$ \sqrt n \log \sqrt n$ is lower than $n$ ?! I see an example in asymptotic notation that not very sensible for me.
what is the intuitive idea to understand $ \sqrt n \log \sqrt n$ is lower than $n$ in concept of asymptotic view?
 A: It's similar to $n \log n$ being less than $n^2$.
A: $$\lim\limits_{x \to +\infty}\frac{\ln \sqrt{x}}{\sqrt{x}} = \lim\limits_{y \to +\infty}\frac{\ln y}{y} = \lim\limits_{y \to +\infty}\frac{1}{y} = 0$$
Addition:
Without using L'Hôpital's rule we can show, that for all $\boldsymbol n$ $\forall n \in \mathbb{N}$ holds $ \log _{2} n < n^{\alpha} $ for $  \alpha \in  \left[  \frac{1}{e \cdot ln2}, 1\right)  $: as it is same with $ \dfrac{\ln(\log _{2} n)}{\ln(n)} < \alpha $, then we need only to find maximum of function $ \dfrac{\ln(\log _{2} x)}{\ln(x)} $ which is in $ x = 2^{e} $. Now, knowing  $ \frac{1}{e \cdot ln2} \approx  0.530737$ we have result for $\alpha=\frac{1}{2}$.
We can obtain without L'Hôpital also $ \log _{2} n < n^{\alpha} $ for $  \alpha \in  \left(  0, 1\right)  $, but only on some subset of $\mathbb{N}$, which needs little more subtle evaluation.
A: Intuitively, you should already see that $\log(\sqrt{n})$ is asymptotically less than $\sqrt{n}$. Therefore $\sqrt{n}\log(\sqrt{n})$ is asymptotically less than $n$.
A: The point of asymptotic behavior is to abstract away implementation details and look at the "big picture" of how long an algorithm takes.
An $O(n)$ algorithm has to (roughly) read (most of) the input.  A $O(\sqrt n \log \sqrt n)$ algorithm doesn't even have to read the entire input, it reads a significantly smaller amount.  That's not a small detail, so you should expect them to be different.
