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3.3 不定式
2025-12-03
高等数学
数学

§3.3\S 3.3 不定式#

定义：两个无穷小量/大量之比的极限称为不定式/未定式

一、00\frac{0}{0} 型极限#

定理（洛必达法则 L’Hôpital Rule）：若f(x),g(x)f(x),g(x) 满足

  1. lim⁡x→x0f(x)=lim⁡x→x0g(x)=0\displaystyle \lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0
  2. 在 U(x0)∘\overset{\circ}{U(x_0)} 内可导，g′(x)≠0g'(x)\ne 0
  3. lim⁡x→x0f′(x)g′(x)=A\displaystyle \lim_{x\to x_0}\dfrac{f'(x)}{g'(x)}=A(A 可以是实数或正负无穷)

则 lim⁡x→x0f(x)g(x)=lim⁡x→x0f′(x)g′(x)=⋯=lim⁡x→x0f(n)(x)g(n)(x)=A\displaystyle \lim_{x\to x_0}\dfrac{f(x)}{g(x)}=\lim_{x\to x_0}\dfrac{f'(x)}{g'(x)}=\cdots =\lim_{x\to x_0}\dfrac{f^{(n)}(x)}{g^{(n)}(x)}=A

证明

令 F(x)={f(x)x≠x00x=x0F(x)=\begin{cases}f(x)\quad x\ne x_0\\ 0\quad x=x_0\end{cases}

f(x)g(x)=F(x)G(x)=F(x)−F(x0)G(x)−G(x0)=F′(ξ)(ξ∈(x0,x))\dfrac{f(x)}{g(x)}=\dfrac{F(x)}{G(x)}=\dfrac{F(x)-F(x_0)}{G(x)-G(x_0)}=F'(\xi)(\xi \in(x_0,x))

lim⁡x→x0f(x)g(x)=lim⁡x→x0f′(ξ)g′(ξ)=lim⁡ξ→x0f′(ξ)g′(ξ)=A\displaystyle \lim_{x\to x_0}\dfrac{f(x)}{g(x)}=\lim_{x\to x_0}\dfrac{f'(\xi)}{g'(\xi)}=\lim_{\xi \to x_0}\dfrac{f'(\xi)}{g'(\xi)}=A

例题#

  1. lim⁡x→0sin⁡xx=cos⁡x=1\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=\cos x=1
lim⁡x→1x3−3x+2x3−x2−x+1=lim⁡x→13x2−33x2−2x−1=lim⁡x→16x6x−2=64=32\begin{align*} \lim_{x\to 1}\dfrac{x^3-3x+2}{x^3-x^2-x+1}&=\lim_{x\to 1}\dfrac{3x^2-3}{3x^2-2x-1}\\ &=\lim_{x\to 1}\dfrac{6x}{6x-2}\\ &=\dfrac{6}{4}=\dfrac{3}{2} \end{align*}
lim⁡x→π1+cos⁡xtan⁡2x=lim⁡x→π−sin⁡x2tan⁡x1cos⁡2x=12\begin{align*} \lim_{x\to \pi}\dfrac{1+\cos x}{\tan^2 x}&=\lim_{x\to \pi}\dfrac{-\sin x}{2\tan x\frac{1}{\cos ^2x}}\\ &=\dfrac{1}{2} \end{align*}
lim⁡x→0x−sin⁡xx3=lim⁡x→01−cos⁡x3x2=lim⁡x→0x223x2=16\begin{align*} \lim_{x\to 0}\dfrac{x-\sin x}{x^3}&=\lim_{x\to 0}\dfrac{1-\cos x}{3x^2}\\ &=\lim_{x\to 0}\dfrac{\frac{x^2}{2}}{3x^2}=\dfrac{1}{6} \end{align*}
lim⁡x→0x−tan⁡xx3=lim⁡x→01−1cos⁡2x3x2=lim⁡x→0(−sin⁡2x3x2⋅1cos⁡2x)=−13×1=−13\begin{align*} \lim_{x\to 0}\dfrac{x-\tan x}{x^3}&=\lim_{x\to 0}\dfrac{1-\frac{1}{\cos^2x}}{3x^2}\\ &=\lim_{x\to 0}(-\dfrac{\sin^2x}{3x^2}\cdot \dfrac{1}{\cos ^2x})\\ &=-\dfrac{1}{3}\times 1=-\dfrac{1}{3} \end{align*}
lim⁡x→0+x1−ex=lim⁡x→0+12⋅1x−ex⋅12x=−1\begin{align*} \lim_{x\to 0^+}\dfrac{\sqrt{x}}{1-e^{\sqrt{x}}}&=\lim_{x\to 0^+}\dfrac{\frac{1}{2}\cdot \frac{1}{\sqrt{x}}}{-e^{\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}}=-1 \end{align*}
lim⁡x→0e−(1+x)1xx=lim⁡x→0(1+x)1x[−1x2ln⁡(1+x)+1x⋅11+x]=lim⁡x→0(−e)[1x(x+1)−1x2ln⁡(1+x)]=−elim⁡x→0x−(1+x)ln⁡(1+x)x2(1+x)=−elim⁡x→0x−(x+x22+o(x2))x2=e2\begin{align*} \lim_{x\to 0}\dfrac{e-(1+x)^{\frac{1}{x}}}{x}&=\lim_{x\to 0}(1+x)^{\frac{1}{x}}[-\dfrac{1}{x^2}\ln(1+x)+\dfrac{1}{x}\cdot\dfrac{1}{1+x}]\\ &=\lim_{x\to 0}(-e)[\dfrac{1}{x(x+1)}-\dfrac{1}{x^2}\ln(1+x)]\\ &=-e\lim_{x\to 0}\dfrac{x-(1+x)\ln(1+x)}{x^2(1+x)}\\ &=-e\lim_{x\to 0}\dfrac{x-(x+\frac{x^2}{2}+o(x^2))}{x^2}\\ &=\dfrac{e}{2} \end{align*}

