三部曲法求未定式极限中的1无穷次方型
文章目录
- 1 ∞ 1^\infin 1∞型幂指函数的极限👺
- 分离常数变形
- 分子分母同时除以变化最快项
- 速算结论👺(三部曲)
- 证明
- 应用
- 例
- 逐步演算
- 例
- 例
- 例
- 例
- 例
- 补充:极限含参形式
1 ∞ 1^\infin 1∞型幂指函数的极限👺
- 这类极限问题属于未定式,若极限存在,则应该为 e e e相关的式子,也可能极限不存在(尽管大多数我们遇到的都是极限存在的情形)
分离常数变形
- 有时,需要使用分离常数的技巧将函数的形式转换为
f
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x
)
=
(
1
+
α
(
x
)
)
β
(
x
)
{f(x)=(1+\alpha (x))^{\beta(x)}}
f(x)=(1+α(x))β(x)的形式,
- 例如: ( x + 1 x − 3 ) x = ( x − 3 + 3 + 1 x − 3 ) x = ( 1 + 4 x − 3 ) x (\frac{x+1}{x-3})^x=(\frac{x-3+3+1}{x-3})^x=(1+\frac{4}{x-3})^{x} (x−3x+1)x=(x−3x−3+3+1)x=(1+x−34)x
分子分母同时除以变化最快项
- 这种方法有时比分离常数更加方便和直接
- 通常用在含
∞
\infin
∞的情形下
- lim n → ∞ n n ( n + 1 ) n + 1 \lim\limits_{n\to{\infin}}\frac{n^{n}}{(n+1)^{n+1}} n→∞lim(n+1)n+1nn= lim n → ∞ 1 n + 1 ( n n + 1 ) n \lim\limits_{n\to{\infin}}\frac{1}{n+1}(\frac{n}{n+1})^{n} n→∞limn+11(n+1n)n= lim n → ∞ 1 n + 1 ( 1 1 + 1 n ) n \lim\limits_{n\to{\infin}}\frac{1}{n+1}(\frac{1}{1+\frac{1}{n}})^{n} n→∞limn+11(1+n11)n= lim n → ∞ 1 n + 1 ( 1 ( 1 + 1 n ) n ) \lim\limits_{n\to{\infin}}\frac{1}{n+1}(\frac{1}{(1+\frac{1}{n})^{^{n}}}) n→∞limn+11((1+n1)n1)= lim n → ∞ 1 n + 1 ⋅ lim n → ∞ ( 1 ( 1 + 1 n ) n ) \lim\limits_{n\to{\infin}}\frac{1}{n+1} \cdot \lim\limits_{n\to{\infin}}(\frac{1}{(1+\frac{1}{n})^{^{n}}}) n→∞limn+11⋅n→∞lim((1+n1)n1)= 0 ⋅ 1 e 0\cdot{\frac{1}{e}} 0⋅e1=0
速算结论👺(三部曲)
-
如果判断出 f ( x ) = ( 1 + α ( x ) ) β ( x ) {f(x)=(1+\alpha (x))^{\beta(x)}} f(x)=(1+α(x))β(x)的某个过程的极限属于 1 ∞ 1^\infin 1∞型的情况下求极限 S = lim f ( x ) S=\lim f(x) S=limf(x),则可以按如下步骤求解
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先计算出 A = lim ( α ( x ) β ( x ) ) A=\lim(\alpha(x)\beta(x)) A=lim(α(x)β(x))
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那么: S = e A S=e^A S=eA,也即是说,结果是 e e e的幂的形式
-
证明
- 设 α ( x ) , β ( x ) \alpha(x),\beta(x) α(x),β(x)分别极限过程 x → ∗ x\to{*} x→∗的无穷小量和无穷大量,即 lim α ( x ) = 0 \lim{\alpha(x)}=0 limα(x)=0, lim β ( x ) = ∞ \lim\beta(x)=\infin limβ(x)=∞
- γ ( x ) = α ( x ) β ( x ) \gamma(x)=\alpha(x)\beta(x) γ(x)=α(x)β(x); β ( x ) = 1 α ( x ) α ( x ) β ( x ) \beta(x)=\frac{1}{\alpha(x)}\alpha(x)\beta(x) β(x)=α(x)1α(x)β(x)= 1 α ( x ) γ ( x ) \frac{1}{\alpha(x)}\gamma(x) α(x)1γ(x)
