浅谈洛必达法则

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洛必达法则

当 \(\lim\limits_{x\to a}\operatorname{f}(x)=\lim\limits_{x\to a}\operatorname{g}(x)=0\ 或\ \infty\) 时，\(\lim\limits_{x\to a}\dfrac{\operatorname{f}(x)}{\operatorname{g}(x)}=\lim\limits_{x\to a}\dfrac{\operatorname{f}^\prime(x)}{\operatorname{g}^\prime(x)}\)

实例探究

一. \(\color{green}\dfrac{0}{0}型\)

1. \(\color{red}实例\)

(1) \(\color{blue}求\ \lim\limits_{x\to0}\dfrac{-1+\sqrt{1+4x}}{2x}\)

\(\color{purple}方法一:\)

\(\color{orange}令\operatorname{f}(x)=-1+\sqrt{1+4x},\operatorname{g}(x)=2x\)

\(\color{orange}则\lim\limits_{x\to0}\dfrac{-1+\sqrt{1+4x}}{2x}=\lim\limits_{x\to0}\dfrac{\operatorname{f}(x)}{\operatorname{g}(x)}=\lim\limits_{x\to0}\dfrac{\operatorname{f}^\prime(x)}{\operatorname{g}^\prime(x)}=\lim\limits_{x\to0}\dfrac{\dfrac{1}{2\sqrt{1+4x}}\times4}{2}=1\)

\(\color{purple}方法二:\)

\(\color{orange}\dfrac{-1+\sqrt{1+4x}}{2x}\ 为关于\ t\ 的方程\ xt^2+t-1=0\ 一根\)

\(\color{orange}当\ x\to0\ 时为关于\ t\ 的方程\ t-1=0\ 根，为\ 1\)

二. \(\color{green}\dfrac{\infty}{\infty}型\)

1. \(\color{red}实例\)

(1) \(\color{blue}求\lim\limits_{x\to\infty}\dfrac{3x+1}{4x+2}\)

\(\color{orange}令\ \operatorname{f}(x)=3x+1,\operatorname{g}(x)=4x+2\)

\(\color{orange}则\ \lim\limits_{x\to \infty}\dfrac{3x+1}{4x+2}=\lim\limits_{x\to \infty}\dfrac{\operatorname{f}(x)}{\operatorname{g}(x)}=\lim\limits_{x\to \infty}\dfrac{\operatorname{f}^\prime(x)}{\operatorname{g}^\prime(x)}=\dfrac{3}{4}\)

三. \(\color{green}0\times\infty\ 型\)

1. \(\color{red}探究\left(只是不严谨的示意\right)\)

(1) \(\color{blue}0\times\infty=0\times\dfrac{1}{0}=\dfrac{0}{0}\)

(2) \(\color{blue}0\times\infty=\dfrac{1}{\infty}\times\infty=\dfrac{\infty}{\infty}\)

2. \(\color{red}实例\)

(1) \(\color{blue}求\ \lim\limits_{x\to0^+}x\ln x\)

\(\color{purple}正确示例:\)

\(\color{orange}令\ \operatorname{f}(x)=\ln x,\operatorname{g}(x)=\dfrac{1}{x}\)

\(\color{orange}则\ \lim\limits_{x\to0^+}x\ln x=\lim\limits_{x\to0^+}\dfrac{\ln x}{\dfrac{1}{x}}=\lim\limits_{x\to0^+}\dfrac{f(x)}{g(x)}=\lim\limits_{x\to0^+}\dfrac{f^\prime(x)}{g^\prime(x)}=\lim\limits_{x\to0^+}\dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^2}}=\lim\limits_{x\to0^+}-x=0\)

\(\color{purple}错误示例:\)

\(\color{orange}令\ \operatorname{f}(x)=x,\operatorname{g}(x)=\dfrac{1}{\ln x},\operatorname{h}(x)=1,\operatorname{i}(x)=\ln x\)

\(\color{orange}则\ \operatorname{g}^\prime(x)=\left(\dfrac{\operatorname{h}(x)}{\operatorname{i}(x)}\right)^\prime=\dfrac{\operatorname{i}(x)\operatorname{h}^\prime(x)-\operatorname{h}(x)\operatorname{i}^\prime(x)}{\operatorname{i}^2(x)}=\dfrac{\ln x\times0-1\times\dfrac{1}{x}}{\ln^2x}=-\dfrac{1}{x\ln^2x}\)

\(\color{orange}从而\ \lim\limits_{x\to0^+}x\ln x=\lim\limits_{x\to0^+}\dfrac{x}{\dfrac{1}{\ln x}}=\lim\limits_{x\to0^+}\dfrac{\operatorname{f}(x)}{\operatorname{g}(x)}=\lim\limits_{x\to0^+}\dfrac{\operatorname{f}^\prime(x)}{\operatorname{g}^\prime(x)}=\lim\limits_{x\to0^+}\dfrac{1}{-\dfrac{1}{x\ln^2x}}=\lim\limits_{x\to0^+}-x\ln^2x\)

我们发现自己将式子复杂化了，应该停止计算去想一想，虽然这种方法不一定存在本质的错误，但很有可能是无法解决这个问题的，应该换一种方法

四. \(\color{green}\infty-\infty\ 型\)

1. \(\color{red}探究\left(只是不严谨的示意\right)\)

(1) \(\color{blue}\infty-\infty=\dfrac{1}{0}-\dfrac{1}{0}=\dfrac{0-0}{0\times0}=\dfrac{0}{0}\)

(2) \(\color{blue}\infty-\infty=\dfrac{有限}{0}-\dfrac{有限}{0}=\dfrac{有限-有限}{0}=\dfrac{0}{0}\left(仅个别情况可用\right)\)

2. \(\color{red}实例\)

(1) \(\color{blue}求\ \lim\limits_{x\to\frac{\pi}{2}}\left(\sec x-\tan x\right)\)

\(\color{orange}原式=\lim\limits_{x\to\frac{\pi}{2}}\left(\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}\right)=\lim\limits_{x\to\frac{\pi}{2}}\dfrac{1-\sin x}{\cos x}\)

\(\color{orange}令\ \operatorname{f}(x)=1-\sin x,\operatorname{g}(x)=\cos x\)

\(\color{orange}\operatorname{f}^\prime(x)=-\cos x,\operatorname{g}^\prime(x)=-\sin x\)

\(\color{orange}原式=\lim\limits_{x\to\frac{\pi}{2}}\dfrac{\operatorname{f}(x)}{\operatorname{g}(x)}=\lim\limits_{x\to\frac{\pi}{2}}\dfrac{\operatorname{f}^\prime(x)}{\operatorname{g}^\prime(x)}=\lim\limits_{x\to\frac{\pi}{2}}\tan x=0\)

五. \(\color{green}1^\infty\ 型\)

1. \(\color{red}探究\left(只是不严谨的示意\right)\)

\(\color{blue}1^\infty=e^{\ln 1^\infty}=e^{\infty\ln1}=e^{\infty\times0}\)

2. \(\color{red}实例\)

(1) \(\color{blue}求\ \lim\limits_{x\to0}\left(\dfrac{a^x+b^x+c^x}{3}\right)^{\dfrac{1}{x}}\left(a,b,c>0\right)\)

\(\color{orange}\lim\limits_{x\to0}\left(\dfrac{a^x+b^x+c^x}{3}\right)^{\dfrac{1}{x}}=\lim\limits_{x\to0}e^{\ln\left(\dfrac{a^x+b^x+c^x}{3}\right)^{\dfrac{1}{x}}}=\lim\limits_{x\to0}e^{\dfrac{1}{x}\ln\left(\dfrac{a^x+b^x+c^x}{3}\right)}=\lim\limits_{x\to0}e^{\dfrac{\ln\left(\dfrac{a^x+b^x+c^x}{3}\right)}{x}}=e^{\lim_{\Large x\to0}\dfrac{\ln\dfrac{a^x+b^x+c^x}{3}}{x}}\)

\(\color{orange}令\ \operatorname{f}(x)=\ln\dfrac{a^x+b^x+c^x}{3},\operatorname{g}(x)=x\)

\(\color{orange}\operatorname{f}^\prime(x)=\dfrac{3}{a^x+b^x+c^x}\times\dfrac{a^x\ln a+b^x\ln b+c^x\ln c}{3}=\dfrac{a^x\ln a+b^x\ln b+c^x\ln c}{a^x+b^x+c^x},\operatorname{g}(x)=1\)

\(\color{orange}所以\ 原式=e^{\lim_{\Large x\to0}\dfrac{a^x\ln a+b^x\ln b+c^x\ln c}{a^x+b^x+c^x}}=e^{\dfrac{\ln a+\ln b+\ln c}{3}}=e^{\ln\sqrt[3]{abc}}=\sqrt[3]{abc}\)

\(\left(此结论推广:\lim\limits_{x\to0}\left(\dfrac{\sum\limits_{k=1}^{n}a_k^x}{n}\right)^{\dfrac{1}{x}}=\sqrt[n]{\prod\limits_{k=1}^{n}a_k}\right)\)

未完待续