I am trying to solve this issue but I do not know how to handle the division by $O(\sqrt{n})$.
Prove the following using limits and L’Hôpital’s Rule: That $\log n$ is
$O(\sqrt{n})$.
$$
\begin{align*}
\lim_{n \rightarrow \infty}\frac{\log n}{\sqrt{n}} &= \lim_{n \rightarrow \infty}\frac{\log n}{10^{1/2 \log n}}\\
\end{align*}
$$
Best Answer
You need to prove that $\log x \le c \sqrt x$ for some $c>0$ and for $x$ sufficienty large.
Consider $f(x)=\sqrt x - \log x$. Then $f'(x)=\frac{1}{2\sqrt x}-\frac{1}{x}=\frac{\sqrt x -2}{2x}\ge 0$ for $x\ge4$.
Since $f(4)=2-\log 4>0$, we have $f(x)\ge 0$ for $x\ge4$.