- S = lim ( 1 + α ( x ) ) β ( x ) S=\lim(1+\alpha(x))^{\beta(x)} S=lim(1+α(x))β(x)= lim ( 1 + α ( x ) ) 1 α ( x ) α ( x ) β ( x ) \lim(1+\alpha(x))^{\frac{1}{\alpha(x)}\alpha(x)\beta(x)} lim(1+α(x))α(x)1α(x)β(x)= lim ( ( ( 1 + α ( x ) ) 1 α ( x ) ) γ ( x ) \lim{(((1+\alpha(x))^\frac{1}{\alpha(x)}})^{\gamma(x)} lim(((1+α(x))α(x)1)γ(x)= [ lim ( ( ( 1 + α ( x ) ) 1 α ( x ) ) ] γ ( x ) [\lim{(((1+\alpha(x))^\frac{1}{\alpha(x)}})]^{\gamma(x)} [lim(((1+α(x))α(x)1)]γ(x)= e γ ( x ) e^{\gamma(x)} eγ(x)
- 记 A = lim γ ( x ) A=\lim{\gamma(x)} A=limγ(x),则 S = e A S=e^{A} S=eA
应用
例
- 以下3个的
1
∞
1^\infin
1∞型极限都可以用
e
A
e^A
eA模型法来计算,先确定
α
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x
)
和
β
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x
)
\alpha{(x)}和\beta{(x)}
α(x)和β(x)
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S 1 = lim x → ∞ ( 1 − 1 x ) x S_1=\lim\limits_{x\to \infin}(1-\frac{1}{x})^x S1=x→∞lim(1−x1)x
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S 2 = lim x → ∞ ( 1 + a x ) b x S_2=\lim\limits_{x\to \infin}{(1+\frac{a}{x})^{bx}} S2=x→∞lim(1+xa)bx
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S 3 = lim x → ∞ ( 1 + a x ) b x + c S_3=\lim\limits_{x\to \infin}(1+\frac{a}{x})^{bx+c} S3=x→∞lim(1+xa)bx+c
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- 分别计算
A
1
,
A
2
,
A
3
A_1,A_2,A_3
A1,A2,A3
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A 1 = lim x → ∞ − 1 x x = − 1 A_1=\lim\limits_{x\to \infin} \frac{-1}{x}x=-1 A1=x→∞limx−1x=−1
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A 2 = lim x → ∞ a x b x = a b A_2=\lim\limits_{x\to \infin} \frac{a}{x}bx=ab A2=x→∞limxabx=ab
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A 3 = lim x → ∞ a x ( b x + c ) = a b A_3=\lim\limits_{x\to \infin} \frac{a}{x}(bx+c)=ab A3=x→∞limxa(bx+c)=ab
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- S 1 = e − 1 S_1=e^{-1} S1=e−1, S 2 = e a b S_2=e^{ab} S2=eab, S 3 = e a b S_3=e^{ab} S3=eab
逐步演算
