I would like to prove that $\log^2(n) = O(n)$.
My attempt so far is:
Since $\lim_{n \to \infty} \log^2(n) = \infty \text{ and } \lim_{n \to \infty} n = \infty$ we get from L'Hôpital's rule that (let $f(n) = \log^2n$ and $g(n) = n$)
$$\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{g^{\prime}(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{1} = \lim _{n \rightarrow \infty} \frac{n}{\log n} \cdot \frac{\ln 2}{2} = \infty$$
Hence $\log^2 n = O(n)$
Is this valid, and if not, where is it breaking?
EDIT: $$\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{g^{\prime}(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{1} = \lim _{n \rightarrow \infty} \frac{2log n}{n} = 0$$