Can I prove that $\sqrt n\log{\sqrt n}=\Theta(n)$? I'd like to  prove that $\sqrt n\log{\sqrt n}=\Theta(n)$ but I'm not sure how to do it.

This is my attempt:
$$
\lim_{n \to \infty}\frac{\log{\sqrt n}}{\sqrt n}=0 \Rightarrow \log{\sqrt n}=o(\sqrt n) \text{(Little o)}
$$

Therefore:
$$
\sqrt n\log{\sqrt n}=o(\sqrt n\cdot \sqrt n)=o(n)
$$
But this is too much as I need theta.
$$\lim_{n\to\infty} \frac{\sqrt n \log\sqrt n}{n} = 0$$

Since this $\Theta(\sqrt n \log \sqrt n) = o(n) &lt; \Theta(n)$.

Two functions $f(x), g(x)$ are asymptotically equal iff

$$\lim_{x\to\infty} \frac{f(x)}{g(x)} = c,\ c \in \mathbb{R},\ c &gt; 0$$