二、∞∞\frac{\infty}{\infty} 型极限#

定理：若 f(x),g(x)f(x),g(x) 满足：

  1. lim⁡x→x0f(x)=∞lim⁡x→x0g(x)=∞\displaystyle \lim_{x\to x_0}f(x)=\infty\quad \lim_{x\to x_0}g(x)=\infty
  2. 在 U(x0)∘\overset{\circ}{U(x_0)} 内可导，g′(x)≠0g'(x)\ne 0
  3. lim⁡x→x0f′(x)g′(x)=A\displaystyle \lim_{x\to x_0}\dfrac{f'(x)}{g'(x)}=A(A 可以是实数或正负无穷)

则 lim⁡x→x0f(x)g(x)=lim⁡x→x0f′(x)g′(x)\displaystyle \lim_{x\to x_0}\dfrac{f(x)}{g(x)}=\lim_{x\to x_0}\dfrac{f'(x)}{g'(x)}

例题#

lim⁡x→+∞x2+1x=lim⁡x→+∞12x2+1⋅2x1=lim⁡x→+∞xx2+1=1\begin{align*} \lim_{x\to +\infty}\dfrac{\sqrt{x^2+1}}{x}&=\lim_{x\to +\infty}\dfrac{\frac{1}{2\sqrt{x^2+1}}\cdot 2x}{1}\\ &=\lim_{x\to +\infty}\dfrac{x}{\sqrt{x^2+1}}=1 \end{align*}
  1. lim⁡x→+∞log⁡axxa=0(a>0)\displaystyle \lim_{x\to +\infty}\dfrac{\log_a x}{x^a}=0(a \gt 0)
lim⁡x→+∞π2−arctan⁡x1x=lim⁡x→+∞−11+x2−1x2=lim⁡x→+∞x2x2+1=1\begin{align*} \lim_{x\to +\infty}\dfrac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}&=\lim_{x\to +\infty}\dfrac{-\frac{1}{1+x^2}}{-\frac{1}{x^2}}\\ &=\lim_{x\to +\infty}\dfrac{x^2}{x^2+1}=1 \end{align*}

注意 lim⁡x→−∞π2−arctan⁡x1x=−∞\displaystyle \lim_{x\to -\infty}\dfrac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}=-\infty

三、其他不定式#

  • 0⋅∞→取倒数00/∞∞0\cdot \infty \overset{取倒数}{\rightarrow} \frac{0}{0}/\frac{\infty}{\infty}
  • 1∞→取对数0⋅∞1^\infty \overset{取对数}{\rightarrow} 0\cdot \infty
  • 00→0⋅∞0^0 \rightarrow 0\cdot \infty
  • ∞0→0⋅∞\infty^0 \rightarrow 0\cdot \infty
  • ∞−∞\infty-\infty，通分

