Is this claim valid as a proof?

Question

I have a function $g(x)$ and I need to prove that $\; g(x)-x g'(x)\neq0$ on domain $0<x<a$, where $a$ is a real positive number. Using Taylor series for the mentioned expression around $x=0$, we obtain $ \; g(x)-x g'(x)=a x^3+O(x^5)$. So, does this mean that $\; g(x)-x g'(x)$ is always non-zero over the domain $(0,a)$? Can it be considered as a proof?

Answer 1

No, this is not a proof, since any constant possible could be contained in the $O$, and thus $O(x^5)$ can be arbitrarily large on $(0,a)$. Namely, it can be equal to $ax^3$ wherever we want it to be.

Answer 2

Your claim is true but not at all the interval $(0,a)$, it is true at a certain neighborwohood of $ 0$.

$$(\forall x\in (0,a))\; g(x)-xg'(x)=ax^3+O(x^5)\implies $$

$$(\forall x\in (0,a)) g(x)-xg'(x)=ax^3(1+x\epsilon(x))$$ with $$\lim_{x\to 0,x\ne 0}\epsilon(x)=0$$

For $ x $ close to zero, $ -\frac 12<x\epsilon(x)<\frac 12$, thus

$$(\exists \eta>0)\;: (\forall x\in (0,\eta))$$ $$g(x)-xg'(x)=ax^3(1+x\epsilon(x))>\frac{ax^3}{2}>0$$