I have a few questions regarding the theorem attached below. Why does functions $\frac{f(x)}{(x-a)^n}$ and $\frac{f^{(n)}(a)}{n!}$ have an equal sign if $x$ is sufficiently close to $a$? Is it only because the limit is $0$, so if they would be of different sign then it wouldn't make a zero?
Then relating to image below: My understanding is that if $(x-a)^n$ has a different signs if $x<a$ and $x>0$, so that $f(x)$ has to have different sign so that the expression is equal to $\frac{f^{(n)}(a)}{n!}$. Is it correct or there is different reasoning? Then, it is not explicitly said why $\frac{f(x)}{(x-a)^n}$ has the same sign, when $x$ is sufficiently close to $a$.
Because $$\lim_{x\to a}\bigg[\frac{f(x)}{(x-a)^n}-\frac{f^{(n)}(a)}{n!}\bigg]=0$$
We'll say that this is $\lim [A-B]=0$
Assume $A>0$, then for $A-B\to0$, we require $0\leq A-B< A\implies B\in(0,A]$
Equally assume $A<0$, then for $A-B\to0$ we require $A<A-B\leq0\implies B\in[A,0)$
clearly in both cases $sign(A)=sign(B)$