If I have that $f(x)=T_n(x)+o(x^n)$ ($T_n$ is the Taylor polynomial of $n-$th order ) and $f(x)=p(x)+o(x^n)$, is this $o(x^n)$ the same "thing" in both terms?
So why I'm asking this question. The reason is that I will like to show that $p=T_n$.
$f(x)-p=o(x^n)$
I know that $o(x^n)=q$ (where as $q$ is the zero polynomial) in the limit $x\to 0$.
When we write $g(x)= o(x^n) $, we must precise where, so we have to write $$g(x)=o(x^n) \Bigl( x\to a, x\in A\Bigr)$$
with $$A\subset D_g \text{ and } a\in \overline{A}$$
this will mean that $$g(x)=x^n\epsilon(x)$$ with $$\lim_{x\to a,x\in A}\epsilon(x)=0$$ For example, if $n\ge 1$,
$$e^{-x}=o(x^n) (x\to +\infty)$$ $$\ln(x)=o(x^n) (x\to +\infty)$$