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lim x → ∞ ( 1 − 1 x ) x \lim\limits_{x\to \infin}{(1-\frac{1}{x})}^x x→∞lim(1−x1)x= lim x → ∞ ( 1 − 1 x ) − ( − x ) \lim\limits_{x\to \infin}{(1-\frac{1}{x})}^{-(-x)} x→∞lim(1−x1)−(−x)= lim x → ∞ 1 ( 1 − 1 x ) − x \lim\limits_{x\to \infin}\frac{1}{{{(1-\frac{1}{x})}^{-x}}} x→∞lim(1−x1)−x1= lim x → ∞ 1 lim x → ∞ ( 1 − 1 x ) − x \frac{\lim\limits_{x\to \infin}1}{\lim\limits_{x\to \infin}(1-\frac{1}{x})^{-x}} x→∞lim(1−x1)−xx→∞lim1= 1 e \frac{1}{e} e1
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lim x → ∞ ( 1 + a x ) b x \lim\limits_{x\to \infin}{(1+\frac{a}{x})^{bx}} x→∞lim(1+xa)bx= lim x → ∞ ( 1 + a x ) x a a b \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{\frac{x}{a}ab} x→∞lim(1+xa)axab= lim x → ∞ \lim\limits_{x\to \infin} x→∞lim ( ( 1 + a x ) x a ) a b \left ({(1+\frac{a}{x})}^{\frac{x}{a}}\right)^{ab} ((1+xa)ax)ab= e a b e^{ab} eab
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lim x → ∞ ( 1 + a x ) b x + c \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{bx+c} x→∞lim(1+xa)bx+c= lim x → ∞ ( 1 + a x ) b x \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{bx} x→∞lim(1+xa)bx ⋅ \cdot ⋅ lim x → ∞ ( 1 + a x ) c \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{c} x→∞lim(1+xa)c= e a b ⋅ 1 c e^{ab}\cdot 1^c eab⋅1c= e a b ⋅ 1 e^{ab}\cdot 1 eab⋅1= e a b e^{ab} eab
例
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f
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x
)
f(x)
f(x)=
(
x
+
2
x
)
2
x
(x+2^{x})^{\frac{2}{x}}
(x+2x)x2,
S
=
lim
x
→
0
f
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x
)
S=\lim\limits_{x\to{0}}{f(x)}
S=x→0limf(x)=?
- 这是一个 1 ∞ 1^{\infin} 1∞的未定式
- 方法1
- f ( x ) f(x) f(x)变形: f ( x ) f(x) f(x)= ( 1 + x + 2 x − 1 ) 2 x (1+x+2^{x}-1)^{\frac{2}{x}} (1+x+2x−1)x2
- 根据上述结论,求
A
A
A=
lim
x
→
0
(
x
+
2
x
−
1
)
⋅
2
x
\lim\limits_{x\to{0}}(x+2^{x}-1)\cdot{\frac{2}{x}}
x→0lim(x+2x−1)⋅x2=
2
(
lim
x
→
0
(
x
+
2
x
−
1
)
⋅
1
x
)
2(\lim\limits_{x\to{0}}(x+2^{x}-1)\cdot{\frac{1}{x}})
2(x→0lim(x+2x−1)⋅x1)=
2
(
1
+
lim
x
→
0
2
x
−
1
x
)
2(1+\lim\limits_{x\to{0}}\frac{2^{x}-1}{x})
2(1+x→0limx2x−1)=
2
(
1
+
ln
2
)
2(1+\ln{2})
2(1+ln2)
- lim x → 0 2 x − 1 x \lim\limits_{x\to{0}}\frac{2^{x}-1}{x} x→0limx2x−1= lim x → 0 x ln 2 x \lim\limits_{x\to{0}}\frac{x\ln{2}}{x} x→0limxxln2= ln 2 \ln{2} ln2
