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!

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 $1/6$.
