洛必达法则讲解和例题

洛必达法则

若满足00,∞∞\dfrac 00,\dfrac \infty\infty00,∞∞型，则lim⁡f(x)g(x)=lim⁡f′(x)g′(x)\lim\dfrac{f(x)}{g(x)}=\lim \dfrac{f'(x)}{g'(x)}limg(x)f(x)=limg′(x)f′(x)

00,∞∞\dfrac 00,\dfrac \infty\infty00,∞∞可直接使用洛必达，∞−∞,0⋅∞,1∞,∞0,00\infty-\infty,0\cdot\infty,1^\infty,\infty^0,0^0∞−∞,0⋅∞,1∞,∞0,00则需转化成00,∞∞\dfrac 00,\dfrac \infty\infty00,∞∞才能使用
若lim⁡f′(x)g′(x)\lim \dfrac{f'(x)}{g'(x)}limg′(x)f′(x)仍满足00,∞∞\dfrac 00,\dfrac \infty\infty00,∞∞型，则可继续用lim⁡f′(x)g′(x)=lim⁡f′′(x)g′′(x)\lim\dfrac{f'(x)}{g'(x)}=\lim \dfrac{f''(x)}{g''(x)}limg′(x)f′(x)=limg′′(x)f′′(x)
洛必达不是万能的，求极限首选无穷小替换，再用洛必达。

00\dfrac 0000型未定式

求lim⁡x→0x−sin⁡xx3\lim\limits_{x\rightarrow0}\dfrac{x-\sin x}{x^3}x→0limx3x−sinx

解：原式=lim⁡x→01−cos⁡x3x2=lim⁡x→0sin⁡x6x=16lim⁡x→0sin⁡xx=16=\lim\limits_{x\rightarrow0}\dfrac{1-\cos x}{3x^2}=\lim\limits_{x\rightarrow0}\dfrac{\sin x}{6x}=\dfrac16 \lim\limits_{x\rightarrow0}\dfrac{\sin x}{x}=\dfrac 16=x→0lim3x21−cosx=x→0lim6xsinx=61x→0limxsinx=61

如果一开始将sin⁡x\sin xsinx替换为xxx，则原式为000，这样是错误的。无穷小替换一般在乘除时可用，但在加减时要慎用。

∞∞\dfrac{\infty}{\infty}∞∞型未定式

题1： 求lim⁡x→0+ln⁡sin⁡3xln⁡sin⁡2x\lim\limits_{x\rightarrow0^+}\dfrac{\ln\sin 3x}{\ln\sin 2x}x→0+limlnsin2xlnsin3x
解：原式=lim⁡x→0+1sin⁡3x⋅cos⁡3x⋅31sin⁡2x⋅cos⁡2x⋅2=lim⁡x→0+3cos⁡3x2cos⁡2x⋅lim⁡x→0+sin⁡2xsin⁡3x=32⋅lim⁡x→0+2x3x=32×23=1=\lim\limits_{x\rightarrow 0^+}\dfrac{\frac{1}{\sin 3x}\cdot \cos 3x \cdot 3}{\frac{1}{\sin 2x}\cdot \cos 2x \cdot 2}=\lim\limits_{x\rightarrow 0^+}\dfrac{3\cos 3x}{2\cos 2x}\cdot \lim\limits_{x\rightarrow 0^+}\dfrac{\sin 2x}{\sin 3x}=\dfrac 32\cdot \lim\limits_{x\rightarrow 0^+}\dfrac{2x}{3x}=\dfrac 32\times\dfrac 23=1=x→0+limsin2x1⋅cos2x⋅2sin3x1⋅cos3x⋅3=x→0+lim2cos2x3cos3x⋅x→0+limsin3xsin2x=23⋅x→0+lim3x2x=23×32=1

x→0x\rightarrow0x→0，aaa为常数，则cos⁡ax\cos axcosax趋近于111

题2： 求lim⁡x→+∞xex2x+ex\lim\limits_{x\rightarrow+\infty}\dfrac{xe^{\frac x2}}{x+e^x}x→+∞limx+exxe2x

解：原式=lim⁡x→+∞ex2+x2ex21+ex=lim⁡x→+∞2ex2+xex22+2ex=lim⁡x→+∞2ex2+x2ex22ex=lim⁡x→+∞4+x4ex2=lim⁡x→+∞12ex2=0=\lim\limits_{x\rightarrow+\infty}\dfrac{e^{\frac x2}+\frac x2e^{\frac x2}}{1+e^x}=\lim\limits_{x\rightarrow+\infty}\dfrac{2e^{\frac x2}+xe^{\frac x2}}{2+2e^x}=\lim\limits_{x\rightarrow+\infty}\dfrac{2e^{\frac x2}+\frac x2e^{\frac x2}}{2e^x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4+x}{4e^{\frac x2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{2e^{\frac x2}}=0=x→+∞lim1+exe2x+2xe2x=x→+∞lim2+2ex2e2x+xe2x=x→+∞lim2ex2e2x+2xe2x=x→+∞lim4e2x4+x=x→+∞lim2e2x1=0

由于(ex)′=ex(e^x)'=e^x(ex)′=ex，所以我们可以用洛必达法则将不含eee的部分通过求导去掉，留下含有eee的部分再求值。

∞−∞\infty-\infty∞−∞型未定式

求lim⁡x→0(1x2−1xsin⁡x)\lim\limits_{x\rightarrow0}(\dfrac{1}{x^2}-\dfrac{1}{x\sin x})x→0lim(x21−xsinx1)

