由两个重要极限推导常见等价无穷小以及常见导数公式

在学习高等数学的极限和导数时，发现老师给了很多等价无穷小公式以及求导法则，记忆力有限，死记硬背容易背叉，所以尝试动手推导其过程，加深理解。

推导原则：利用两个重要极限和无穷小替换推导，尽可能不用后面的知识，如洛必达法则，泰勒公式等。后面公式的推导可能会用到前面已经推导的结论（因为按照教程顺序还没学到后面知识~）。

这两个重要极限不做推导，只有死记硬背，作为后续基础。

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

lim ⁡ x → + ∞ ( 1 + 1 x ) x = e \lim_{x\rightarrow+\infty}(1+\frac{1}{x})^x=e x→+∞lim​(1+x1​)x=e 在我当前的理解里，后续的洛必达法则其实就是求导，对指数及对数函数来说，求导和无穷小替换都会用到这个重要极限，这也是为什么我在之前的文章说：在证明这个重要极限时不能使用洛必达法则

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 limx→0​g(x)f(x)​=1，那么说g(x)和f(x)为等价无穷小，记作f(x)~g(x)

lim ⁡ x → 0 t a n x x = lim ⁡ x → 0 s i n x c o s x ∗ x = lim ⁡ x → 0 1 c o s x = 1 \lim_{x\rightarrow0}\frac{tanx}{x}=\lim_{x\rightarrow0}\frac{sinx}{cosx*x}=\lim_{x\rightarrow0}\frac{1}{cosx}=1 limx→0​xtanx​=limx→0​cosx∗xsinx​=limx→0​cosx1​=1

lim ⁡ x → 0 a r c s i n x x = lim ⁡ t → 0 t s i n t = 1 \lim_{x\rightarrow0}\frac{arcsinx}{x}=\lim_{t\rightarrow0}\frac{t}{sint}=1 limx→0​xarcsinx​=limt→0​sintt​=1

原函数不好理解，可以考虑换元法，即转为反函数。令arcsinx=t,则sint=x。后续不再单独解释

lim ⁡ x → 0 a r c t a n x x = lim ⁡ t → 0 t t a n t = 1 \lim_{x\rightarrow0}\frac{arctanx}{x}=\lim_{t\rightarrow0}\frac{t}{tant}=1 limx→0​xarctanx​=limt→0​tantt​=1

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 x→0lim​xln(1+x)​=x→0lim​ln(1+x)x1​=ln(x→+∞lim​(1+x1​)x)=lne=1

令 e x − 1 = t e^x-1=t ex−1=t，则x=ln(1+t)。因为 x → 0 x\rightarrow0 x→0,所以 t → 0 t\rightarrow0 t→0 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 x→0lim​xex−1​=t→0lim​ln(1+t)t​=1

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)=m，lna=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=>m∗n=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=m∗n=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 x→0lim​xlnaax−1​=t→0lim​loga​(t+1)lnat​=t→0lim​ln(t+1)t​=1

因为 lim ⁡ x → 0 l n ( 1 + x ) x = 1 \lim_{x\rightarrow0}\frac{ln(1+x)}{x}=1 limx→0​xln(1+x)​=1 所以 lim ⁡ t → 1 l n ( t ) t − 1 = 1 \lim_{t\rightarrow1}\frac{ln(t)}{t-1}=1 limt→1​t−1ln(t)​=1

所以下面令t=(1+bx)^a， x → 0 , 则 t → 1 x\rightarrow0,则t\rightarrow1 x→0,则t→1。 所以利用等价无穷小 lim ⁡ t → 1 l n ( t ) t − 1 = 1 \lim_{t\rightarrow1}\frac{ln(t)}{t-1}=1 limt→1​t−1ln(t)​=1 替换： lim ⁡ x → 0 ( 1 + b x ) a − 1 a b x = lim ⁡ x → 0 l n ( 1 + b x ) a a b x = lim ⁡ x → 0 l n ( 1 + b x ) b x = 1 \lim_{x\rightarrow0}\frac{(1+bx)^a-1}{abx}=\lim_{x\rightarrow0}\frac{ln(1+bx)^a}{abx}=\lim_{x\rightarrow0}\frac{ln(1+bx)}{bx}=1 limx→0​abx(1+bx)a−1​=limx→0​abxln(1+bx)a​=limx→0​bxln(1+bx)​=1

