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$?$

This is a nice case for composition of Taylor series. Using
$$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\tan(\tan(x))=x+\frac{2 x^3}{3}+\frac{3 x^5}{5}+O\left(x^7\right)$$
$$\sin(\sin(x))=x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^7\right)$$ then 
$$\frac{\tan(\tan( x)) - \sin (\sin( x)}{ \tan (x) - \sin (x)}=\frac {x^3+\frac{x^5}{2}+O\left(x^7\right) } {\frac{x^3}{2}+\frac{x^5}{8}+O\left(x^7\right) }=2+\frac{x^2}{2}+O\left(x^4\right)$$ which shows the limit and also how it is approached.
