What is the point of the remainder term in the Taylor series Take the first order taylor approximation:
We say $f$ is differentiable at $x_0 \in S^o$ if $\exists \space \nabla f(x_0)$ and a function $R: \mathbb{R}^n \to \mathbb{R}$ such that
$f(x) = f(x_0) + \langle \nabla f(x_0), x-x_0 \rangle + R(x-x_0)$
I don't understand what the point of the Remainder term is? Is it just trying to say there is a degree of error involved in the taylor approximation (which is the first two terms if I am not mistaken?)
What does this limit: $\lim_{x \to x_0} \frac{R(x-x_0)}{\| x-x_0 \|}=0$ convey in terms of information? If this implies that $R(x-x_0)$ grows slower than $\|x-x_0\|$ why is that useful?

Any function approximates another. For instance, here are two approximations of $f$:
$$f(x) \approx f(x_0),$$
$$f(x) \approx f(x_0) + \langle \nabla f(x_0), x-x_0 \rangle.$$
If we don't know how close we approximate $f$, these formulas are of little use. The remainder answers this question.
In addition, if you had a remainder that is as large as the some term in the development (which generalizes to higher orders), it would be useless to consider this term. So we require that the remainder decreases faster (when you get closer to $x_0$) than the smallest term.