1 + x n − 1 \sqrt[n]{1+x}-1 n1+x ​−1 ~ 1 n x \frac{1}{n}x n1​x 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 an−1=an+an−1+an−2+...+a−(an−1+an−2+...+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+an−1+an−2+...+a−(an−1+an−2+...+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∗(an−1+an−2+...+a+1)−(an−1+an−2+...+a+1) = ( a − 1 ) ( a n − 1 + a n − 2 + . . . + a + 1 ) =(a-1)(a^{n-1}+a^{n-2}+...+a+1) =(a−1)(an−1+an−2+...+a+1) 正式推导 lim ⁡ x → 0 1 + x n − 1 x n \lim_{x\rightarrow0}\frac{\sqrt[n]{1+x}-1}{\frac{x}{n}} limx→0​nx​n1+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}} =limx→0​nx​(1+x)n1​−1​

将 ( 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 )} limx→0​x∗((1+x)nn−1​+(1+x)nn−2​+...+(1+x)n1​+1)n∗(1+x−1)​

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

lim ⁡ x → 0 log ⁡ a ( 1 + x ) x ln ⁡ a = lim ⁡ x → 0 ln ⁡ a ∗ log ⁡ a ( 1 + x ) x = lim ⁡ x → 0 l n ( 1 + x ) x = 1 \lim_{x\rightarrow0}\frac{\log_a(1+x)}{\frac{x}{\ln a}}=\lim_{x\rightarrow0}\frac{\ln a*\log_a(1+x)}{x}=\lim_{x\rightarrow0}\frac{ln(1+x)}{x}=1 limx→0​lnax​loga​(1+x)​=limx→0​xlna∗loga​(1+x)​=limx→0​xln(1+x)​=1

lim ⁡ x → 0 t a n x − s i n x 1 2 ∗ x 3 = lim ⁡ x → 0 s i n x ∗ ( 1 − c o s x ) c o s x 1 2 ∗ x 3 = lim ⁡ x → 0 2 ∗ s i n x ∗ ( 1 − c o s x ) x 3 ∗ c o s x = lim ⁡ x → 0 2 ∗ ( 1 − c o s x ) sin ⁡ 2 x ∗ c o s x = lim ⁡ x → 0 2 ∗ ( 1 − c o s x ) ( 1 + c o s x ) ∗ ( 1 − c o s x ) ∗ c o s x = lim ⁡ x → 0 2 ( 1 + c o s x ) ∗ c o s x = 1 \lim_{x\rightarrow0}\frac{tanx-sinx}{\frac{1}{2}*x^3}=\lim_{x\rightarrow0}\frac{\frac{sinx*(1-cosx)}{cosx}}{\frac{1}{2}*x^3}=\lim_{x\rightarrow0}\frac{2*sinx*(1-cosx)}{x^3*cosx}=\lim_{x\rightarrow0}\frac{2*(1-cosx)}{\sin^2x*cosx}=\lim_{x\rightarrow0}\frac{2*(1-cosx)}{(1+cosx)*(1-cosx)*cosx}=\lim_{x\rightarrow0}\frac{2}{(1+cosx)*cosx}=1 limx→0​21​∗x3tanx−sinx​=limx→0​21​∗x3cosxsinx∗(1−cosx)​​=limx→0​x3∗cosx2∗sinx∗(1−cosx)​=limx→0​sin2x∗cosx2∗(1−cosx)​=limx→0​(1+cosx)∗(1−cosx)∗cosx2∗(1−cosx)​=limx→0​(1+cosx)∗cosx2​=1

lim ⁡ x → 0 1 − c o s x x 2 2 = lim ⁡ x → 0 2 ∗ ( 2 ∗ sin ⁡ 2 x 2 ) x 2 = lim ⁡ x → 0 x 2 x 2 = 1 \lim_{x\rightarrow0}\frac{1-cosx}{\frac{x^2}{2}}=\lim_{x\rightarrow0}\frac{2*(2*\sin^2\frac{x}{2})}{x^2}=\lim_{x\rightarrow0}\frac{x^2}{x^2}=1 limx→0​2x2​1−cosx​=limx→0​x22∗(2∗sin22x​)​=limx→0​x2x2​=1

