题目

求lim⁡x→01−cos⁡x(cos⁡2x)12(cos⁡3x)13x2求 \lim_{x\rightarrow 0}\frac{1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3}}{x^2}

解

引论一

若f,a,b,c为自由变量f=abc，有f′=(a′a+b′b+c′c)(abc)若f,a,b,c为自由变量f=abc，有\\ f'=(\frac{a'}{a}+\frac{b'}{b}+\frac{c'}{c})(abc)

引论二

d(cos⁡tx)1t(cos⁡tx)1tdx=−tan⁡tx\frac{d(\cos tx)^{\frac{1}{t}}}{(\cos tx)^{\frac{1}{t}}dx}=-\tan tx

解

lim⁡x→01−cos⁡x(cos⁡2x)12(cos⁡3x)13x2=lim⁡x→0d(1−cos⁡x(cos⁡2x)12(cos⁡3x)13)2xdx洛必达=lim⁡x→0−(dcos⁡xcos⁡xdx+d(cos⁡2x)12(cos⁡2x)12dx+d(cos⁡3x)13(cos⁡3x)13dx)(cos⁡x(cos⁡2x)12(cos⁡3x)13)2x引论一=lim⁡x→0(tan⁡x+tan⁡2x+tan⁡3x)(cos⁡x(cos⁡2x)12(cos⁡3x)13)2x引论二=lim⁡x→0x+o(x)+2x+o(x)+3x+o(x)2x泰勒展开=3\begin{aligned} &\lim_{x\rightarrow 0}\frac{1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3}}{x^2}\\ =&\lim_{x\rightarrow 0} \frac{d(1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2xdx}\qquad 洛必达 \\ =&\lim_{x\rightarrow 0} \frac{-(\frac{d\cos x}{\cos xdx}+\frac{d(\cos 2x)^\frac{1}{2}}{(\cos 2x)^\frac{1}{2}dx}+\frac{d(\cos 3x)^\frac{1}{3}}{(\cos 3x)^\frac{1}{3}dx})(\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2x} \qquad 引论一\\ =&\lim_{x\rightarrow 0} \frac{(\tan x+\tan 2x+ \tan 3x)(\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2x}\qquad 引论二\\ =&\lim_{x\rightarrow 0} \frac{x+o(x)+2x+o(x)+3x+o(x)}{2x}\qquad 泰勒展开\\ =&3 \end{aligned}