- 于是 S = e A S=e^{A} S=eA= ( e 1 + ln 2 ) 2 (e^{1+\ln{2}})^{2} (e1+ln2)2= ( e ⋅ 2 ) 2 (e\cdot{2})^2 (e⋅2)2= 4 e 2 4e^{2} 4e2
- 方法2
- f ( x ) f(x) f(x)= [ 2 x ( x 2 x + 1 ) ] 2 x [2^{x}(\frac{x}{2^{x}}+1)]^{\frac{2}{x}} [2x(2xx+1)]x2
-
S
S
S=
4
⋅
lim
x
→
0
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1
+
x
2
x
)
2
x
4\cdot{\lim\limits_{x\to{0}}}(1+\frac{x}{2^{x}})^{\frac{2}{x}}
4⋅x→0lim(1+2xx)x2
- A = lim x → 0 x 2 x ⋅ 2 x A=\lim\limits_{x\to{0}}\frac{x}{2^{x}}\cdot{\frac{2}{x}} A=x→0lim2xx⋅x2= 2 2 2
- S S S= 4 e 2 4e^{2} 4e2
例
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lim x → 0 ( ∫ 0 x 2 3 e 1 2 t 2 d t − x 2 3 + 1 ) 1 x 2 = 1 \lim\limits_{x\to{0}}\left( \int_{0}^{\sqrt[3]{x^2}}{e^{\frac{1}{2}t^2}}\mathrm{d}t-x^{\frac{2}{3}}+1 \right)^{\large{\frac{1}{x^2}} }=1 x→0lim(∫03x2e21t2dt−x32+1)x21=1
- 这是一个 1 ∞ 1^{\infin} 1∞型未定式,令 α ( x ) \alpha(x) α(x)= ∫ 0 x 2 3 e 1 2 t 2 d t − x 2 3 \int_{0}^{\sqrt[3]{x^2}}{e^{\frac{1}{2}t^2}}\mathrm{d}t-x^{\frac{2}{3}} ∫03x2e21t2dt−x32, β \beta β= 1 x 2 \frac{1}{x^2} x21
- 则 A = lim x → 0 α ( x ) β ( x ) A=\lim\limits_{x\to{0}} \alpha(x)\beta(x) A=x→0limα(x)β(x)= 0 0 0,从而原式等于 e A e^{A} eA= e 0 e^{0} e0= 1 1 1
例
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lim
x
→
0
(
e
x
+
e
2
x
+
⋯
+
e
n
x
n
)
1
x
\lim\limits_{x\to{0}}(\frac{e^{x}+e^{2x}+\cdots+e^{nx}}{n})^{\frac{1}{x}}
x→0lim(nex+e2x+⋯+enx)x1
- 分析这是一个 1 ∞ 1^{\infin} 1∞未定式
- 令 g ( x ) g(x) g(x)= e x + e 2 x + ⋯ + e n x n \frac{e^{x}+e^{2x}+\cdots+e^{nx}}{n} nex+e2x+⋯+enx,则 g ( x ) g(x) g(x)= e x + e 2 x + ⋯ + e n x − n n + 1 \frac{e^{x}+e^{2x}+\cdots+e^{nx}-n}{n}+1 nex+e2x+⋯+enx−n+1
- 记 a ( x ) a(x) a(x)= e x + e 2 x + ⋯ + e n x − n n \frac{e^{x}+e^{2x}+\cdots+e^{nx}-n}{n} nex+e2x+⋯+enx−n; b ( x ) b(x) b(x)= 1 x \frac{1}{x} x1,则 A ( x ) = a ( x ) b ( x ) A(x)=a(x)b(x) A(x)=a(x)b(x)= ( e x − 1 ) + ( e 2 x − 1 ) + ⋯ + ( e n x − 1 ) n x \frac{(e^{x}-1)+(e^{2x}-1)+\cdots+(e^{nx}-1)}{nx} nx(ex−1)+(e2x−1)+⋯+(enx−1)
- lim x → 0 A ( x ) \lim\limits_{x\to{0}}A(x) x→0limA(x)= lim x → 0 x + 2 x + ⋯ + n x n x \lim\limits_{x\to{0}}\frac{x+2x+\cdots+n{x}}{nx} x→0limnxx+2x+⋯+nx= 1 2 n ( n + 1 ) 1 n \frac{1}{2}n(n+1)\frac{1}{n} 21n(n+1)n1= n + 1 2 \frac{n+1}{2} 2n+1
