Can the limit definition of Big O equal $0$?

Question

Say I have a function:

$$f(x)=x^2+4x+17$$

To see if this function is Big $O(x^3)$ by taking the limit of $\frac{f(x)}{g(x)}$ as $x$ goes to infinity we get:

$$\lim\limits_{x \to \infty}\frac{x^2+4x+17}{x^3} = 0$$

With the answer of $0$ for the limit can we still say $f(x) \in O(x^3)$ or does the limit needs to be a nonzero number?

Thanks!

Answer 1

Yes. When the limit $\lim_{x\to\infty}f(x)/g(x)=0$ then we in fact have $f=o(g)$, which always implies $f=O(g)$.