I am reading a book and it says that if $T_n(x)$ is the Taylor polynomial of $f$ of order $n$ at $x=a$ then $\lim_{x\rightarrow a} \frac{T_n(x)-f(x)}{(x-a)^n} = 0$. In other words, the error is negligible when compared with $(x-a)^n$.
May I ask why the error was compared to $(x-a)^n$ and not with any other function? What's the significance of $(x-a)^n?$
In Taylor formula polynomials are used. $(x-a)^n$ is such a polynomial, that its behaviour near $a$ is well-known.
As a result, we can easy obtain important results. For example, if $f'(a)=0$ and $f''(a)>0$ we know, that the function behaves near $a$ as $C(x-a)^2$, where $C$ is positive, hence there is a local minimum in $a$.