- 所以原式= e n + 1 2 e^{\frac{n+1}{2}} e2n+1
例
-
lim
x
→
∞
(
x
n
(
x
+
1
)
(
x
+
2
)
⋯
(
x
+
n
)
)
x
\lim\limits_{x\to{\infin}} (\frac{x^{n}}{(x+1)(x+2)\cdots(x+n)})^{x}
x→∞lim((x+1)(x+2)⋯(x+n)xn)x
- 分析可知原式是 1 ∞ 1^{\infin} 1∞的未定式
- 令 f ( x ) f(x) f(x)= x n ( x + 1 ) ( x + 2 ) ⋯ ( x + n ) \frac{x^{n}}{(x+1)(x+2)\cdots(x+n)} (x+1)(x+2)⋯(x+n)xn
- 利用三部曲结论, ( f ( x ) − 1 + 1 ) x (f(x)-1+1)^{x} (f(x)−1+1)x可以做
- 但是这里可以使用通项思维,将原来的形式分解成形式相近的项,对这些项进行研究,或许能使得计算过程变得简单,稍加变形 f ( x ) f(x) f(x)= x ( 1 + x ) x ( x + 2 ) ⋯ x ( x + n ) \frac{x}{(1+x)}\frac{x}{(x+2)}\cdots\frac{x}{(x+n)} (1+x)x(x+2)x⋯(x+n)x,然后 g ( x ) g(x) g(x)= [ f ( x ) ] x [f(x)]^{x} [f(x)]x= [ x ( 1 + x ) ] x [ x ( x + 2 ) ] x ⋯ [ x ( x + n ) ] x [\frac{x}{(1+x)}]^{x}[\frac{x}{(x+2)}]^{x}\cdots[\frac{x}{(x+n)}]^{x} [(1+x)x]x[(x+2)x]x⋯[(x+n)x]x
- 这就将问题分解为 n n n个 [ x x + k ] x [\frac{x}{x+k}]^{x} [x+kx]x的 1 ∞ 1^{\infin} 1∞问题,这些处理起来就简单了(分离常数为 ( 1 − k x + k ) x (1-\frac{k}{x+k})^{x} (1−x+kk)x)去处理,求 x → ∞ x\to{\infin} x→∞的极限为 e − k e^{-k} e−k
- 更进一步,我们可以把 [ x x + k ] x [\frac{x}{x+k}]^{x} [x+kx]x= [ x + k x ] − x [\frac{x+k}{x}]^{-x} [xx+k]−x= ( 1 + k x ) − x (1+\frac{k}{x})^{-x} (1+xk)−x,那么也可以起到分离常数的效果,求 x → ∞ x\to{\infin} x→∞的极限为 e − k e^{-k} e−k
- 从而原式: lim x → ∞ g ( x ) \lim\limits_{x\to{\infin}}g(x) x→∞limg(x)= e − n ( n + 1 ) 2 e^{-\frac{n(n+1)}{2}} e−2n(n+1)
例
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lim n → ∞ ( ( 1 + x ) 1 x e ) 1 x \lim\limits_{n\to\infin} (\frac{(1+x)^{\frac{1}{x}}}{e})^{\frac{1}{x}} n→∞lim(e(1+x)x1)x1
-
分析
- 分析可知该极限使一个 1 ∞ 1^{\infin} 1∞型未定式
- 令 f ( x ) f(x) f(x)= ( 1 + x ) 1 x e \frac{(1+x)^{\frac{1}{x}}}{e} e(1+x)x1; g ( x ) g(x) g(x)= [ f ( x ) ] 1 x [f(x)]^{\frac{1}{x}} [f(x)]x1
- 求解本题的重要要求是
- 分析极限类型(未定式)
- 掌握幂指函数指数化或者三部曲法
- 掌握等价无穷小:
ln
(
1
+
x
)
−
x
∼
−
1
2
x
2
\ln(1+x)-x\sim{-\frac{1}{2}x^2}
ln(1+x)−x∼−21x2
(1)
-
方法1:
- 直接使用三部曲法
- 将 f ( x ) f(x) f(x)变形为 f ( x ) f(x) f(x)= 1 + ( 1 + x ) 1 x − e e 1+\frac{(1+x)^{\frac{1}{x}}-e}{e} 1+e(1+x)x1−e; a ( x ) = ( 1 + x ) 1 x − e e a(x)=\frac{(1+x)^{\frac{1}{x}}-e}{e} a(x)=e(1+x)x1−e; b ( x ) = 1 x b(x)=\frac{1}{x} b(x)=x1
