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$?$

@Surb identified your error with the choice$$f_1=\tan(\tan x),\,g_1=\tan x,\,f_2=\sin(\sin x),\,g_2=\sin x.$$One method that would work is to use$$\tan x=x+\frac13 x^3+o(x^3),\,\sin x=x-\frac16 x^3+o(x^3)$$together with$$x+cx^3+c(x+cx^3)^3=x+2cx^3+o(x^3),$$viz.$$\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\tan(x+\frac13 x^3+o(x^3))-\sin(x-\frac16 x^3+o(x^3))}{\frac12x^3+o(x^3)}\\=\frac{(x+\frac23 x^3)-(x-\frac13 x^3)+o(x^3)}{\frac12x^3+o(x^3)}=2+o(1).$$
