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由两个重要极限推导常见等价无穷小以及常见导数公式

人间四月天i 2022-03-11 阅读 169

两个重要极限

第一个重要极限

lim ⁡ x → 0 x s i n x = 1 \lim_{x\rightarrow0}\frac{x}{sinx}=1 x0limsinxx=1

第二个重要极限

lim ⁡ x → + ∞ ( 1 + 1 x ) x = e \lim_{x\rightarrow+\infty}(1+\frac{1}{x})^x=e x+lim(1+x1)x=e

等价无穷小

定义,当 f ( x ) → 0 , g ( x ) → 0 f(x)\rightarrow0,g(x)\rightarrow0 f(x)0,g(x)0时,如果 lim ⁡ x → 0 f ( x ) g ( x ) = 1 \lim_{x\rightarrow0}\frac{f(x)}{g(x)}=1 limx0g(x)f(x)=1,那么说g(x)和f(x)为等价无穷小,记作f(x)~g(x)

1. ln(1+x)~x

lim ⁡ x → 0 l n ( 1 + x ) x = lim ⁡ x → 0 l n ( 1 + x ) 1 x = l n ( lim ⁡ x → + ∞ ( 1 + 1 x ) x ) = l n e = 1 \lim_{x\rightarrow0}\frac{ln(1+x)}{x}= \lim_{x\rightarrow0}ln(1+x)^\frac{1}{x}= ln( \lim_{x\rightarrow+\infty}(1+\frac{1}{x})^x)= lne=1 x0limxln(1+x)=x0limln(1+x)x1=ln(x+lim(1+x1)x)=lne=1

2.e^x-1~x

e x − 1 = t e^x-1=t ex1=t,则x=ln(1+t)。因为 x → 0 x\rightarrow0 x0,所以 t → 0 t\rightarrow0 t0
lim ⁡ x → 0 e x − 1 x = lim ⁡ t → 0 t l n ( 1 + t ) = 1 \lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{t\rightarrow0}\frac{t}{ln(1+t)}=1 x0limxex1=t0limln(1+t)t=1

3. a^x-1~xlna

预备知识
对数函数不好理解,转为指数函数就好了
假设 log ⁡ a ( t + 1 ) = m , ln ⁡ a = n \log_a(t+1)=m,\ln a=n loga(t+1)=mlna=n,那么有:
a m = t + 1 , e n = a a^m=t+1,e^n=a am=t+1,en=a,所以, ( e n ) m = t + 1 (e^n)^m=t+1 (en)m=t+1,即: e m n = t + 1 = > m ∗ n = l n ( t + 1 ) e^{mn}=t+1=>m*n=ln(t+1) emn=t+1=>mn=ln(t+1)
所以有结论: log ⁡ a ( t + 1 ) ∗ ln ⁡ a = m ∗ n = ln ⁡ ( t + 1 ) \log_a(t+1)*\ln a=m*n=\ln(t+1) loga(t+1)lna=mn=ln(t+1)

正式推导:

lim ⁡ x → 0 a x − 1 x ln ⁡ a = lim ⁡ t → 0 t log ⁡ a ( t + 1 ) ln ⁡ a = lim ⁡ t → 0 t ln ⁡ ( t + 1 ) = 1 \lim_{x\rightarrow0}\frac{a^x-1}{x\ln a}=\lim_{t\rightarrow0}\frac{t}{\log_a(t+1)\ln a}=\lim_{t\rightarrow0}\frac{t}{\ln(t+1)}=1 x0limxlnaax1=t0limloga(t+1)lnat=t0limln(t+1)t=1

1 + x n − 1 \sqrt[n]{1+x}-1 n1+x 1 ~ 1 n x \frac{1}{n}x n1x

预备知识:
a n − 1 = a n + a n − 1 + a n − 2 + . . . + a − ( a n − 1 + a n − 2 + . . . + a ) − 1 a^n-1=a^n+a^{n-1}+a^{n-2}+...+a-(a^{n-1}+a^{n-2}+...+a)-1 an1=an+an1+an2+...+a(an1+an2+...+a)1
= a n + a n − 1 + a n − 2 + . . . + a − ( a n − 1 + a n − 2 + . . . + a + 1 ) =a^n+a^{n-1}+a^{n-2}+...+a-(a^{n-1}+a^{n-2}+...+a+1) =an+an1+an2+...+a(an1+an2+...+a+1)
= a ∗ ( a n − 1 + a n − 2 + . . . + a + 1 ) − ( a n − 1 + a n − 2 + . . . + a + 1 ) =a*(a^{n-1}+a^{n-2}+...+a+1)-(a^{n-1}+a^{n-2}+...+a+1) =a(an1+an2+...+a+1)(an1+an2+...+a+1)
= ( a − 1 ) ( a n − 1 + a n − 2 + . . . + a + 1 ) =(a-1)(a^{n-1}+a^{n-2}+...+a+1) =(a1)(an1+an2+...+a+1)
正式推导
lim ⁡ x → 0 1 + x n − 1 x \lim_{x\rightarrow0}\frac{\sqrt[n]{1+x}-1}{x} limx0xn1+x 1
= lim ⁡ x → 0 ( 1 + x ) 1 n − 1 x n =\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{n}}-1}{\frac{x}{n}} =limx0nx(1+x)n11

