Proving that $\log x$ is big $O$ of $x^k$ for every positive $k$

Question

Can I know a way to prove the above condition purely by the definition (and may be Taylor Series) and without using De l'Hôpital's rule? It is obvious for $k$ greater than or equal to $1$ but how can you show for $k$ is less than $1$?

Answer 1

It is $O(x^k)$ because it is $o(x^k)$ for any $k>0$. This results from the basic high school limit $$\lim_{x\to\infty}\frac{\log x}{x}=0$$ and the functional properties of the logarithm.