Which of $\log{\sqrt{n\log{n}}}$ and $\sqrt{\log{n}}$ grows faster? Which of the following functions grows faster:
$\log{\sqrt{n\log{n}}}$ or $\sqrt{\log{n}}$?
I feel the second one should be the answer, but I find it difficult to prove as the derivatives get very complex. Does anyone know any idea?
 A: The first is (ignoring very small $n$) greater than $\log (n^{1/2})= \frac12 \log n$. 
So setting $a = \log n$ the first is greater than $a/2$ while the latter is $\sqrt{a}$. Whence your intuition was not correct and the first grows faster.
A: For $n >> 1$:
$\log{\sqrt{n\log{n}}} = \frac{1}{2} \log{(n\log{n})} > \frac{1}{2} \log{n} > \frac{1}{2} \sqrt{\log{n}}$.
Thus the first term grows faster.
A: 
The first one (blue in the graph) growrs faster


A: First note that if $f(n)$ and $g(n)$ are both positive functions and
$$\lim\limits_{n\to\infty}\frac{f(n)}{g(n)}=\infty$$
Then we can say that $f(n)$ grows faster than $g(n)$ as $n$ approaches $\infty$. Symbolically this is written as $f(n)\gg g(n)$. So now we have
$$\lim\limits_{n\to\infty}\frac{\log\left(\sqrt{n\log n}\right)}{\sqrt{\log n}}$$
Let $k=\sqrt{\log n}$, then
$$\lim\limits_{k\to\infty}\frac{\log\left(\sqrt{e^{k^2}}k\right)}{k}$$
$$=\lim\limits_{k\to\infty}\frac{\log\left(\sqrt{e^{k^2}}\right)+\log k}{k}$$
$$=\lim\limits_{k\to\infty}\frac{\frac12 k^2+\log k}{k}$$
$$=\frac12\lim\limits_{k\to\infty}\frac{k^2+2\log k}{k}$$
$$=\frac12\lim\limits_{k\to\infty}\left(k+\frac{2\log k}{k}\right)=\infty$$
Therefore, as $n\to\infty$,
$$\log\left(\sqrt{n\log n}\right) \gg \sqrt{\log n} $$