例题#

  1. [0⋅∞0\cdot \infty 型]
lim⁡x→0+xαln⁡x(α>0)=lim⁡x→0+ln⁡xx−α=lim⁡x→0+1x−α(xα−1)=lim⁡x→0+(−1α⋅xα)=0\begin{align*} \lim_{x\to 0^+}x^\alpha \ln x(\alpha \gt 0)&=\lim_{x \to 0^+}\dfrac{\ln x}{x^{-\alpha}}\\ &=\lim_{x\to 0^+}\dfrac{\frac{1}{x}}{-\alpha (x^{\alpha-1})}\\ &=\lim_{x\to 0^+}(-\dfrac{1}{\alpha}\cdot x^\alpha)=0 \end{align*}
  1. [1∞1^\infty 型]
lim⁡x→0(cos⁡x)1x2=lim⁡x→0eln⁡(cos⁡x)1x2=elim⁡x→0ln⁡cos⁡xx2=elim⁡x→01cos⁡x⋅(−sin⁡x)2x=e−12\begin{align*} \lim_{x\to 0}(\cos x)^{\frac{1}{x^2}}&=\lim_{x\to 0}e^{\ln (\cos x)^{\frac{1}{x^2}}}\\ &=e^{\lim_{x\to 0}\frac{\ln \cos x}{x^2}}\\ &=e^{\lim_{x\to 0}\frac{\frac{1}{\cos x}\cdot (-\sin x)}{2x}}\\ &=e^{-\frac{1}{2}} \end{align*}
  1. [000^0 型]
lim⁡x→0+xx=lim⁡x→0+eln⁡xx=lim⁡x→0+exln⁡x=e0=1\begin{align*} \lim_{x\to 0^+}x^x&=\lim_{x\to 0^+}e^{\ln x^x}\\ &=\lim_{x\to 0^+}e^{x\ln x}\\ &=e^0=1 \end{align*}
  1. [∞0\infty ^0 型]
lim⁡x→+∞(x+1+x2)1ln⁡x=elim⁡x→+∞ln⁡(x+1+x2)ln⁡x=elim⁡x→+∞11+x21x=e1=e\begin{align*} \lim_{x \to +\infty}(x+\sqrt{1+x^2})^{\frac{1}{\ln x}}&=e^{\lim_{x \to +\infty}\frac{\ln (x+\sqrt{1+x^2})}{\ln x}}\\ &=e^{\lim_{x \to +\infty}\frac{\frac{1}{\sqrt{1+x^2}}}{\frac{1}{x}}}\\ &=e^1=e \end{align*}
  1. [∞−∞\infty-\infty 型]
lim⁡x→π2(sec⁡x−tan⁡x)=lim⁡x→π21−sin⁡xcos⁡x=lim⁡x→π2−cos⁡x−sin⁡x=−0−1=0\begin{align*} \lim_{x\to \frac{\pi}{2}}(\sec x-\tan x)&=\lim_{x\to \frac{\pi}{2}}\dfrac{1-\sin x}{\cos x}\\ &=\lim_{x\to \frac{\pi}{2}}\dfrac{-\cos x}{-\sin x}\\ &=\dfrac{-0}{-1}=0 \end{align*}
lim⁡x→+∞x32(x+2−2x+1+x)=t=1xlim⁡t→0+1+2tt−21+t+1tt32=lim⁡t→0+1+2t−21+t+1t2=lim⁡t→0+12×121+2t−2×12×11+t×12t=lim⁡t→0+−1+t−1+2t2t1+t1+2t=lim⁡t→0+−t(1+t+1+2t)⋅2t=−14\begin{align*} \lim_{x \to +\infty}x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})&\overset{t=\frac{1}{x}}{=}\lim_{t\to 0^+}\dfrac{\sqrt{\frac{1+2t}{t}}-2\sqrt{1+t}+\sqrt{\frac{1}{t}}}{t^{\frac{3}{2}}}\\ &=\lim_{t \to 0^+}\dfrac{\sqrt{1+2t}-2\sqrt{1+t}+1}{t^2}\\ &=\lim_{t \to 0^+}\dfrac{\frac{1}{2}\times\frac{1}{2\sqrt{1+2t}}-2\times\frac{1}{2}\times{\frac{1}{\sqrt{1+t}}\times 1}}{2t}\\ &=\lim_{t \to 0^+}\dfrac{-\sqrt{1+t}-\sqrt{1+2t}}{2t\sqrt{1+t}\sqrt{1+2t}}\\ &=\lim_{t \to 0^+}\dfrac{-t}{(\sqrt{1+t}+\sqrt{1+2t})\cdot 2t}\\ &=-\dfrac{1}{4} \end{align*}
3.3 不定式
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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