解：原式=lim⁡x→0sin⁡x−xx2sin⁡x=lim⁡x→0sin⁡x−xx3=−lim⁡x→01−cos⁡x3x2=−lim⁡x→0sin⁡x6x=−16lim⁡x→0sin⁡xx=−16=\lim\limits_{x\rightarrow0}\dfrac{\sin x-x}{x^2\sin x}=\lim\limits_{x\rightarrow0}\dfrac{\sin x-x}{x^3}=-\lim\limits_{x\rightarrow0}\dfrac{1-\cos x}{3x^2}=-\lim\limits_{x\rightarrow0}\dfrac{\sin x}{6x}=-\dfrac16 \lim\limits_{x\rightarrow0}\dfrac{\sin x}{x}=-\dfrac 16=x→0limx2sinxsinx−x=x→0limx3sinx−x=−x→0lim3x21−cosx=−x→0lim6xsinx=−61x→0limxsinx=−61

此处将∞−∞\infty-\infty∞−∞型转换为00\dfrac 0000型

0⋅∞0\cdot\infty0⋅∞型未定式

求lim⁡x→0+x2ln⁡x\lim\limits_{x\rightarrow0^+}x^2\ln xx→0+limx2lnx

解：原式=lim⁡x→0+ln⁡x1x2=lim⁡x→0+1x−2x3=lim⁡x→0+−x32x=lim⁡x→0+−x22=0=\lim\limits_{x\rightarrow0^+}\dfrac{\ln x}{\frac{1}{x^2}}=\lim\limits_{x\rightarrow0^+}\dfrac{\frac 1x}{-\frac{2}{x^3}}=\lim\limits_{x\rightarrow0^+}-\dfrac{x^3}{2x}=\lim\limits_{x\rightarrow0^+}-\dfrac{x^2}{2}=0=x→0+limx21lnx=x→0+lim−x32x1=x→0+lim−2xx3=x→0+lim−2x2=0

此处将0⋅∞0\cdot\infty0⋅∞型转换为∞∞\dfrac{\infty}{\infty}∞∞型

1∞1^\infty1∞型未定式

求lim⁡x→1(2−x)1ln⁡x\lim\limits_{x\rightarrow1}(2-x)^{\frac{1}{\ln x}}x→1lim(2−x)lnx1

解：原式=elim⁡x→1ln⁡(2−x)1ln⁡x=elim⁡x→1ln⁡(2−x)ln⁡x=elim⁡x→1−12−x1x=elim⁡x→1−x2−x=e−1=e^{\lim\limits_{x\rightarrow1}\ln(2-x)^\frac{1}{\ln x}}=e^{\lim\limits_{x\rightarrow1}\frac{\ln(2-x)}{\ln x}}=e^{\lim\limits_{x\rightarrow1}\frac{-\frac{1}{2-x}}{\frac1x}}=e^{\lim\limits_{x\rightarrow1}-\frac{x}{2-x}}=e^{-1}=ex→1limln(2−x)lnx1=ex→1limlnxln(2−x)=ex→1limx1−2−x1=ex→1lim−2−xx=e−1

此处将1∞1^\infty1∞型转换为00\dfrac 0000型

当x→1x\rightarrow1x→1时，−x2−x-\dfrac{x}{2-x}−2−xx不满足洛必达法则的要求，不能用洛必达法则，直接等于−1-1−1

000^000未定式

求lim⁡x→0+xx\lim\limits_{x\rightarrow0^+}x^xx→0+limxx

解：原式=elim⁡x→0+ln⁡xx=elim⁡x→0+xln⁡x=elim⁡x→0+ln⁡x1x=elim⁡x→0+1x−1x2=elim⁡x→0+−x=e0=1=e^{\lim\limits_{x\rightarrow0^+}\ln x^x}=e^{\lim\limits_{x\rightarrow0^+}x\ln x}=e^{\lim\limits_{x\rightarrow0^+}\frac{\ln x}{\frac 1x}}=e^{\lim\limits_{x\rightarrow0^+}\frac{\frac 1x}{-\frac{1}{x^2}}}=e^{\lim\limits_{x\rightarrow0^+}-x}=e^0=1=ex→0+limlnxx=ex→0+limxlnx=ex→0+limx1lnx=ex→0+lim−x21x1=ex→0+lim−x=e0=1

此处将000^000型转换为∞∞\dfrac{\infty}{\infty}∞∞型

∞0\infty^0∞0型未定式

求lim⁡x→0(cot⁡x)x\lim\limits_{x\rightarrow0}(\cot x)^xx→0lim(cotx)x

解：原式=elim⁡x→0ln⁡(cot⁡x)x=elim⁡x→0xln⁡cot⁡x=elim⁡x→0ln⁡cot⁡x1x=elim⁡x→0−1cot⁡x1sin⁡2x−1x2=elim⁡x→0x2cot⁡xsin⁡2x=elim⁡x→01cot⁡x=elim⁡x→0tan⁡x=e0=1=e^{\lim\limits_{x\rightarrow0}\ln(\cot x)^x}=e^{\lim\limits_{x\rightarrow0}x\ln\cot x}=e^{\lim\limits_{x\rightarrow0}\frac{\ln\cot x}{\frac 1x}}=e^{\lim\limits_{x\rightarrow0}\frac{-\frac{1}{\cot x}\frac{1}{\sin^2 x}}{-\frac{1}{x^2}}}=e^{\lim\limits_{x\rightarrow0}\frac{x^2}{\cot x\sin^2 x}}=e^{\lim\limits_{x\rightarrow0}\frac{1}{\cot x}}=e^{\lim\limits_{x\rightarrow0}\tan x}=e^0=1=ex→0limln(cotx)x=ex→0limxlncotx=ex→0limx1lncotx=ex→0lim−x21−cotx1sin2x1=ex→0limcotxsin2xx2=ex→0limcotx1=ex→0limtanx=e0=1

此处将∞0\infty^0∞0型转换为∞∞\dfrac{\infty}{\infty}∞∞型