<s>[INST] Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$ 
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$

In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it is easy. However how to prove that $r(\sin x -x)=o(x^5)$? I tried to do it but then I have: $$\frac{r(\sin x -x)}{\sin x -x}\cdot \frac{\sin x -x}{x^5}=\frac{r(\sin x -x)}{\sin x -x}\cdot(\frac{\sin x}{x}-1)\cdot \frac{1}{x^4} \rightarrow 0\cdot(1-1)\cdot(-\infty)$$Can you help me how can I prove it?
 [/INST] I would be stupid about it and work formally, in the ring $\Bbb Q[[x]]/(x^6)$.
Now, in that ring, $\sin(x)=x-\frac{x^3}6+\frac{x^5}{120}$. Let me write $g(x)=\sin(x)-x$, which is divisible by $x^3$. Then $\sin(g(x))=g(x)+g(g(x))$, but since $g(g(x))$ is divisible by $x^9$, we can ignore it. Thus in $\Bbb Q[[x]]/(x^6)$, we get $\sin(\sin(x)-x)=-\frac{x^3}6+\frac{x^5}{120}$, just the same as $\sin(x)-x$.
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