- 从而
c
(
x
)
c(x)
c(x)=
(
1
+
x
)
1
x
−
e
e
⋅
1
x
\frac{(1+x)^{\frac{1}{x}}-e}{e}\cdot{\frac{1}{x}}
e(1+x)x1−e⋅x1;
- ( e ( 1 x ln ( 1 + x ) ) − e ({e^{(\frac{1}{x}\ln(1+x)})-e} (e(x1ln(1+x))−e= e ξ ( 1 x ln ( 1 + x ) − 1 ) e^{\xi}(\frac{1}{x}\ln(1+x)-1) eξ(x1ln(1+x)−1); ξ ∈ ( 1 x ln ( 1 + x ) , 1 ) \xi\in(\frac{1}{x}\ln(1+x),1) ξ∈(x1ln(1+x),1)
- 当 x → 0 x\to{0} x→0时, 1 x ln ( 1 + x ) → 1 \frac{1}{x}\ln(1+x)\to{1} x1ln(1+x)→1,从而由夹逼准则 ξ → 1 \xi\to{1} ξ→1,即 e ξ → e e^{\xi}\to{e} eξ→e
- lim x → 0 c ( x ) \lim\limits_{x\to{0}}c(x) x→0limc(x)= lim x → 0 e ξ ( 1 x ln ( 1 + x ) − 1 ) e x \lim\limits_{x\to{0}}\frac{e^{\xi}(\frac{1}{x}\ln(1+x)-1)}{ex} x→0limexeξ(x1ln(1+x)−1)= lim x → 0 ( 1 x ln ( 1 + x ) − 1 ) x \lim\limits_{x\to{0}}\frac{(\frac{1}{x}\ln(1+x)-1)}{x} x→0limx(x1ln(1+x)−1)= lim x → 0 ( ln ( 1 + x ) − x ) x 2 \lim\limits_{x\to{0}}\frac{(\ln(1+x)-x)}{x^2} x→0limx2(ln(1+x)−x)
- 再根据等价无穷小(1),替换: lim x → 0 c ( x ) \lim\limits_{x\to{0}}c(x) x→0limc(x)= lim x → 0 − 1 2 x 2 x 2 \lim\limits_{x\to{0}}\frac{-\frac{1}{2}x^2}{x^2} x→0limx2−21x2= − 1 2 -\frac{1}{2} −21
- 原式= e − 1 2 e^{-\frac{1}{2}} e−21
-
方法2:
- 使用幂指型化为指数型的手法变形(复合函数求导法)
-
f
(
x
)
f(x)
f(x)=
e
1
x
ln
(
1
+
x
)
e
\frac{e^{\frac{1}{x}\ln{(1+x)}}}{e}
eex1ln(1+x),从而
g
(
x
)
g(x)
g(x)=
e
1
x
2
ln
(
1
+
x
)
e
1
x
\huge\frac{e^{\frac{1}{x^2}\ln{(1+x)}}}{e^{\frac{1}{x}}}
ex1ex21ln(1+x)=
e
[
1
x
2
ln
(
1
+
x
)
−
1
x
]
\huge e^{[{\frac{1}{x^2}\ln{(1+x)}}-\frac{1}{x}]}
e[x21ln(1+x)−x1]
- 而 lim x → 0 1 x 2 [ ln ( 1 + x ) − x ] \lim\limits_{x\to{0}} {\frac{1}{x^2}[\ln{(1+x)}}-x] x→0limx21[ln(1+x)−x]= lim x → 0 − 1 2 x 2 x 2 \lim\limits_{x\to{0}}- \frac{\frac{1}{2}x^2}{x^2} x→0lim−x221x2= − 1 2 -\frac{1}{2} −21
- lim x → 0 g ( x ) \lim\limits_{x\to{0}}g(x) x→0limg(x)= e − 1 2 e^{-\frac{1}{2}} e−21
补充:极限含参形式
- lim n → ∞ ( 1 + 1 n ) n 2 \lim\limits_{n\to\infin}(1+\frac{1}{n})^{n^2} n→∞lim(1+n1)n2,也是一个 1 ∞ 1^{\infin} 1∞的未定式
- 利用上述结论,得 A = lim 1 n n 2 A=\lim{\frac{1}{n}n^2} A=limn1n2= n n n, lim n → ∞ ( 1 + 1 n ) n 2 \lim\limits_{n\to\infin}(1+\frac{1}{n})^{n^2} n→∞lim(1+n1)n2= e A e^{A} eA= e n e^{n} en,由于 n → ∞ n\to{\infin} n→∞所以 e n → ∞ e^{n}\to\infin en→∞,所以极限不存在