<s>[INST] Evaluating $\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$ 
Evaluate:
  $$\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$$

I have been trying to solve this for $15$ minutes but sin(sin(x)) part has me stuck.
My attempt:
I tried multiplying with $x$ inside the $\sin$ as $\sin{(\frac{x\sin{x}}{x})}$. No leads.
 [/INST] Note that by Taylor's expansion


*

*$x\sin(\sin x)=x^2-\frac13x^4+\frac1{10}x^6+o(x^6)$

*$\sin^2x=x^2-\frac13x^4+\frac2{45}x^6+o(x^6)$


thus
$$\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}=\frac{x^2-\frac13x^4+\frac1{10}x^6-x^2+\frac13x^4-\frac2{45}x^6+o(x^6)}{x^6}=\frac1{18}+o(1)\to\frac1{18}$$
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