x,则总存在整数
n
n
n,使得
n
≤
x
≤
n
+
1
n \le x \le n+1
n≤x≤n+1,则有
(
1
+
1
n
+
1
)
n
<
(
1
+
1
x
)
x
<
(
1
+
1
n
)
n
+
1
(1+\frac{1}{n+1})^n < (1+\frac{1}{x})^x<(1+\frac{1}{n})^{n+1}
(1+n+11)n<(1+x1)x<(1+n1)n+1
lim
n
→
∞
(
1
+
1
n
+
1
)
n
=
lim
n
→
∞
(
1
+
1
n
+
1
)
1
+
1
n
+
1
=
lim
n
→
∞
(
1
+
1
n
+
1
)
n
+
1
lim
n
→
∞
(
1
+
1
n
+
1
)
=
e
1
=
e
\lim_{n \rightarrow \infty}(1+\frac{1}{n+1})^n = \lim_{n \rightarrow \infty}\frac{(1+\frac{1}{n+1})}{1 + \frac{1}{n+1}} = \frac{\lim_{n \rightarrow \infty}(1+\frac{1}{n+1})^{n+1}}{\lim_{n \rightarrow \infty}(1 + \frac{1}{n+1})} = \frac{e}{1} = e
n→∞lim(1+n+11)n=n→∞lim1+n+11(1+n+11)=limn→∞(1+n+11)limn→∞(1+n+11)n+1=1e=e
同理,
lim
n
→
∞
(
1
+
1
n
)
n
+
1
=
lim
n
→
∞
(
1
+
1
n
)
(
1
+
1
n
)
n
=
lim
n
→
∞
(
1
+
1
n
)
lim
n
→
∞
(
1
+
1
n
)
n
=
e
\lim_{n \rightarrow \infty}(1+\frac{1}{n})^{n+1} = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})(1 + \frac{1}{n})^n = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = e
n→∞lim(1+n1)n+1=n→∞lim(1+n1)(1+n1)n=n→∞lim(1+n1)n→∞lim(1+n1)n=e
故,根据夹逼定理,函数
f
(
x
)
=
lim
n
→
∞
f
r
a
c
(
1
+
1
x
)
x
f(x) = \lim_{n \rightarrow \infty}frac(1 + \frac{1}{x})^x
f(x)=limn→∞frac(1+x1)x 的极限存在,为
e
e
e。