Proof that $\log^2 n = O(n)$

Question

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: $$\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{g^{\prime}(n)}=\lim _{n \rightarrow \infty} \frac{f^{\prime}(n)}{1} = \lim _{n \rightarrow \infty} \frac{2log n}{n} = 0$$

Answer 1

We have that

$$\lim _{x \rightarrow \infty} \frac{\log^2 x}{x}\stackrel{H.R.}=\lim _{x \rightarrow \infty} \frac{2\log x}{x}\stackrel{H.R.}=\lim _{x \rightarrow \infty} \frac{2}{x}=0$$

therefore $\log^2 n = O(n)$ and $\log^2 n = o(n)$.

Answer 2

In light of the actual limit it is better (i.e. more accurate) to use asymptotic relations other than $O(\cdot)$. Since the limit, as @user showed, is $0$, the relationship is in fact $$ f(n) = o(g(n))\\ g(n) = \omega(f(n)) $$ Here $f(n) = \log^2 n$ and $g(n) = n, \ \omega(f(n))$ means that the ratio diverges (tends to infinity), $o(\cdot)$, as explained above, means that the ratio converges to $0$