I am new to Taylor expansions and I would like to calculate the Taylor polynomial of the function $x^{1/x}=e^{(1/x)\log x}$. Since the function is not defined at $x=0$, how should I choose the point which I will calculate the expansion at? I would like to understand the following result
$$x^{1/x}=1-\frac1x \log\frac1x+O(x^{-2})\qquad (1)$$
The first derivative of my function is $$f'(x)=x^{1/x-2}\cdot(1-\log x)$$ Given that the Taylor polynomial is $f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots$, I can't get how this first derivative is used in $(1)$. For which $a$ is $(1)$ calculated? Moreover, why I cannot obtain the same first two terms using the derivative?
Should I use the Taylor expansion of $e^u$ substituting $u$ with my $u(x)$? Is this possible? Anyway, I should obtain the same result in every situation.
Could you please guide me to understand these points? What's my problem?
Thank you for your help.