暂时无法仅利用极限或等价无穷小知识证明，必须用到洛必达法则。 洛必达法则理解参考：https://blog.csdn.net/qq_31076523/article/details/104737300 在运用洛必达法则的过程中会用到求导知识，本例来说用到了以下三类，这三类公式的证明参考后面的求导法则证明 C ′ = 0 ( C 为 常 数 ) C'=0(C为常数) C′=0(C为常数) ( x n ) ′ = ( n − 1 ) ∗ x (x^n)'=(n-1)*x (xn)′=(n−1)∗x s i n ′ ( x ) = c o s x sin'(x)=cosx sin′(x)=cosx c o s ′ ( x ) = − s i n x cos'(x)=-sinx cos′(x)=−sinx

lim ⁡ x → 0 x − s i n x x 3 6 = lim ⁡ x → 0 1 − c o s x 3 x 2 6 = lim ⁡ x → 0 s i n x x = 1 \lim_{x\rightarrow0}\frac{x-sinx}{\frac{x^3}{6}}=\lim_{x\rightarrow0}\frac{1-cosx}{\frac{3x^2}{6}}=\lim_{x\rightarrow0}\frac{sinx}{x}=1 limx→0​6x3​x−sinx​=limx→0​63x2​1−cosx​=limx→0​xsinx​=1

运用洛必达法则时用到了求导法则 t a n x = 1 cos ⁡ 2 x tanx=\frac{1}{\cos^2x} tanx=cos2x1​

lim ⁡ x → 0 t a n x − x x 3 3 = lim ⁡ x → 0 1 c o s 2 x − 1 x 2 = lim ⁡ x → 0 tan ⁡ 2 x x 2 = 1 \lim_{x\rightarrow0}\frac{tanx-x}{\frac{x^3}{3}}=\lim_{x\rightarrow0}\frac{\frac{1}{cos^2x}-1}{x^2}=\lim_{x\rightarrow0}\frac{\tan^2x}{x^2}=1 limx→0​3x3​tanx−x​=limx→0​x2cos2x1​−1​=limx→0​x2tan2x​=1

运用洛必达法则时用到了求导法则 l n ( 1 + x ) = 1 x + 1 ln(1+x)=\frac{1}{x+1} ln(1+x)=x+11​

lim ⁡ x → 0 x − l n ( 1 + x ) x 2 2 = lim ⁡ x → 0 1 − 1 1 + x x = lim ⁡ x → 0 x ( 1 + x ) ∗ x = lim ⁡ x → 0 1 ( 1 + x ) = 1 \lim_{x\rightarrow0}\frac{x-ln(1+x)}{\frac{x^2}{2}}=\lim_{x\rightarrow0}\frac{1-\frac{1}{1+x}}{x}=\lim_{x\rightarrow0}\frac{x}{(1+x)*x}=\lim_{x\rightarrow0}\frac{1}{(1+x)}=1 limx→0​2x2​x−ln(1+x)​=limx→0​x1−1+x1​​=limx→0​(1+x)∗xx​=limx→0​(1+x)1​=1

最后三例均通过洛必达法则证明，可以看到简化了很多计算。

求导用定义证明就是求极限。 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。

( x n ) ′ = lim ⁡ t → 0 ( x + t ) n − x n t = lim ⁡ t → 0 ( C n 0 ∗ x n + C n 1 ∗ x n − 1 t + C n 2 ∗ x n − 2 ∗ t 2 + . . . + C n n t n ) − x n t = lim ⁡ t → 0 ( C n 1 ∗ x n − 1 t + C n 2 ∗ x n − 2 ∗ t 2 + . . . + C n n t n ) t (x^n)'=\lim_{t\rightarrow 0}\frac{(x+t)^n-x^n}{t}=\lim_{t\rightarrow 0}\frac{(C_n^0*x^n+C_n^1*x^{n-1}t+C_n^2*x^{n-2}*t^2+...+C_n^nt^n)-x^n}{t} =\lim_{t\rightarrow 0}\frac{(C_n^1*x^{n-1}t+C_n^2*x^{n-2}*t^2+...+C_n^nt^n)}{t} (xn)′=limt→0​t(x+t)n−xn​=limt→0​t(Cn0​∗xn+Cn1​∗xn−1t+Cn2​∗xn−2∗t2+...+Cnn​tn)−xn​=limt→0​t(Cn1​∗xn−1t+Cn2​∗xn−2∗t2+...+Cnn​tn)​ = lim ⁡ t → 0 ( C n 1 ∗ x n − 1 + C n 2 ∗ x n − 2 t + . . . + C n n ∗ t n − 1 =\lim_{t\rightarrow 0}(C_n^1*x^{n-1}+C_n^2*x^{n-2}t+...+C_n^n*t^{n-1} =limt→0​(Cn1​∗xn−1+Cn2​∗xn−2t+...+Cnn​∗tn−1) = n ∗ x n − 1 =n*x^{n-1} =n∗xn−1

x + t − x 2 \frac{x+t-x}{2} 2x+t−x​ 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​+2a−b​)+sin(2a+b​−2a−b​) = 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(2a−b​)+cos(2a+b​)∗sin(2a−b​)+sin(2a+b​)∗cos(2a−b​)−cos(2a+b​)∗sin(2a−b​) = 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(2a−b​)

