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 x→0limsinxx=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 limx→0g(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 x→0limxln(1+x)=x→0limln(1+x)x1=ln(x→+∞lim(1+x1)x)=lne=1
2.e^x-1~x
令
e
x
−
1
=
t
e^x-1=t
ex−1=t，则x=ln(1+t)。因为
x
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0
x\rightarrow0
x→0,所以
t
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0
t\rightarrow0
t→0
lim
x
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e
x
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x
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lim
t
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l
n
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t
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\lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{t\rightarrow0}\frac{t}{ln(1+t)}=1
x→0limxex−1=t→0limln(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
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t
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1
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=
m
，
ln
a
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n
\log_a(t+1)=m，\ln a=n
loga(t+1)=m，lna=n，那么有：
a
m
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t
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1
,
e
n
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a
a^m=t+1,e^n=a
am=t+1,en=a，所以，
(
e
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m
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t
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1
(e^n)^m=t+1
(en)m=t+1，即:
e
m
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1
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>
m
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n
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e^{mn}=t+1=>m*n=ln(t+1)
emn=t+1=>m∗n=ln(t+1)
所以有结论：
log
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ln
a
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m
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ln
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\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→0limxlnaax−1=t→0limloga(t+1)lnat=t→0limln(t+1)t=1
4. 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
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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
=
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=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)
=
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=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)
=
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(
a
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=(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
\lim_{x\rightarrow0}\frac{\sqrt[n]{1+x}-1}{x}
limx→0xn1+x
−1
=
lim
x
→
0
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1
+
x
)
1
n
−
1
x
n
=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{n}}-1}{\frac{x}{n}}
=limx→0nx(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
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\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→0x∗((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→0x∗(n−1+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+t−x
预备知识：
s
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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)
=
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=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
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=2sin(\frac{a+b}{2})*cos(\frac{a-b}{2})
=2sin(2a+b)∗cos(2a−b)
求导推导：
s
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lim
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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→0tsin(x+t)−sin(x)=limt→0tsin(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→0t2sin(2x+t−x)∗cos(2x+t+x)=limt→0t2sin2tcos22x+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→0tt∗cos(x+2t)=cosx