How to expand $x^n$ as $n \to 0$?

Question

I am trying to expand $x^n$ in small $n$ using Taylor series.

Using wolfram alpha, I found that it is $1+ n\log(x) + \cdots$

I tried to Taylor expand $x^n$ around $n=0$ but I cannot get this result.

Answer 1

Hint. One may recall that $$ e^z=1+z+\cdots+\frac{z^k}{k!}+\cdots,\quad z \in \mathbb{C}, $$ then, for $x>0$ and for $0<n<1$, $$ \begin{align} x^n&=e^{n\ln x}=1+n\ln x+\frac{n^2(\ln x)^2}{2!}+\cdots+\frac{n^k(\ln x)^k}{k!}+\cdots, \\\\x^n&=1+n\ln x+\cdots+\mathcal{O}\left(n^k\right), \end{align} $$ as announced.