Proof

We give only a sketch of the proof:

The polynomial function x -> F(x) is the first part of the Taylor series expansion of f. From analysis it follows that there exists a real-valued function h satisfying

f(x) = F(x) + x^{n + 1}h(x)    for x -> 0.

From this we conclude that F is an n-th-order approximation for f around 0.

Considering the second part of the theorem, suppose

g(x) = G(x) + O(x^{n + 1})    for x -> 0.

Then we have

(fg - FG)(x) = (fg - Fg)(x) + (Fg - FG)(x) =

g(x)O(x^{n + 1}) + F(x)O(x^{n + 1}) = O(x^{n + 1})    for x -> 0.  

So FG is indeed the n-th-order approximation of fg about 0.

The proof for f + g is simpler. Do it yourself.