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$?$

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 $2$.
