<s>[INST] $\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$ Can one help finding this limit
$$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$$
L'Hospital's rule is permited.
(Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$)
 [/INST] Hint. Recall that, for $x$ near $0$, you have
$$\tan x = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\mathcal{O}(x^9)$$
$$ \sin x = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{17x^7}{720}+\mathcal{O}(x^9)$$ to obtain (using only relevant terms) $$\tan(\sin x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+\mathcal{O}(x^9)$$ $$\sin (\tan x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+\mathcal{O}(x^9)$$
$$x^{3}-\sin^{2}x\tan x = -\frac{x^7}{15}+\mathcal{O}(x^9)$$
Then the desired limit is equal to $-2$.
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