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?
 A: 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
$$\lim\limits_{x \to a}\dfrac{E(x)}{x-a} = f^{\prime}(a)-f^{\prime}(a) = 0\text{.}$$
A: 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 $f'(a)$.
A: A longer answer: If we had a local linear approximation for $f$ at $a$, $L_f(x) = c\,x + d$ which we can write as $f(x)=c_1(x-a) + c_0$, then any local approximation should be asymptotically correct as $x\to a$: $\frac{c_0}{f(a)} = \lim_{x\to a}\frac{L_f(x)}{f(x)} = 1$, or $c_0=f(a)$. Otherwise we aren't capturing the local information at any order.
On the other hand, for a local linear approximation to work, we'd want the linearization to be asymptotically correct as you approach $a$ as well. Since we already constrained $f\to L_f$ as $x\to a$, we want
$$
\lim_{x\to a}\frac{L_f(x)-L_f(a)}{f(x)-f(a)} = 1.
$$
But by L'Hospital's rule (assuming continuity and differentiability of $f$ at $a$)
$$
\lim_{x\to a}\frac{L_f(x)}{f(x)} = \lim_{x\to a}\frac{c_1(x-a)}{f(x)-f(a)} = \lim_{x\to a}\frac{c_1}{f'(x)} = \frac{c_1}{f'(a)} = 1.
$$
if and only if $c_1=f'(a)$.
Regarding the error term $E(x) = f(x) - L_f(x)$, this all implies by L'Hospital's rule again
$$
\lim_{x\to a}\frac{E(x)}{x-a} = \lim_{x\to a}\frac{f(x) - L_f(x)}{x-a} = \lim_{x\to a}\frac{f'(x) - L_f'(x)}{1} = f'(a) - L_f'(a) = f'(a) - c_1 = 0.
$$
In summary, if you have an asymptotically correct linear approximation to a function $f$ then the error is small relative to the linear term, or else you didn't do a good job in the first place.
A: You can divide the equation by $(x-a)$
$E(x) = f(x) - f(a) - f^{'}(a)(x-a)$
$\frac{E(x)}{x-a}=\frac{f(x) - f(a)}{x-a}-f^{'}(a)$
$\frac{f(x) - f(a)}{x-a}$ is a difference quotient. If we calculate the limit it becomes a differential quotient. And by definition we get
$$\lim\limits_{x \to a} \frac{f(x) - f(a)}{x-a}=f^{'}(a)$$
$f(x)$ has to be differentiable on its domain. 
A: The idea is this: We want to approximate a function $f$ (which it is possible to approximate thus) by a linear function $\ell$ near a point $a.$ Therefore, it must be the case that $f(a)$ coincides with $\ell(a).$ Furthermore, we must require that $f'(a)=\ell'(a).$ Since higher derivatives of $\ell$ vanish, there is nothing more to be gained in this direction.
Now the question is: Do these conditions suffice to give us the best $\ell$ that does the job? By definition, we have that $$\ell(x)=f(a)+f'(a)(x-a),$$ so that $$f(x)=\ell(x)+E(x).$$ Now, whether or not $\ell$ does the job well depends on how $E(x)$ compares to the other summand in the equation $f(x)=\ell(x)+E(x).$ If you can see this, then we're not far from the goal, for the other summand is just a linear function, whose significant part is clearly the term $x-a.$ In particular, the error $E(x)$ must be smaller than the term $\ell(x)$ for the approximation to be sensible. There is simply no other quantity to compare the error to except the term $\ell(x)$ if we want our approximation to be the best linear one. This is why we require that near $a,$ the term $E(x)$ should always be dominated by $x-a$ (for clearly if this is the case then we have what we want, since $|x-a|<|\ell(x)|$), for otherwise we could extract more linear information from it, as it were, which would mean that our approximation wasn't the best possible, after all, which we do not want.
