Evaluate $\lim\limits_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$

The key here is obtaining the expansions for $x^{\sin x} $ and $\sin^xx$ but it is simpler to deal with their logarithms. Consider \begin{align} f(x) &=\sin x \log x-x\log\sin x\notag\\ &=(x\log x)\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots \right)-x\log x-x\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots\right)\notag\\ &=\notag\\ &=-\frac{x^3\log x} {6}+\frac{x^3}{6}+o(x^3)\notag \end{align} And then the expression under limit is $$\sin^xx\cdot\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}$$ which can be further written as $$\sin^xx \left(\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}\right) +\frac{\log x} {6}\cdot(1-\sin^xx) $$ From the expansion of $f(x) $ it is clear that $f(x) =o(x^2)$ and hence $$\frac{e^{f(x)} - 1}{x^3}=-\frac{\log x} {6}+\frac{1}{6}+o(1)$$ The desired limit is $1/6$ if we can prove that $\sin^xx \to 1$ and $(\log x) (1-\sin^xx) \to 0$ and this is not difficult to prove.

We just need to note that $$(\log x) \cdot(1-\sin^xx)=-\frac{\exp(x\log\sin x) - 1}{x\log\sin x} \cdot x\log \sin x\cdot\log x$$ and the fraction above tends to $1$ so that the limit of above expression is equal to the limit of $$-x\log x\left(\log x+\log\frac{\sin x} {x} \right) $$ which is same as $$-x(\log x) ^2-(x\log x) \log\frac{\sin x} {x} $$ and the above clearly tends to $0$.