( 1 + x ) 1 n (1+x)^{\frac{1}{n}} (1+x)n1看作预备知识中的a,则
原式= lim ⁡ x → 0 n ∗ ( 1 + x − 1 ) x ∗ ( ( 1 + x ) n − 1 n + ( 1 + x ) n − 2 n + . . . + ( 1 + x ) 1 n + 1 ) \lim_{x\rightarrow0}\frac{n*(1+x-1)}{x*((1+x)^{\frac{n-1}{n}}+(1+x)^{\frac{n-2}{n}}+...+(1+x)^{\frac{1}{n}}+1 )} limx0x((1+x)nn1+(1+x)nn2+...+(1+x)n1+1)n(1+x1)

= lim ⁡ x → 0 n x x ∗ ( n − 1 + 1 ) = 1 =\lim_{x\rightarrow0}\frac{nx}{x*(n-1+1)}=1 =limx0x(n1+1)nx=1

求导法则

求极限时,形式为 0 0 \frac{0}{0} 00 型时,可以根据上述的等价无穷小进行替换

s i n ′ ( x ) = c o s x sin'(x)=cosx sin(x)=cosx

x + t − x 2 \frac{x+t-x}{2} 2x+tx
预备知识:
s i n a + s i n b = s i n ( a + b 2 + a − b 2 ) + s i n ( a + b 2 − a − b 2 ) sina+sinb=sin(\frac{a+b}{2}+\frac{a-b}{2})+sin(\frac{a+b}{2}-\frac{a-b}{2}) sina+sinb=sin(2a+b+2ab)+sin(2a+b2ab)
= s i n ( a + b 2 ) ∗ c o s ( a − b 2 ) + c o s ( a + b 2 ) ∗ s i n ( a − b 2 ) + s i n ( a + b 2 ) ∗ c o s ( a − b 2 ) − c o s ( a + b 2 ) ∗ s i n ( a − b 2 ) =sin(\frac{a+b}{2})*cos(\frac{a-b}{2})+cos(\frac{a+b}{2})*sin(\frac{a-b}{2})+sin(\frac{a+b} {2})*cos(\frac{a-b}{2})-cos(\frac{a+b}{2})*sin(\frac{a-b}{2}) =sin(2a+b)cos(2ab)+cos(2a+b)sin(2ab)+sin(2a+b)cos(2ab)cos(2a+b)sin(2ab)
= 2 s i n ( a + b 2 ) ∗ c o s ( a − b 2 ) =2sin(\frac{a+b}{2})*cos(\frac{a-b}{2}) =2sin(2a+b)cos(2ab)

求导推导:
s i n ′ x = lim ⁡ t → 0 s i n ( x + t ) − s i n ( x ) t = lim ⁡ t → 0 s i n ( x + t ) + s i n ( − x ) t sin'x=\lim_{t\rightarrow 0}\frac{sin(x+t)-sin(x)}{t} =\lim_{t\rightarrow 0}\frac{sin(x+t)+sin(-x)}{t} sinx=limt0tsin(x+t)sin(x)=limt0tsin(x+t)+sin(x)

= lim ⁡ t → 0 2 s i n ( x + t − x 2 ) ∗ c o s ( x + t + x 2 ) t = lim ⁡ t → 0 2 s i n t 2 c o s 2 x + t 2 t =\lim_{t\rightarrow 0}\frac{2sin(\frac{x+t-x}{2})*cos(\frac{x+t+x}{2})}{t} =\lim_{t\rightarrow 0}\frac{2sin\frac{t}{2}cos\frac{2x+t}{2}}{t} =limt0t2sin(2x+tx)cos(2x+t+x)=limt0t2sin2tcos22x+t

= lim ⁡ t → 0 t ∗ c o s ( x + t 2 ) t = c o s x =\lim_{t\rightarrow 0}\frac{t*cos(x+\frac{t}{2})}{t}=cosx =limt0ttcos(x+2t)=cosx

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