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.

  • 7
    $\begingroup$ You can't prove it since it's not true. $\endgroup$ Jun 7, 2017 at 18:15
  • $\begingroup$ "I calculated 1+1=2, but that's too much as I need 1" -- see the problem here? $\endgroup$
    – Raphael
    Jun 7, 2017 at 21:43

1 Answer 1


$$\lim_{n\to\infty} \frac{\sqrt n \log\sqrt n}{n} = 0$$

Since this $\Theta(\sqrt n \log \sqrt n) = o(n) < \Theta(n)$.

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

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

  • $\begingroup$ but this is exactly what I wrote in my question $\endgroup$
    – Yos
    Jun 7, 2017 at 18:22
  • $\begingroup$ @Yos, no, you have divided $\log \sqrt n$ instead of $\sqrt n \log \sqrt n$. $\endgroup$
    – rus9384
    Jun 7, 2017 at 18:23
  • $\begingroup$ but you're getting the same result I got, what does you answer contribute? $\endgroup$
    – Yos
    Jun 7, 2017 at 18:25
  • 2
    $\begingroup$ @Yos, yes, result is the same, because the statement itself is wrong. It would be correct if limit would be finite non-zero. $\endgroup$
    – rus9384
    Jun 7, 2017 at 18:27
  • $\begingroup$ So you should say this explicitly in your answer. Otherwise you're just providing an almost identical calculation and identical result. $\endgroup$
    – Yos
    Jun 7, 2017 at 18:28

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