Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc

One way without MacLaurin or L'Hospital: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} = \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 - \cos(\sin x)]}{\cos(\sin x)(\tan x -\sin x)}=$$ $$=1+\lim_{x\to0} \frac{\sin(\sin x)}{\sin x}\lim_{x\to0} \frac{1 - \cos(\sin x)}{\sin^2 x}\lim_{x\to0}\frac{x^2}{1-\cos x}\lim_{x\to0}\frac{\sin^2x}{x^2} = 1+1.\frac{1}{2}.2.1 =2  $$.
We used $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}$$ $$\lim_{x\to0}\frac{\sin x}{x}= \lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\cos x=1, \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2} $$
