$$\lim\_{x\to 0}\frac{\textstyle x^{\textstyle(\sin x)^{\textstyle x}}-(\textstyle \sin x)^{\textstyle x^{\textstyle \sin x}}}{\textstyle x^3}=\frac{1}{6}$$

The limit is easy to get results, but how to rigorously prove it without using Taylor formula?

At first, I guess the numerator is equivalent to $x-\sin x$

And I also find the following several limits have the same results:

$$\lim\_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle x}}-(\textstyle \sin x)^{\textstyle (\sin x)^{\textstyle \sin x}}}{\textstyle x^3}=\frac{1}{6}$$

$$\lim\_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle \tan x}}-(\textstyle \sin x)^{\textstyle x^{\textstyle \tan x}}}{\textstyle x^3}=\frac{1}{6}$$

I guess if $f(x)\sim g(x)\sim h(x)\sim O(x)^k$ when $x\to 0$ , then $$\lim\_{x\to 0}\textstyle f(x)^{\textstyle g(x)^{\textstyle h(x)}}=f(x)$$