Why should the error of a local linearization for $x$ near $a$ be small relative to $(x-a)$? I am reading a calculus book. It says

The fact that the graph of $f$ looks like a line as we zoom in means that not only is $E(x)$ small for $x$ near $a$, but also that $E(x)$ is small relative to $(x−a)$

where $E(x) = f(x) - f(a) - f(a)^{'}(x-a)$ is an error of the local linearization, and then introduces a theorem that $$\lim_{x\to a}{\frac{E(x)}{x-a}}=0$$
I just can't understand why $E(x)$ should be small relative to $(x-a)$. What is the need to be so if we want a function to be linear when we zoom in?
P.S. My question is not about why the theorem holds. Rather I can't understand the quoted statement above. Why relative to $(x-a)$ not other value. What is special about this relation?

Short answer: It means that $f'(a)\cdot(x-a)+f(a)$ is strictly the best linear approximation to $f$ close to $a$. Thus justifying that $f$ has a single slope at $a$, called $f'(a)$.
