<s>[INST] Finding $\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}$ Problem
$$\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}.$$
Attempt
\begin{align*}
&\lim_{x \to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}\\
=&\lim_{x \to 0}\frac{\exp [(\sin x)^x\ln x]-\exp [{x^{\sin x}\ln(\sin x)]}}{x^3}
\\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]}(\exp [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]-1)}{x^3}
\\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]} [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]}{x^3}
\end{align*}
This will help?
 [/INST] My Solution
Recall the following formulas
$$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x-\cdots\tag 1$$
$$e^x=1+x+\frac{1}{2}x^2+\cdots\tag 2$$
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \tag 3$$
which are the starting point we need to rely on.
From$(1)$,
\begin{align}
\frac{\sin x-x}{x}=\dfrac{x-\dfrac{1}{3!}x^3+o(x^4)-x}{x}=-\frac{1}{6}x^2+o(x^3).\tag{4}
\end{align}
From $(3)(4)$,
\begin{align}
\ln\sin x&=\ln\left(x\cdot \frac{\sin x}{x}\right)=\ln x+\ln\left(1+\frac{\sin x-x}{x}\right)\notag\\
&=\ln x+\frac{\sin x-x}{x}-\frac{1}{2}\left(\frac{\sin x-x}{x}\right)^2+o(x^4)\notag\\
&=\ln x-\frac{1}{6}x^2+o(x^3).\tag{5}
\end{align}
From $(5)$,
$$x\ln\sin x=x(\ln x-\frac{1}{6}x^2+o(x^3))=x\ln x-\frac{1}{6}x^3+o(x^3).\tag{6}$$
From $(2)(6)$,
\begin{align*}
(\sin x)^x&=\exp(x\ln\sin x)=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\frac{1}{6}(x\ln\sin x)^3+o(x^3\ln^3 x)\notag\\
&=1+x\ln x+\frac{1}{2}x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{7}
\end{align*}
From $(7)$, we obtain a vital result that
\begin{align*}
{\color{red}{f(x):=(\sin x)^x\ln x}}=\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x+o(x^3\ln^4 x).\tag{8}
\end{align*}
Morover, by $(1)(2)$, we have
\begin{align*}
x^{\sin x}&=\exp(\sin x\ln x)=1+\sin x\ln x+\frac{1}{2}\sin^2 x\ln^2 x+\frac{1}{6}\sin^3 x\ln^3 x+o(x^3\ln^3 x)\notag\\
&=1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{9}
\end{align*}
From $(5)(9)$, we oabtain another vital result that
\begin{align*}
&{\color{red}{g(x):=x^{\sin x}\ln\sin x}}\\
=&(1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x))(\ln x-\frac{1}{6}x^2+o(x^3))\\
=&\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x-\frac{1}{6}x^2+o(x^3\ln^4 x).\tag{10}
\end{align*}
Now,notice that, by$(8)(10)$,
$$f(x)-g(x)=\frac{1}{6}x^2+o(x^3\ln^4 x)\to 0(x \to 0^+),$$
and, by $(10)$,
$$\frac{e^g(x)}{x}=e^{g(x)-\ln x}=e^{x\ln^2 x+o(x\ln^2 x)}=e^0=1.$$
It follows that
\begin{align*}
\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}&=\lim_{x \to 0+}\frac{e^{f(x)}-e^{g(x)}}{x^3}\\
&=\lim_{x \to 0+}\frac{e^{g(x)}(e^{f(x)-g(x)}-1)}{x^3}\\
&=\lim_{x \to 0+}\frac{e^{g(x)}}{x}\cdot\lim_{x \to 0+}\frac{e^{f(x)-g(x)}-1}{x^2}\\
&=1 \cdot\lim_{x \to 0+}\frac{f(x)-g(x)}{x^2}\\
&=\lim_{x \to 0+}\left(\frac{1}{6}+o(x\ln^4 x)\right)\\
&=\frac{1}{6}.
\end{align*}
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