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?

I'm going to focus on the intuition. I make no claims that my answer is rigorous.
Write $$E(x) = f(x)-f(a)-f^{\prime}(a)(x-a)\text{.}$$
$f^{\prime}(a)$ is the "slope" of $f$ at $x = a$. Make sure you understand that derivatives are just slopes. For sake of convenience, let's call this $m$.
Way back in your Pre-Calculus days, remember that the computational formula for the slope of a linear equation (hence why there's so much focus on linearity) is
$$m = \dfrac{y_2 -y_1}{x_2 - x_1}\text{.}$$
Now, you can think of $x$ and $a$ as corresponding to $x_2$ and $x_1$ respectively. Hence we have
$$m(x_2 - x_1) = y_2-y_1 \implies m(x-a)=y_2-y_1\text{.}$$
For a linear function $f$, $y_2$ is the $y$-value of $f$ at $x$ and $y_1$ is the $y$-value of $f$ at $a$. Hence, we have
$$m(x-a) = f(x)-f(a)$$
when $f$ is linear, but when $f$ is non-linear, this is only an approximation. So,
$$m(x-a) \approx f(x)-f(a)$$
and hence, for $x$ near $a$, we would hope that
$$E(x) \approx f(x)-f(a)-[f(x)-f(a)] = 0\text{.}$$
This explains the first claim. 
I think @callculus explained the second claim sufficiently well: note that in order to even begin to compute $f^{\prime}(a)$, $f$ must be differentiable at $a$. Notice that
$$\begin{align}
\lim\limits_{x \to a}\dfrac{E(x)}{x-a} &= \lim_{x \to a}\dfrac{f(x)-f(a)-f^{\prime}(a)(x-a)}{x-a} \\
&= \lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}-\lim\limits_{x \to a}\dfrac{(x-a)f^{\prime}(a)}{x-a} \\
&=\lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}-f^{\prime}(a)
\end{align}$$
The limit $$\lim\limits_{x \to a}\dfrac{f(x)-f(a)}{x-a}$$
must exist for $f$ to be differentiable at $a$ (this is by the definition of a derivative); otherwise, you wouldn't be able to compute $f^{\prime}(a)$. 
As long as $f$ is differentiable at $a$, the resulting limit above is
$$\lim\limits_{x \to a}\dfrac{E(x)}{x-a} = f^{\prime}(a)-f^{\prime}(a) = 0\text{.}$$
