函数的等价无穷小

发布时间:2018-08-10 作者:大扑棱蛾子 阅读次数:
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如果limβα=1,那么就说βα是等价无穷小,记作αβ\text{如果} lim\frac{\beta}{\alpha}=1, \text{那么就说}\beta \text{与} \alpha \text{是等价无穷小,记作} \alpha \sim \beta

几个常用的等价无穷小关系式

ln(1+x)x(x0)ex1x(x0)(1+x)α1αx(x0)\begin{array}{rr} ln(1+x) \sim x \qquad (x \to 0) \\ e^x-1 \sim x \qquad (x \to 0) \\ (1+x)^\alpha -1 \sim \alpha x \qquad (x \to 0) \end{array}

证明过程

例1:求limx0loga(1+x)x\text{例1:求}\quad \lim\nolimits_{x \to 0} \frac{\log_a(1+x)}{x}

limx0loga(1+x)x=limx0loga(1+x)1xf(x)=logax,g(x)=(1+x)1xlimx0loga(1+x)x=limx0fg=f(limx0g(x))=f(limx0(1+x)1x)a=1x,那么x0    a所以limx0(1+x)1x=lima(1+1a)a=elimx0fg=logaelimx0loga(1+x)x=limx0logae=limx01logea=limx01lna\begin{array}{cc} \displaystyle \lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0}\log_a(1+x)^\frac{1}{x} \\\\ \text{令} \quad f(x)=\log_ax, \quad g(x)=(1+x)^\frac{1}{x} \\\\ \displaystyle \lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0} f \circ g= f(\lim\limits_{x \to 0}g(x))=f(\lim\limits_{x \to 0}(1+x)^\frac{1}{x})\\\\ \text{令} a = \frac{1}{x} \text{,那么} x \to 0 \iff a \to \infty \text{所以} \\\\ \lim\limits_{x \to 0}(1+x)^\frac{1}{x}=\lim\limits_{a \to \infty}(1+\frac{1}{a})^{a}=e \\\\ \lim\limits_{x \to 0} f \circ g = \log_ae \\\\ \displaystyle \lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0}\log_ae=\lim\limits_{x \to 0}\frac{1}{\log_ea}=\lim\limits_{x \to 0}\frac{1}{\ln\:a} \end{array}

将上面证明过程中的aa换成ee就有

limx0ln(1+x)x=1lne=1\lim\limits_{x \to 0}\frac{\ln(1+x)}{x}=\frac{1}{\ln\:e}=1


例2:求limx0ax1x\text{例2:求}\quad \lim\nolimits_{x \to 0} \frac{a^x-1}{x}

Letax1=t,sox=loga(1+t),Whenx0thent0,SothatLet \quad a^x-1=t, so \quad x=\log_a(1+t),When \quad x \to 0 \quad then \quad t \to 0,\quad So \quad that

limx0ax1x=limt0tloga(1+t)=limt0t1loga(1+t)=limt0tlog(1+t)a=limt0log(1+t)1ta=log(limx0(1+t)1t)a=logea=lna\begin{array}{ll} \displaystyle \lim\limits_{x \to 0} \frac{a^x-1}{x} &\displaystyle = \lim\limits_{t \to 0} \frac{t}{\log_a(1+t)} \\\\ &\displaystyle = \lim\limits_{t \to 0}\: t \cdot \frac{1}{\log_a(1+t)} = \lim\limits_{t \to 0}\: t \cdot \log_{(1+t)}a \\\\ &\displaystyle = \lim\limits_{t \to 0} \: \log_{(1+t)^{\frac{1}{t}}}a = \log_{(\lim\nolimits_{x \to 0}(1+t)^{\frac{1}{t}})}a\\\\ &\displaystyle = \log_e a = \ln a \end{array}

将上面证明过程中的$$a$$换成$$e$$就有

limx0ex1x=lne=1\lim\limits_{x \to 0} \frac{e^x-1}{x} = \ln e=1

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