How to calculate $\lim_{x\to 0^+} \frac{x^x- (\sin x)^x}{x^3}$ As I asked, I don't know how to deal with $x^x- (\sin x)^x$.
Please give me a hint. Thanks!
 A: The given expression can be written as $$(\sin x) ^{x} \cdot\frac{\exp(x \log x- x\log \sin x) - 1}{x(\log x - \log \sin x)} \cdot\frac{\log x - \log \sin x} {x^{2}}$$ and first two factors tend to $1$. The last factor can be written as $$\frac{\log(1+(x/\sin x) - 1)}{(x/\sin x) - 1}\cdot\frac{x-\sin x} {x^{3}}\cdot\frac{x}{\sin x} $$ and first and last factors are tending to $1$ and the middle one famously tends to $1/6$ (via L'Hospital's Rule or Taylor's theorem). Thus the final answer to your question is $1/6$.
A: Note that
$$\frac{x^x - (\sin x)^x}{x^3} = x^x\frac{1 - \left(\frac{\sin x}{x}\right)^x}{x^3}$$
We can find the Taylor expansion 
$$\left(\frac{ \sin x}{x} \right)^x = \exp\left(x \log \left(\frac{\sin x}{x} \right) \right) = 1 - \frac{x^3}{6} + O(x^5),$$
using
$$\frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4)\\ \log (1 - y) = -y + O(y^2)\\ \exp(z) = 1 + z + O(z^2)$$
Thus,
$$\frac{1 - \left(\frac{\sin x}{x}\right)^x}{x^3} = \frac{1}{6} + O(x^2)$$
Since $\lim_{x \to 0+} x^x = \lim_{x \to 0+} \exp(x \log x) = 1$ we find the desired  limit is $1/6$.
A: $$\lim_{x\rightarrow0^+}\frac{\ln{x}-\ln{\sin{x}}}{x^2}=\lim_{x\rightarrow0^+}\left(\frac{\ln\left(1+\frac{x}{\sin{x}}-1\right)}{\frac{x}{\sin{x}}-1}\cdot\frac{\frac{x}{\sin{x}}-1}{x^2}\right)=$$
$$=\lim_{x\rightarrow0}\left(\frac{x-\sin{x}}{x^3}\cdot\frac{x}{\sin{x}}\right)=\lim_{x\rightarrow0}\frac{1-\cos{x}}{3x^2}=\frac{1}{6}\lim_{x\rightarrow0}\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}=\frac{1}{6}$$
and since
$$\frac{x^x-(\sin{x})^x}{x^3}=\frac{1+x\ln{x}+\frac{x^2\ln^2x}{2!}+...-1-x\ln\sin{x}-\frac{x^2\ln^2\sin{x}}{2!}-...}{x^3}=$$
$$=\frac{\ln{x}-\ln{\sin{x}}}{x^2}+O(x^2),$$
we obtain that our limit is $\frac{1}{6}$.
