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重要极限及其推论

Kamimika...大约 2 分钟学习笔记

重要极限及其推论

三角函数 00\frac{0}{0}

limx0sinxx=1\lim_{x \to 0} \dfrac{\sin x}{x} = 1, (limxsinxx=0\lim_{x \to \infty} \dfrac{\sin x}{x} = 0)

推论:

  1. limxx0sinφ(x0)φ(x0)=1,(limxx0φ(x)=0,φ(x)0)\lim_{x \to x_0} \dfrac{\sin \varphi(x_0)}{\varphi(x_0)} = 1, (\lim_{x \to x_0} \varphi(x)=0, \varphi(x) \neq 0)
  2. limx0tanxx=limx0sinxxcosx=1\lim_{x \to 0} \dfrac{\tan x}{x} = \lim_{x \to 0} \dfrac{\sin x}{x\cos x} = 1
  3. limx01cosxx2=limx02sin2x2(x2)24=12\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \lim_{x \to 0} \dfrac{2\sin^2 \dfrac{x}{2}}{(\dfrac{x}{2})^2 \cdot 4} = \dfrac{1}{2}
  4. limx0arcsinxx=limx0arcsinxsin(arcsinx)=1\lim_{x \to 0} \dfrac{\arcsin x}{x} = \lim_{x \to 0} \dfrac{\arcsin x}{\sin(\arcsin x)} = 1
  5. limx0arctanxx=limx0arctanxtan(arctanx)=1\lim_{x \to 0} \dfrac{\arctan x}{x} = \lim_{x \to 0} \dfrac{\arctan x}{\tan(\arctan x)} = 1

对数/指数/幂 11^\infty

limx(1+1x)x=e,limx0(1+x)1x=e\lim_{x \to \infty} (1 + \dfrac{1}{x})^x = e, \lim_{x \to 0} (1 + x)^\dfrac{1}{x} = e    limx(11x)x=limx(1+1x)x=1e\implies \lim_{x \to \infty} (1 - \dfrac{1}{x})^x = \lim_{x \to \infty} (1 + \dfrac{1}{x})^{-x} = \dfrac{1}{e}

推论:

  1. limxx0(1+φ(x))1φ(x)=e\lim_{x \to x_0} (1 + \varphi(x))^\dfrac{1}{\varphi(x)} = e, (limxx0φ(x)=)(\lim_{x \to x_0} \varphi(x) = \infty)
  2. limx0ln(1+x)x=limx0ln(1+x)1x=lne=1\lim_{x \to 0} \dfrac{\ln (1+x)}{x} = \lim_{x \to 0} \ln (1 + x)^\dfrac{1}{x} = \ln e = 1
  3. limx0ex1x=limt0tln(1+t)=1\lim_{x \to 0} \dfrac{e^x-1}{x} = \lim_{t \to 0} \dfrac{t}{\ln (1+t)} = 1, (t=ex1,x=ln(t+1))(t=e^x - 1, x=\ln(t+1)) 换元法 (x0x \to 0t0t \to 0)
  4. limx0ax1x=limt0tln(1+t)lna=lna\lim_{x \to 0} \dfrac{a^x - 1}{x} = \lim_{t \to 0} \dfrac{t}{\ln(1+t)}\ln a = \ln a, (t=ax1,x=ln(1+t)lna)(t=a^x-1, x=\dfrac{\ln(1+t)}{\ln a}) 换元法 (x0x \to 0t0t \to 0)
  5. limx0(1+x)α1x=limx0,t0tln(1+t)αln(1+x)x=α\lim_{x \to 0} \dfrac{(1+x)^\alpha - 1}{x} = \lim_{x \to 0, t \to 0} \dfrac{t}{\ln (1+t)} \cdot \dfrac{\alpha \ln(1+x)}{x} = \alpha, (t=(1+x)α1,ln(1+t)=αln(1+x))(t=(1+x)^\alpha - 1, \ln(1+t)=\alpha \ln(1+x)) 换元法 (x0x \to 0t0t \to 0)

应用:幂函数 f(x)g(x)=eg(x)lnf(x)=eg(x)ln(1+f(x)1)f(x)1[f(x)1]eg(x)[f(x)1]f(x)^{g(x)} = e^{g(x) \ln f(x)} = e^{g(x) \frac{\ln(1+f(x)-1)}{f(x)-1} [f(x)-1]} \to e^{g(x)[f(x)-1]}, (f(x)1)(f(x) \to 1)limf(x)g(x)=elimg(x)lnf(x)=[limf(x)]limg(x)\lim f(x)^{g(x)} = e^{\lim g(x) \ln f(x)} = [\lim f(x)]^{\lim g(x)}, (f(x)A>0)(f(x) \to A>0)

警告

注意: f(x),g(x)f(x), g(x) 必须同时取极限, limf(x)g(x)[limf(x)]g(x)\lim f(x)^{g(x)} \neq [\lim f(x)]^{g(x)} (特别是 limf(x)=e\lim f(x) = e 时易错)

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