I'm not sure about this.
Consider a function $f$ and it's $n$ degree Taylor Polynomial in $x_0$ $T_{f,x_0,n}$.
Considered the remainder function
$R(x)=f(x)-T_{f,x_0,n}=o(x-x_0)^{n}$
Can I conclude from this that as $x \to x_0$ $f(x) \sim T_{f,x_0,n}$?
As $x \to x_0$ we have
$f(x)-T_{f,x_0,n}=o(x-x_0)^{n} \implies f(x)-T_{f,x_0,n}=o(T_{f,x_0,n}) \implies f(x) \sim T_{f,x_0,n} $
Is this correct?
Edit: Considered two functions $f$ and $g$ with "$f\sim g$ as $x\to x_0$" I mean that $lim_{x\to x_0} \frac{f(x)}{g(x)}=1$ or equivalently that $f(x)-g(x)=o(g(x))$