Expanding $x^x$ to series with $o(x)$ polynomials I have some doubts in finding $x^x$ series.  I know that from Taylor theorem I have
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$
I want to have $o(x)$ polynomials so:
Let $$ f(x) = x^x $$
$$ f'(x) = (e^{x\cdot \ln x}) = (\ln x + 1)x^x$$
In use of taylor I should
$$ f(x) = f(0) + \frac{f'(0)(x)}{1!} + o(x) $$
but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined...
But I found informations that it should be 
$$  1+ \ln x \cdot x + o(x)$$
I don't know why :-(

Although $f(0)$ and $f'(0)$ are not defined, one can write$$\lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} x^x=e^{\lim_{x\to 0^+} x\ln x}=e^0=1$$In that case $${f(x)=x^x\sim 1\\f'(x)=x^x(\ln x+1)\sim \ln x+1}$$both when $x\to0^+.$ Here is a plot of the function:
