Why does the error term tend to zero faster than the difference? In the differential approximation for f

we have an error term "E" which tends to zero faster than the difference on the x axis "h" thus we can use the little o notation 

My question is how do we know that the error term approaches zero faster
than the difference when x (or "a" int the first picture) is approaching x0? And why do we bother to use the little o notation? Is it easier to use it in multi-variable functions?
 A: A function $f$ is called differentiable at a point $a$ with derivative $f'(a)$ if
$$ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} - f'(a) = 0,$$
or equivalently if
$$ \frac{f(a+h) - f(a)}{h} - f'(a)= E(h),$$
where $E(h)$ is a function tending to $0$ as $h \to 0$. We do not know that $E(h)$ vanishes as quickly as $h$ vanishes in general.
I suspect you mean that $h E(h)$ vanishes faster than $h$ alone, which is clear as $h$ vanishes exactly as fast as $h$, and $E(h)$ vanishes also. This is enough to say that
$$ f(a+h) = f(a) + hf'(a) + o(h),$$
as you've written at the end.
A: We know that because, for positive numbers $a$ and $b$ such that $a>b$ and a number $x<1$ we have $x^a < x^b$, as opposed to the "usual" $x^a > x^b$ for $x>1$
now, say we have $x^b+100x^a$ and $x<1$, even though $x^a$ is less than $x^b$, since there's a factor of 100 multiplying $x^a$ it would seem unwise to ignore it. However, as $x$ tends to zero, i.e. if we take a linear approximation at a point sufficiently close to our base point, we get $x^a<<<x^b$ i.e. $x^a$ tends to zero a lot faster than $x^b$. for example if we take a linear approximation at a distance $.01$ to our base point, we would multiply the second term by $.01^2=.0001$, which even when multiplied by a big coefficient is typically so small that it can be ignored without any concern.
