give the definition of $g(x) = o(x^3)$

Question

The exercise says to give the definition of $g(x) = o(x^3)$ for $x\to 0$ :

what I know is that "$o$" means :

$$\lim_{x\to u} {f(x)\over g(x)} = 0$$

and that :

$$f = o(g)$$

does that mean that :

$$\lim_{x\to 0} {g(x)\over x^3} = 0?$$

Answer 1

Yes. By definition, as $x \to 0$, one has $$ g(x)=o(x^3) $$ if and only if $$ \lim_{x \to 0}\frac{g(x)}{x^3}=0. $$