两个重要极限定理推导

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} x0limxsinx=1(1)

lim ⁡ x → ∞ ( 1 + 1 x ) x = e (2) \lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e \tag{2} xlim(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} { 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^* N0N ,有 Y n ≤ X n ≤ Z n Y_n \le X_n \le Z_n YnXnZn

(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 nlimXn=a
定义二:

F ( x ) F(x) F(x) G ( x ) G(x) G(x) 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 xX0 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) 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)

A ≤ l i m f ( x ) ≤ A A \le lim f(x) \le A Alimf(x)A

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:

两个重要极限定理推导_第1张图片

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 x x x。图片很直观地看出 sin ⁡ x ≤ x ≤ tan ⁡ x \sin x \le x \le \tan x sinxxtanx,并在 x → 0 x \rightarrow 0 x0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} sinxxtanx1sinxxsinxtanx1sinxxcosx1
由上式取倒数得:
cos ⁡ x ≤ sin ⁡ x x ≤ 1 \cos x \le \frac{\sin x}{x} \le 1 cosxxsinx1
因为,
lim ⁡ x → 0 cos ⁡ x = 1 \lim_{x \rightarrow 0} \cos x = 1 x0limcosx=1
所以,
lim ⁡ x → 0 sin ⁡ x x = 1 \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 x0limxsinx=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+Cn11n1(n1)1+Cn21n2(n1)2++n!n(n1)(n2)110(n1)n=1+1+2!1(1n1)+3!1(1n1)(1n2)++n!1(1n1)(1n2)(1nn1)<2+2!1+3!1+n!1<2+21+221+231++2n11=32n11<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(1n1)++n!1(1n1)(1n2)(1nn1)+(n+1)!1(1n+11)(1n+12))(1n+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+1xn=(n+1)!1(1n+11)(1n+12))(1n+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 nxn+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 nlim(1+n+11)n=nlim1+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 nlim(1+n1)n+1=nlim(1+n1)(1+n1)n=nlim(1+n1)nlim(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)=limnfrac(1+x1)x 的极限存在,为 e e e

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