一道极限题的计算

题目

$$求 \lim_{x\rightarrow 0}\frac{1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3}}{x^2}$$

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

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

解

$$\begin{aligned} &\lim_{x\rightarrow 0}\frac{1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3}}{x^2}\\ =&\lim_{x\rightarrow 0} \frac{d(1-\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2xdx}\qquad 洛必达 \\ =&\lim_{x\rightarrow 0} \frac{-(\frac{d\cos x}{\cos xdx}+\frac{d(\cos 2x)^\frac{1}{2}}{(\cos 2x)^\frac{1}{2}dx}+\frac{d(\cos 3x)^\frac{1}{3}}{(\cos 3x)^\frac{1}{3}dx})(\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2x} \qquad 引论一\\ =&\lim_{x\rightarrow 0} \frac{(\tan x+\tan 2x+ \tan 3x)(\cos x (\cos 2x)^\frac{1}{2}(\cos 3x)^\frac{1}{3})}{2x}\qquad 引论二\\ =&\lim_{x\rightarrow 0} \frac{x+o(x)+2x+o(x)+3x+o(x)}{2x}\qquad 泰勒展开\\ =&3 \end{aligned}$$