一道三角函数极限计算的一题多解

求极限$\lim_{x\rightarrow 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$

法一：

$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\ \lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} =\ \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 - \cos(\sin x)]}{\cos(\sin x)(\tan x -\sin x)}=$ $=1+\lim_{x\to0} \frac{\sin(\sin x)}{\sin x}\lim_{x\to0} \frac{1 - \cos(\sin x)}{\sin^2 x}\lim_{x\to0}\frac{x^2}{1-\cos x}\lim_{x\to0}\frac{\sin^2x}{x^2} = 1+1*\frac{1}{2}21 =2$

法二：

$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) $ $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) $ $\tan(\tan(x))=x+\frac{2 x^3}{3}+\frac{3 x^5}{5}+O\left(x^7\right) $ $\sin(\sin(x))=x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^7\right) $

故$\frac{tan(tan(x)) - sin (sin( x)}{ tan (x) - sin (x)}$

$ =\frac {x^3+\frac{x^5}{2}+O\left(x^7\right) } {\frac{x^3}{2}+\frac{x^5}{8}+O\left(x^7\right) } \= 2+\frac{x^2}{2}+O\left(x^4\right) =2$

法三：

$\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$ $= \lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)\cos(\sin x)}{\tan x - \sin x}$ $=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)+\tan(\sin x)-\tan(\sin x)\cos(\sin x)}{\tan x -\sin x}$ $=\lim_{x\to 0}\left(\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}+\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\right)$

根据拉格朗日中值定理得：

$\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\sec^2 c $ , $c\in[sinx,tanx]$

故$\lim_{x\to 0} \frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x} =1$

而$\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}$ $=\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))\cos x}{\sin x(1 - \cos x)}$ $=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\cdot \frac{1-\cos(\sin x)}{\sin^2 x}\cdot \frac{\sin^2 x}{1-\cos x}\cdot \cos x$ $=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\lim_{x\to 0}\frac{1-\cos(\sin x)}{\sin^2 x}\lim_{x\to 0}(1+\cos x)\lim_{x\to 0}\cos x$ $=1\cdot \frac{1}{2}\cdot 2\cdot 1$ $=1.$

故原极限为2.

法四：

令$A=\frac{\tan (\tan x)-x}{\tan x-\sin x}$，$B=\frac{x-\sin (\sin x)}{\tan x-\sin x}$

通过洛必达可得：

$A=\frac{tan (tan x)-x}{\tan x-\sin x}$

$\sim \frac{\frac{1}{cos ^2x·cos ^2(tan x)}-1}{\frac{1-cos ^3x}{cos ^2x}}$

$\sim \frac{1-cos ^2x·cos ^2(\tan x)}{3(1-\cos x)}$

$\sim \frac{(1-cos ^2x)-cos ^2x(1-cos ^2(\tan x))}{3(1-\cos x)}$

故:

$A\sim \frac{1+\cos x}{3}-\frac{1-\cos (\tan x)}{3(1-\cos x)}\sim \frac{2}{3}+\frac{\frac{1}{cos ^2x}\sin (\tan x)}{3\sin x}\sim \frac{2}{3}+\frac{1}{3}=1$

类似地$B=1$

故原式$=A+B=2$