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?

You can note that $$\sin x - x=-\frac{x^3}{6}+o(x^3)\tag{1}$$ and replacing $x$ with $\sin x - x$ we get $$\sin(\sin x - x) - (\sin x - x) =-\frac{(\sin x - x)^3}{6}+o((\sin x - x) ^3)$$ Using $(1)$ we can see that both the terms in right hand side of above equation are $o(x^5) $ and we are done. 
