<s>[INST] Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan x - \sin x} = 1$$
But the correct answer is $2$. Where am I wrong$?$
 [/INST] Your way is wrong since
$${\tan(\tan x) - \sin (\sin x)}\neq { \tan x - \sin x}$$
indeed this step
$$\frac{\tan x \tan (\tan x)}{\tan x}=\tan x\cdot \frac{ \tan (\tan x)}{\tan x}\color{red}{=\tan x \cdot 1}=\tan x$$
and the similar for $\sin x$ are not allowed (in general we can't evaluate limit only for a part or factor of the whole expression; see also here). 
We can avoid Tayor's expansion and use that (see here):


*

*$\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13$

*$\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac{1}6$
then
$$\dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}=\dfrac{\frac{\tan^3 x}{x^3}\frac{\tan(\tan x)-\tan x}{\tan^3x} - \frac{\sin^3 x}{x^3}\frac{\sin(\sin x)-\sin x}{\sin^3x}+\frac{\tan x-x}{x^3}-\frac{\sin x-x}{x^3}}{ \frac{\tan x-x}{x^3} - \frac{\sin x-x}{x^3}}\\\to\frac{1\cdot \frac13-1\cdot\left(-\frac16\right)+\frac13-\left(-\frac16\right)}{ \frac13-\left(-\frac16\right)}=2$$
</s>