Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$

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.

Your second step is invalid. In general one can't replace a sub-expression by its limit while evaluating limit of a bigger expression in step by step fashion. You can get more details in this answer.

You can approach this problem by adding and subtracting $\tan(\sin x) $ in numerator. The numerator can thus be expressed as $$\{\tan \tan x-\tan \sin x\} +\{\tan \sin x-\sin\sin x\} $$ which can be further rewritten as $$\tan(\tan x - \sin x) (1+\tan\tan x\cdot\tan\sin x)+\tan\sin x-\sin\sin x$$ Now the first term divided by $\tan x - \sin x$ (denominator) tends to $1$ and hence the desired limit is equal to $$1+\lim_{x\to 0}\frac{\tan\sin x - \sin\sin x} {\tan x - \sin x} $$ The expression under limit above can be written as $$\frac{\sin\sin x} {\sin x} \cdot \frac{1-\cos\sin x} {1-\cos x}\cdot\frac{\cos x} {\cos\sin x} $$ The first and last factors tend to $1$ and the middle factor is equal to $$\frac{1-\cos\sin x} {\sin^2x}\cdot\frac{\sin^2x}{x^2}\cdot\frac{x^2}{1-\cos x} $$ and the above clearly tends to $(1/2)\cdot 1\cdot 2=1$. The desired answer is thus $2$.

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

$$\dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}=\dfrac{\frac{\tan^3 x}{x^3}\frac{\tan(\tan x)-\tan x}{\tan^3x} - \frac{\sin^3 x}{x^3}\frac{\sin(\sin x)-\sin x}{\sin^3x}+\frac{\tan x-x}{x^3}-\frac{\sin x-x}{x^3}}{ \frac{\tan x-x}{x^3} - \frac{\sin x-x}{x^3}}\\\to\frac{1\cdot \frac13-1\cdot\left(-\frac16\right)+\frac13-\left(-\frac16\right)}{ \frac13-\left(-\frac16\right)}=2$$