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} sin′x=limt→0​tsin(x+t)−sin(x)​=limt→0​tsin(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} =limt→0​t2sin(2x+t−x​)∗cos(2x+t+x​)​=limt→0​t2sin2t​cos22x+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 =limt→0​tt∗cos(x+2t​)​=cosx

方法一： c o s ′ x = s i n ′ ( π 2 − x ) = c o s ( π 2 − x ) ∗ − 1 = − s i n x cos'x=sin'(\frac{\pi}{2}-x)=cos(\frac{\pi}{2}-x)*-1=-sinx cos′x=sin′(2π​−x)=cos(2π​−x)∗−1=−sinx 方法二： c o s ′ x = ( 1 − 2 s i n 2 x 2 ) ′ = − 4 ∗ s i n ( x 2 ) ∗ c o s ( x 2 ) ∗ 1 2 = − 2 ∗ s i n ( x 2 ) ∗ c o s ( x 2 ) = − s i n x cos'x=(1-2sin^2\frac{x}{2})'=-4*sin(\frac{x}{2})*cos(\frac{x}{2})*\frac{1}{2}=-2*sin(\frac{x} {2})*cos(\frac{x}{2})=-sinx cos′x=(1−2sin22x​)′=−4∗sin(2x​)∗cos(2x​)∗21​=−2∗sin(2x​)∗cos(2x​)=−sinx

t a n x = ( s i n x c o s x ) ′ = c o s 2 x + s i n 2 x cos ⁡ 2 x = 1 c o s 2 x = s e c 2 x tanx=(\frac{sinx}{cosx})'=\frac{cos^2x+sin^2x}{\cos^2x}=\frac{1}{cos^2x}=sec^2x tanx=(cosxsinx​)′=cos2xcos2x+sin2x​=cos2x1​=sec2x

l n ′ ( 1 + x ) = lim ⁡ t → 0 l n ( 1 + x + t ) − l n ( 1 + x ) t = lim ⁡ t → 0 1 t l n ( 1 + x + t 1 + x ) ln'(1+x)=\lim_{t\rightarrow 0}\frac{ln(1+x+t)-ln(1+x)}{t}=\lim_{t\rightarrow 0}\frac{1}{t}ln(\frac{1+x+t}{1+x}) ln′(1+x)=limt→0​tln(1+x+t)−ln(1+x)​=limt→0​t1​ln(1+x1+x+t​) = lim ⁡ t → 0 l n ( 1 + t 1 + x ) 1 + x t ∗ 1 1 + x = l n e 1 1 + x = 1 1 + x =\lim_{t\rightarrow 0}ln(1+\frac{t}{1+x})^{\frac{1+x}{t}*\frac{1}{1+x}}=lne^{\frac{1}{1+x}}=\frac{1}{1+x} =limt→0​ln(1+1+xt​)t1+x​∗1+x1​=lne1+x1​=1+x1​

( e x ) ′ = lim ⁡ t → 0 e x + t − e x t = lim ⁡ t → 0 e x ( e t − 1 ) t = e x (e^x)'=\lim_{t\rightarrow 0}\frac{e^{x+t}-e^x}{t}=\lim_{t\rightarrow 0}\frac{e^x(e^t-1)}{t}=e^x (ex)′=t→0lim​tex+t−ex​=t→0lim​tex(et−1)​=ex

( a x ) ′ = lim ⁡ t → 0 a x + t − a x t = lim ⁡ t → 0 a x ( a t − 1 ) t = lim ⁡ t → 0 a x l n a ∗ t t = a x l n a (a^x)'=\lim_{t\rightarrow 0}\frac{a^{x+t}-a^x}{t}=\lim_{t\rightarrow 0}\frac{a^x(a^t-1)}{t}=\lim_{t\rightarrow 0}\frac{a^xlna*t}{t}=a^xlna (ax)′=t→0lim​tax+t−ax​=t→0lim​tax(at−1)​=t→0lim​taxlna∗t​=axlna